#### NCERT Chapter : 10 – Vector Algebra

Exercise – 10.1

Question 1: Graphically represent a 40 km displacement towards 30 o east of north.

Answer 1: Based on the formula given in Vector Algebra

Vector $$\overrightarrow{OP}$$ represent a 40 km displacement towards 30o east of north.

Question 2: Categorize the following measures as vectors and scalars.

(a) 20 kg (b) 4 meters north – south (c) 80o

(d) 70 watt (e) 10– 17 coulomb (f) 56 m / s– 2

Answer 2: Based on the formula given in Vector Algebra

(a) In 20 kg, only magnitude is involved. So, it is a scalar quantity.

(b) In 4 meters north – south, both the direction and magnitude are involved. So, it is a vector quantity

(c) In 80o, only magnitude is involved. So, it is a scalar quantity.

(d) In 70 watt, only magnitude is involved. So, it is a scalar quantity.

(e) In 10 – 17 coulombs, only magnitude is involved. So, it is a scalar quantity.

(f) In 56 m / s– 2, both the direction and magnitude are involved. So, it is a vector quantity

Question 3: Categorize the following quantities as vector and scalar.

(a) Time period (b) distance (c) force

(d) Velocity (e) work done

Answer 3: Based on the formula given in Vector Algebra

(a) In time period, only magnitude is involved. So, it is a scalar quantity.

(b) In distance, only magnitude is involved. So, it is a scalar quantity.

(c) In force, both the direction and magnitude are involved. So, it is a vector quantity

(d) In velocity, both the direction and magnitude are involved. So, it is a vector quantity

(e) In work done, only magnitude is involved. So, it is a scalar quantity.

Question 4: In the following diagram, recognize the corresponding vectors

(a) Coinitial

(b) Equal

(c) Collinear but not equal

Answer 4: Based on the formula given in Vector Algebra

(a) Coinitial vectors are those vectors which have same initial point. So, $$\overrightarrow{a} \;and\; \overrightarrow{d}$$ vectors are coinitial.

(b) Equal vectors are vectors which have same magnitude and direction. So, $$\overrightarrow{b} \;and\; \overrightarrow{d}$$ vectors are equal.

(c) Collinear but not equal are those vectors which are parallel but has different directions. So, $$\overrightarrow{a} \;and\; \overrightarrow{c}$$ vectors are collinear but not equal.

Question 5: Check whether the following statements are true or false.

(a) $$\overrightarrow{b} \;and\; \overrightarrow{- b}$$ vectors are collinear

(b) The magnitudes of the two collinear are always equal.

(c) Collinear vectors are the two vectors having same magnitude.

Answer 5: Based on the formula given in Vector Algebra

(a). True because the two vectors are parallel .

(b). False because collinear vectors must be parallel.

(c). False.

#### Exercise 10.2

Question 1: For the following vectors, calculate the magnitude of the following.

$$\overrightarrow{m} = \hat{i} + \hat{j} + \hat{k}; \;\;\; \overrightarrow{n} = \hat{2i} – \hat{7j} – \hat{3k}; \;\;\; \overrightarrow{o} = \frac{1}{\sqrt{3}} \;\hat{i} + \frac{1}{\sqrt{3}}\; \hat{j} – \frac{1}{\sqrt{3}}\; \hat{k}$$

Answer 1: Based on the formula given in Vector Algebra

Given, $$\overrightarrow{m} = \hat{i} + \hat{j} + \hat{k}; \;\;\; \overrightarrow{n} = \hat{2i} – \hat{7j} – \hat{3k}; \;\;\; \overrightarrow{o} = \frac{1}{\sqrt{3}} \;\hat{i} + \frac{1}{\sqrt{3}}\; \hat{j} – \frac{1}{\sqrt{3}}\; \hat{k}$$

$$\left | \overrightarrow{m} \right | = \sqrt{(1) ^{2} + (1) ^{2} + (1) ^{2}} = \sqrt{3} \\ \left | \overrightarrow{n} \right | = \sqrt{(2) ^{2} + (- 7) ^{2} + (- 3) ^{2}} = \sqrt{4 + 49 + 9} = \sqrt{62} \\ \left | \overrightarrow{o} \right | = \sqrt{(\frac{1}{\sqrt{3}}) ^{2} + (\frac{1}{\sqrt{3}}) ^{2} + (\frac{1}{\sqrt{3}}) ^{2}} = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = 1 \\$$

Question-2: Mention two dissimilar vectors having similar magnitude?

Answer-2: Based on the formula given in Vector Algebra

$$\overrightarrow{m} = (\hat{2i} – \hat{2j} + \hat{3k}); \;and\; \overrightarrow{n} = (\hat{2i} + \hat{2j} – \hat{3k}) \\ It\; can\; be\; observed\; that:\; \\ \left | \overrightarrow{m} \right | = \sqrt{(2) ^{2} + (- 2) ^{2} + (3) ^{2}} = \sqrt{17} \; and \\ \left | \overrightarrow{n} \right | = \sqrt{(2) ^{2} + (2) ^{2} + (3) ^{2}} = \sqrt{17}$$

Thus, the two dissimilar vectors $$\overrightarrow{m} \;and\; \overrightarrow{n}$$ have similar magnitudes. Because of different directions the two vectors are dissimilar.

Question 3: Mention two dissimilar vectors having similar direction?

Answer 3: Based on the formula given in Vector Algebra

Consider, $$\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}; \;\;\;and\; \overrightarrow{b} = \hat{2i} + \hat{2j} + \hat{2k} \\ The\; direction\; cosines\; of\; \overrightarrow{a}\; are\; given\; by, \\ p = \frac{1}{\sqrt{(1) ^{2} + (1) ^{2} + (1) ^{2}}} = \frac{1}{\sqrt{3}}, \; q = \frac{1}{\sqrt{(1) ^{2} + (1) ^{2} + (1) ^{2}}} = \frac{1}{\sqrt{3}}, \;and\; r = \frac{1}{\sqrt{(1) ^{2} + (1) ^{2} + (1) ^{2}}} = \frac{1}{\sqrt{3}}$$

$$The\; direction\; cosines\; of\; \overrightarrow{b}\; are\; given\; by, \\ p = \frac{2}{\sqrt{(2) ^{2} + (2) ^{2} + (2) ^{2}}} = \frac{2}{2 \sqrt{3}} = \frac{1}{\sqrt{3}}, \;q = \frac{2}{\sqrt{(2) ^{2} + (2) ^{2} + (2) ^{2}}} = \frac{2}{2 \sqrt{3}} = \frac{1}{\sqrt{3}}\; and\; r = \frac{2}{\sqrt{(2) ^{2} + (2) ^{2} + (2) ^{2}}} = \frac{2}{2 \sqrt{3}} = \frac{1}{\sqrt{3}}$$ $$The\; direction\; cosines\; of\; \overrightarrow{a}\; and\; \overrightarrow{b}\; are\; similar.$$

Thus, the direction of the two vectors is similar.

Question 4: $$4 \hat{i} + 5 \hat{j} \;and\; p \hat{i} + q \hat{j}$$ are the vectors and they are equal. Obtain the values of p and q

Answer 4: Based on the formula given in Vector Algebra

Given, $$4 \hat{i} + 5 \hat{j} \;and\; p \hat{i} + q \hat{j}$$ are equal.

The equivalent components are equal.

So, the value of p = 4 and q = 5.

Question 5: The initial point of the vector is (3, 2) and the terminal point of the vector is (- 6, 8). Obtain the vector and scalar components of the given vector.

Answer 5: Based on the formula given in Vector Algebra

Given,

The initial point of the vector A (3, 2) and the terminal point of the vector is B (- 6, 8).

The vector $$\overrightarrow{AB} = (- 6 – 3) \hat{i} + (8 – 2) \hat{j} \\ \overrightarrow{AB} = – 9 \hat{i} + 6 \hat{j}$$

The vector components of the given vector are $$– 9 \hat{i} \;and\; 6 \hat{j}$$.

The scalar components of the given vector are – 9 and 6.

Question 6: The vectors $$\overrightarrow{m} = \hat{i} + \hat{3j} + \hat{k}, \;\;\; \overrightarrow{n} = \hat{- 2i} – \hat{5j} – \hat{3k}, \;\;and \; \overrightarrow{o} = \hat{8i} – \hat{j} – \hat{2k}$$. Obtain the sum.

Answer 6: Based on the formula given in Vector Algebra

Given:

$$\overrightarrow{m} = \hat{i} + \hat{3j} + \hat{k}, \;\;\; \overrightarrow{n} = \hat{- 2i} – \hat{5j} – \hat{3k}, \;\;and \; \overrightarrow{o} = \hat{8i} – \hat{j} – \hat{2k} \\ \overrightarrow{m} + \overrightarrow{n} + \overrightarrow{o} = (1 – 2 + 8) \hat{i} + (3 – 5 – 1) \hat{j} + (1 – 3 – 2) \hat{k} \\ = 7 \hat{i} – 3 \hat{j} – 4 \hat{k}$$

Question 7: Obtain the unit vector of $$\overrightarrow{p} = \hat{i} + \hat{2j} + \hat{k}$$ in the direction of the given vector.

The unit vector $$\hat{p}$$ in the direction of vector $$\overrightarrow{p} = \hat{i} + \hat{2j} + \hat{k}$$

$$\left | \overrightarrow{p} \right | = \sqrt{(1) ^{2} + (2) ^{2} + (1) ^{2}} = \sqrt{1 + 4 + 1} = \sqrt{6} \\ \hat{p} = \frac{\overrightarrow{p}}{\left | \overrightarrow{p} \right |} = \frac{\hat{i} + \hat{2j} + \hat{k}}{\sqrt{6}} = \frac{1}{\sqrt{6}} \hat{i} + \frac{1}{\sqrt{6}} \hat{2j} + \frac{1}{\sqrt{6}} \hat{k} \\ = \frac{1}{\sqrt{6}} \hat{i} + \frac{2}{\sqrt{6}} \hat{j} + \frac{1}{\sqrt{6}} \hat{k}$$

Question 8: For a vector $$\overrightarrow{AB}$$, obtain the unit vector where the point A (2, 3, 4) and point B (5, 6, 7). The unit vector should be in the direction of given vector.

Answer 8: Based on the formula given in Vector Algebra

Given, points A (2, 3, 4) and B (5, 6, 7).

$$\overrightarrow{AB} = (5 – 2) \hat{i} + (6 – 3) \hat{j} + (7 – 4) \hat{k} \\ \overrightarrow{AB} = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \\ \left | \overrightarrow{AB} \right | = \sqrt{(3) ^{2} + (3) ^{2} + (3) ^{2}} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3 \sqrt{3} \\ The\; unit\; vector\; in\; the\; direction\; of\; \overrightarrow{AB}\; is \\ \frac{\overrightarrow{AB}}{\left | \overrightarrow{AB} \right |} = \frac{3 \hat{i} + 3 \hat{j} + 3 \hat{k}}{3 \sqrt{3}} = \frac{1}{\sqrt{3}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k}$$

Question 9: In the direction of $$\overrightarrow{m} + \overrightarrow{n}$$, obtain the unit vector for given vectors $$\overrightarrow{m} = \hat{3i} – \hat{j} + \hat{2k} \; and\; \overrightarrow{n} = \hat{2i} – \hat{3j} – \hat{k}$$.

Answer 9: Based on the formula given in Vector Algebra

Given,

The vectors $$\overrightarrow{m} = \hat{3i} – \hat{j} + \hat{2k} \; and\; \overrightarrow{n} = \hat{2i} – \hat{3j} – \hat{k}$$

$$\overrightarrow{m} = \hat{3i} – \hat{j} + \hat{2k} \\ \overrightarrow{n} = \hat{2i} – \hat{3j} – \hat{k} \\ \overrightarrow{m} + \overrightarrow{n} = (3 + 2) \hat{i} + (- 1 – 3) \hat{j} + (2 – 1) \hat{k} \\ = 5 \hat{i} – 4 \hat{j} + 1 \hat{k} \\ \left | \overrightarrow{m} + \overrightarrow{n} \right | = \sqrt{(5) ^{2} + (- 4) ^{2} + (1) ^{2}} = \sqrt{25 + 16 + 1} = \sqrt{42}$$

Thus, in the direction of $$\overrightarrow{m} + \overrightarrow{n}$$, the vector is,

$$\frac{(\overrightarrow{m} + \overrightarrow{n})}{\left | \overrightarrow{m} + \overrightarrow{n} \right |} = \frac{5 \hat{i} – 4 \hat{j} + 1 \hat{k}}{\sqrt{42}} \\ = \frac{1}{\sqrt{42}} 5 \hat{i} – \frac{1}{\sqrt{42}} 4 \hat{j} + \frac{1}{\sqrt{42}} \hat{k} \\ = \frac{5}{\sqrt{42}} \hat{i} – \frac{4}{\sqrt{42}} \hat{j} + \frac{1}{\sqrt{42}} \hat{k}$$

Question 10: A vector $$6 \hat{i} – 2 \hat{j} + 3 \hat{k}$$ has a magnitude of 8 units. Find the vector in the direction of given vector.

Answer 10: Based on the formula given in Vector Algebra

Suppose, $$\overrightarrow{m} = 6 \hat{i} – 2 \hat{j} + 3 \hat{k} \\ \left | \overrightarrow{m} \right | = \sqrt{6^{2} + (- 2) ^{2} + 3^{2}} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 \\ \hat{m} = \frac{\overrightarrow{m}}{\left | \overrightarrow{m} \right |} = \frac{6 \hat{i} – 2 \hat{j} + 3 \hat{k}}{7}$$.

Thus, the vector in the direction of given vector which has 8 units magnitude is given by,

$$8 \hat{m} = 8 (\frac{6 \hat{i} – 2 \hat{j} + 3 \hat{k}}{7}) = \frac{48}{7} \hat{i} – \frac{16}{7} \hat{j} + \frac{24}{7} \hat{k}$$

Question 11: Prove whether the vectors $$3 \hat{i} – 4 \hat{j} + 5 \hat{k} \;and\; 9 \hat{i} – 12 \hat{j} + 15 \hat{k}$$ are collinear.

Answer 11: Based on the formula given in Vector Algebra

Suppose, $$\overrightarrow{p} = 3 \hat{i} – 4 \hat{j} + 5 \hat{k} \;and\; \overrightarrow{q} = 9 \hat{i} – 12 \hat{j} + 15 \hat{k}$$

The condition for the vectors to be collinear is,

$$\overrightarrow{q} = \lambda \overrightarrow{p}$$

Accordingly,

$$9 \hat{i} – 12 \hat{j} + 15 \hat{k} = 3\; (3 \hat{i} – 4 \hat{j} + 5 \hat{k} )$$, which satisfies the condition with $$\lambda = 3$$

Hence, proved

Question 12: Obtain the direction cosines of the vectors $$2 \hat{i} – 4 \hat{j} + 6 \hat{k}$$

Answer 12: Based on the formula given in Vector Algebra

$$\overrightarrow{m} = 2 \hat{i} – 4 \hat{j} + 6 \hat{k} \\ \left | \overrightarrow{m} \right | = \sqrt{(2) ^{2} + (- 4) ^{2} + (6) ^{2}} = \sqrt{4 + 16 + 36} = \sqrt{56} \\ Thus,\; the\; direction\; cosines\; of\; \overrightarrow{m} \;are\; \left ( \frac{2}{\sqrt{56}}, \frac{- 4}{\sqrt{56}}, \frac{6}{\sqrt{56}} \right )$$

Question 13: P (1, 2, – 3) and Q (- 1, – 2, 1) are the joining points of a vector directed from P to Q. Obtain the direction cosines of the vector.

Answer 13: Based on the formula given in Vector Algebra

P (1, 2, – 3) and Q (- 1, – 2, 1) are the joining points of a vector.

$$\overrightarrow{PQ} = (- 1 – 1) \hat{a} + (- 2 – 2) \hat{b} + (1 – (- 3)) \hat{c} \\ \overrightarrow{PQ} = (- 2) \hat{a} + (- 4) \hat{b} + (4) \hat{c} \\ \left | \overrightarrow{PQ} \right | = \sqrt{(- 2) ^{2} + (- 4) ^{2} + 4^{2}} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$

The direction cosines of the vector $$\overrightarrow{PQ}$$ are $$\left ( – \frac{2}{6}, – \frac{4}{6}, \frac{4}{6} \right ) = \left ( – \frac{1}{3}, – \frac{2}{3}, \frac{2}{3} \right )$$

Question 14: Prove that the $$\hat{i} + \hat{j} + \hat{k}$$ is evenly tending to the axes OX, OY and OZ

Suppose, $$\overrightarrow{m} = \hat{i} + \hat{j} + \hat{k} \\ Then,\; \left | \overrightarrow{m} \right | = \sqrt{(1) ^{2} + (1) ^{2} + (1) ^{2}} = \sqrt{3}$$

The direction cosines of the vector $$\overrightarrow{m}$$ are $$\left ( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right )$$

Now, let α, β, and γ be the angles formed by with the positive directions of x, y, and z axes.

Thus, we obtain,

$$cos \;\alpha = \frac{1}{\sqrt{3}},\; cos \;\beta = \frac{1}{\sqrt{3}} \;and\; cos \;\gamma = \frac{1}{\sqrt{3}}$$

Thus, the vector is evenly tending to the axes OX, OY and OZ

Question 15: The position vectors of a joining points A and B are $$\hat{i} + 2 \hat{j} – \hat{k} \;and\; – \hat{i} + \hat{j} + \hat{k}$$ respectively. Obtain the position vector of C which divides the given points in 2 : 1 ratio.

(a) Internally

(b) Externally

Answer 15: Based on the formula given in Vector Algebra

The position vector of C which divides the given points in m : n ratio is written as:

(a) Internally: $$\frac{m \overrightarrow{b} + n \overrightarrow{a}}{m + n}$$

(b) Externally: $$\frac{m \overrightarrow{b} – n \overrightarrow{a}}{m – n}$$

Given,

Position vectors of a joining points A and B,

$$\overrightarrow{OA} = \hat{i} + 2 \hat{j} – \hat{k} \;and\; \overrightarrow{OB} = \hat{i} + \hat{j} + \hat{k}$$

(a) The position vector of point C which divides the line joining two points A and B internally in the ratio 2:1 is given by,

$$\overrightarrow{OC} = \frac{2 (- \hat{i} + \hat{j} + \hat{k}) + 1 (\hat{i} + 2\hat{j} – \hat{k})}{2 + 1} = \frac{(- 2 \hat{i} + 2 \hat{j} + 2 \hat{k}) + (\hat{i} + 2\hat{j} – \hat{k})}{3} \\ = \frac{- \hat{i} + 4 \hat{j} + \hat{k}}{3} \\ = – \frac{1}{3} \hat{i} + \frac{4}{3} \hat{j} + \frac{1}{3} \hat{k}$$

(b) The position vector of point C which divides the line joining two points A and B externally in the ratio 2:1 is given by,

$$\overrightarrow{OC} = \frac{2 (- \hat{i} + \hat{j} + \hat{k}) – 1 (\hat{i} + 2\hat{j} – \hat{k})}{2 – 1} \\ = (- 2 \hat{i} + 2 \hat{j} + 2 \hat{k}) + (\hat{i} + 2\hat{j} – \hat{k}) \\ = – 3 \hat{i} + 3 \hat{k}$$

Question 16: A (3, 4, 5) and B (5, 2, – 3) are the joining points of a vector. Obtain the midpoint position vector.

Answer 16: Based on the formula given in Vector Algebra

The midpoint position vector with the joining points A (3, 4, 5) and B (5, 2, – 3),

Suppose, $$\overrightarrow{OC}$$ be the required vector, then,

$$\overrightarrow{OC} = \frac{(3 \hat{i} + 4 \hat{j} + 5 \hat{k}) + (5 \hat{i} + 2 \hat{j} + (- 3) \hat{k})}{2} = \frac{(3 + 5) \hat{i} + (4 + 2) \hat{j} + (5 – 3) \hat{k}}{2} \\ = \frac{8 \hat{i} + 6 \hat{j} + 2 \hat{k}}{2} \\ = 4 \hat{i} + 3 \hat{j} + \hat{k})$$

Question 17: Prove that the points P, Q and R with position vectors, $$\overrightarrow{p} = 3 \hat{a} – 4 \hat{b} – 4 \hat{c}, \; \overrightarrow{q} = 2 \hat{a} – \hat{b} + \hat{c}, \;and\; \overrightarrow{r} = \hat{a} – 3 \hat{b} – 5 \hat{c}$$ respectively from the vertices of a right angled triangle.

Answer 17: Based on the formula given in Vector Algebra

Given,

The points P, Q and R with position vectors $$\overrightarrow{p} = 3 \hat{a} – 4 \hat{b} – 4 \hat{c}, \; \overrightarrow{q} = 2 \hat{a} – \hat{b} + \hat{c}, \;and\; \overrightarrow{r} = \hat{a} – 3 \hat{b} – 5 \hat{c}$$

$$\overrightarrow{p} = 3 \hat{a} – 4 \hat{b} – 4 \hat{c}, \; \overrightarrow{q} = 2 \hat{a} – \hat{b} + \hat{c}, \;and\; \overrightarrow{r} = \hat{a} – 3 \hat{b} – 5 \hat{c}$$ $$\overrightarrow{PQ} = \overrightarrow{q} – \overrightarrow{p} = (2 – 3) \hat{a} + (-1 + 4) \hat{b} + (1 + 4) \hat{c} = – \hat{a} + 3 \hat{b} + 5 \hat{c} \\ \overrightarrow{QR} = \overrightarrow{r} – \overrightarrow{q} = (1 – 2) \hat{a} + (- 3 + 1) \hat{b} + (- 5 – 1) \hat{c} = – \hat{a} – 2 \hat{b} – 6 \hat{c} \\ \overrightarrow{RP} = \overrightarrow{p} – \overrightarrow{r} = (3 – 1) \hat{a} + (- 4 + 3) \hat{b} + (- 4 + 5) \hat{c} = 2\hat{a} – \hat{b} + \hat{c} \\$$ $$\left | \overrightarrow{PQ} \right | ^{2} = (- 1) ^{2} + 3^{2} + 5^{2} = 1 + 9 + 25 = 35 \\ \left | \overrightarrow{QR} \right | ^{2} = (-1) ^{2} + (- 2) ^{2} + (- 6) ^{2} = 1 + 4 + 36 = 41 \\ \left | \overrightarrow{RP} \right | ^{2} = 2^{2} + (- 1)^{2} + 1^{2} = 4 + 1 + 1 = 6 \\ \left | \overrightarrow{PQ} \right | ^{2} + \left | \overrightarrow{QR} \right | ^{2} = 35 + 6 = 41 = \left | \overrightarrow{QR} \right | ^{2}$$

Hence, PQR is a right angled triangle.

Question 18: Which of the following is incorrect in the triangle PQR?

$$(i) \overrightarrow{PQ} + \overrightarrow{QR} + \overrightarrow{RP} = 0 \\ (ii) \overrightarrow{PQ} + \overrightarrow{QR} – \overrightarrow{PR} = 0 \\ (iii) \overrightarrow{PQ} + \overrightarrow{QR} – \overrightarrow{RP} = 0 \\ (iv) \overrightarrow{PQ} – \overrightarrow{RQ} + \overrightarrow{RP} = 0$$

Answer 18: Based on the formula given in Vector Algebra

In a given triangle, applying the triangle law of addition, we get,

$$\overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR} …. (1) \\ \overrightarrow{PQ} + \overrightarrow{QR} = – \overrightarrow{RP} \\ \overrightarrow{PQ} + \overrightarrow{QR} + \overrightarrow{RP} = \overrightarrow{0} …. (2) \\ Statement\; (i)\; is\; true \\ \overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR} \\ \overrightarrow{PQ} + \overrightarrow{QR} – \overrightarrow{PR} = \overrightarrow{0} \\ Statement\; (ii)\; is\; true \\$$

From equation (2), we get:

$$\overrightarrow{PQ} – \overrightarrow{RQ} + \overrightarrow{RP} = \overrightarrow{0} \\ Statement\; (iv)\; is\; true \\ Considering\; statement\; (iii) \\ \overrightarrow{PQ} + \overrightarrow{QR} – \overrightarrow{RP} = \overrightarrow{0} \\ \overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{RP} ….. (3)$$

From equations (3) and (1), we get:

$$\overrightarrow{PR} = \overrightarrow{RP} \\ \overrightarrow{PR} = – \overrightarrow{PR} \\ 2 \overrightarrow{PR} = \overrightarrow{0} \\ \overrightarrow{PR} = \overrightarrow{0}, \; is \;not\; true. Statement\; (iii)\; is\; true \\$$

Question 19: Check whether the corresponding statements are true if the two vectors $$\overrightarrow{p} \;and\; \overrightarrow{q}$$ are collinear.

$$(i) \overrightarrow{q} = \lambda \overrightarrow{p}, for\; some\; scalar\; \lambda \\ (ii) \overrightarrow{p} = \pm \overrightarrow{q} \\ (iii) The\; components\; of\; \overrightarrow{p} \;and\; \overrightarrow{q}\; are\; proportional \\ (iv) \overrightarrow{p} \;and\; \overrightarrow{q}\; have\; different\; magnitudes\; and\; have\; similar\; direction.$$

Answer 19: Based on the formula given in Vector Algebra

The two vectors are said to be collinear when they are parallel to each other.

$$\overrightarrow{p} \;and\; \overrightarrow{q}$$ are collinear vectors.

Thus, we have,

$$\overrightarrow{q} = \lambda \overrightarrow{p}, (for\; some\; scalar\; \lambda) \\ Suppose,\; \lambda = \pm 1,\; then\; \overrightarrow{q} = \pm 1 \overrightarrow{p} \\ If,\; \overrightarrow{p} = p _{1} \hat{i} + p _{2} \hat{j} + p _{3} \hat{k} ,\; \overrightarrow{q} = q _{1} \hat{i} + q _{2} \hat{j} + q _{3} \hat{k},\; then \;\overrightarrow{q} = \lambda \overrightarrow{p}.$$ $$q _{1} \hat{i} + q _{2} \hat{j} + q _{3} \hat{k} = \lambda (p _{1} \hat{i} + p _{2} \hat{j} + p _{3} \hat{k}) \\ q _{1} \hat{i} + q _{2} \hat{j} + q _{3} \hat{k} = (\lambda p _{1}) \hat{i} + (\lambda p _{2}) \hat{j} + (\lambda p _{3}) \hat{k} \\ q _{1} = \lambda p _{1}, q _{2} = \lambda p _{2}, q _{3} = \lambda p _{3} \\ => \frac{q _{1}}{p _{1}} = \frac{q _{2}}{p _{2}} = \frac{q _{3}}{p _{3}} = \lambda$$

Hence, the components of $$\overrightarrow{p} \;and\; \overrightarrow{q}$$ are proportional.

Though, vectors $$\overrightarrow{p} \;and\; \overrightarrow{q}$$ can have different directions.

Thus, statement (iv) is incorrect.

#### Exercise 10.3

Q.1 : Find the angle between two vectors $$\vec{ m }$$ and $$\vec{ n }$$ with magnitude $$\sqrt{ 3 }$$and 2 , respectively having $$\vec{ m }.\vec{ n } = \sqrt{ 6 }$$

Solution 1:

It is given that,

$$\left | \vec{ m } \right |= \sqrt{ 3 }$$ ,

$$\left | \vec{ n } \right |$$ = 2

And $$\vec{ m }.\vec{ n } = \sqrt{ 6 }$$

$$\vec{ m }.\vec{ n } = \left | \vec{ m } \right |\left | \vec{ n } \right | \cos \theta$$

Now, we know that

Therefore,

$$\boldsymbol{\Rightarrow }$$ $$\sqrt{ 6 } = \sqrt{ 3 } \times 2 \times cos\theta$$

$$\boldsymbol{\Rightarrow }$$ $$\cos \theta = \frac{ \sqrt{ 6 }}{ \sqrt{ 3 } \times 2 }$$

$$\boldsymbol{\Rightarrow }$$ $$\cos \theta = \frac{ 1 }{ \sqrt{ 2 }}$$

$$\boldsymbol{\Rightarrow }$$ $$\theta = \frac{ \pi }{ 4 }$$

Therefore, the angle between the given vectors $$\vec{ m }$$ and $$\vec{ n }$$ is $$\frac{ \pi }{ 4 }$$

Q 2 : Find the angle between the vectors $$\hat{ a } – 2 \hat{ b } + 3 \hat{ c } and 3 \hat{ a } – 2 \hat{ b } + \hat{ c }$$

Solution 2: Based on the formula given in Vector Algebra

The given vectors are:

$$\vec{ m } = \hat{ a } – 2 \hat{ b } + 3 \hat{ c } and \vec{ n } = 3 \hat{ a } – 2 \hat{ b } + \hat{ c }$$ $$\left |\vec{ m } \right | = \sqrt{ 1 ^{ 2 } + \left ( – 2 \right )^{ 2 } + 3 ^{ 2 }}$$ $$\left |\vec{ m } \right | = \sqrt{ 1 + 4 + 9 }$$ $$\left |\vec{ m } \right | = \sqrt{ 14 }$$ $$\left |\vec{ n } \right | = \sqrt{ 3 ^{ 2 } + \left ( – 2 \right )^{ 2 } + 1 ^{ 2 }}$$ $$\left |\vec{ n } \right | = \sqrt{ 9 + 4 + 1 }$$ $$\left |\vec{ n } \right | = \sqrt{ 14 }$$ $$Now, \vec{ m }.\vec{ n } = \left ( \hat{ a } – 2 \hat{ b } + 3 \hat{ c } \right )\left ( 3 \hat{ a } – 2 \hat{ b } + \hat{ c } \right )$$ $$Now, \vec{ m }.\vec{ n } = 1.3 + \left ( – 2 \right )\left ( – 2 \right ) + 3.1$$ $$Now, \vec{ m }.\vec{ n } = 3 + 4 + 3$$ $$Now, \vec{ m }.\vec{ n } = 10$$

Also, we know that

$$\vec{ m }.\vec{ n } = \left | \vec{ m } \right |\left | \vec{ n } \right | \cos \theta$$

Therefore,

$$10 = \sqrt{ 14 } \sqrt{ 14 }\cos \theta$$ $$\cos \theta = \frac{ 10 }{ 14 }$$ $$\theta = \cos^{-1} \frac{ 5 }{ 7 }$$

Q 3. Find the projection of the vector $$\hat{ a } – \hat{ b }$$ on the vector $$\hat{ a } + \hat{ b }$$.

Solution 3: Based on the formula given in Vector Algebra

Let, $$\hat{ i } = \hat{ a } – \hat{ b }$$

And $$\hat{ j } = \hat{ a } + \hat{ b }$$

Now, projection of vector $$\vec{ i }$$ on $$\vec{ j }$$ is given by,

$$\frac{ 1 }{ \left | \vec{ j} \right |} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 1 }{ \sqrt{ 1 + 1 }} \left \{ 1.1 + \left ( – 1 \right ) \left ( 1 \right ) \right \} = \frac{ 1 }{ \sqrt{ 2 }}\left ( 1 – 1 \right ) = 0$$

Hence the projection of vector $$\vec{ i }$$ on $$\vec{ j }$$ is 0

Q 4. Find the projection of the vector $$\hat{ a } + 3 \hat{ b } + 7 \hat{ c }$$ on the vector $$7\hat{ a } – \hat{ b } + 8 \hat{ c }$$

Solution 4: Based on the formula given in Vector Algebra

Let $$\hat{ i } = \hat{ a } + 3 \hat{ b } + 7 \hat{ c }$$ and $$\hat{ j } = 7\hat{ a } – \hat{ b } + 8 \hat{ c }$$

Now, projection of vector $$\vec{ i }$$ on $$\vec{ j }$$ is given by,

$$\frac{ 1 }{ \vec{ j }} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 1 }{ \sqrt{ 7^{ 2 } + \left ( – 1 \right )^{ 2 } + 8 ^{ 2 }}} \left \{ 1 \left ( 7 \right ) + 3 \left ( – 1 \right ) + 7 \left ( 8 \right )\right \}$$ $$\frac{ 1 }{ \vec{ j }} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 7 – 3 + 56 }{ \sqrt{ 49 + 1 + 64 }}$$ $$\frac{ 1 }{ \vec{ j }} \left ( \vec{ i }.\vec{ j } \right ) = \frac{ 60 }{ \sqrt{ 114 }}$$

Q 5: Show that each of the given three vectors is a unit vector :

$$\frac{ 1 }{ 7 } \left ( 2 \hat{ a } + 3 \hat{ b } + 6 \hat{ c }\right ) , \frac{ 1 }{ 7 } \left ( 3 \hat{ a } – 6 \hat{ b } + 2 \hat{ c }\right ) , \frac{ 1 }{ 7 } \left ( 6 \hat{ a } + 2 \hat{ b } – 3 \hat{ c }\right )$$

Also, show that they are mutually perpendicular to each other.

Solution 5: Based on the formula given in Vector Algebra

$$Let \vec{ i } = \frac{ 1 }{ 7 } \left ( 2 \hat{ a } + 3 \hat{ b } + 6 \hat{ c }\right ) = \frac{ 2 }{ 7 } \hat{ a } + \frac{ 3 }{ 7 } \hat{ b } + \frac{ 6 }{ 7 } \hat{ c } \\ Let \vec{ j } = \frac{ 1 }{ 7 } \left ( 3 \hat{ a } – 6 \hat{ b } + 2 \hat{ c }\right ) = \frac{ 3 }{ 7 } \hat{ a } – \frac{ 6 }{ 7 } \hat{ b } + \frac{ 2 }{ 7 } \hat{ c } \\ Let \vec{ k } = \frac{ 1 }{ 7 } \left ( 6 \hat{ a } + 2 \hat{ b } – 3 \hat{ c }\right ) = \frac{ 6 }{ 7 } \hat{ a } + \frac{ 2 }{ 7 } \hat{ b } – \frac{ 3 }{ 7 } \hat{ c }$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{ i } \right | = \sqrt{ \left (\frac{ 2 }{ 7 }\right )^{ 2 } + \left (\frac{ 3 }{ 7 }\right )^{ 2 } + \left (\frac{ 6 }{ 7 }\right )^{ 2 }}$$

$$\left | \vec{ i } \right | = \sqrt{ \frac{ 4 }{ 49 } + \frac{ 9 }{ 49 } + \frac{ 36 }{ 49 }}$$ $$\left | \vec{ i } \right | = 1$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{ j } \right | = \sqrt{ \left (\frac{ 3 }{ 7 }\right )^{ 2 } + \left (\frac{ – 6 }{ 7 }\right )^{ 2 } + \left (\frac{ 2 }{ 7 }\right )^{ 2 }}$$

$$\left | \vec{ j } \right | = \sqrt{ \frac{ 9 }{ 49 } + \frac{ 36 }{ 49 } + \frac{ 4 }{ 49 }}$$ $$\left | \vec{ j } \right | = 1$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{ k } \right | = \sqrt{ \left (\frac{ 6 }{ 7 }\right )^{ 2 } + \left (\frac{ 2 }{ 7 }\right )^{ 2 } + \left (\frac{ – 3 }{ 7 }\right )^{ 2 }}$$

$$\left | \vec{ k } \right | = \sqrt{ \frac{ 36 }{ 49 } + \frac{ 4 }{ 49 } + \frac{ 9 }{ 49 }}$$ $$\left | \vec{ k } \right | = 1$$

Thus, each of the given three vectors is a unit vector.

$$\boldsymbol{\Rightarrow }$$ $$\vec{ i }.\vec{ j } = \frac{ 2 }{ 7 }\times \frac{ 3 }{ 7 } + \frac{ 3 }{ 7 } \times \left ( \frac{ -6 }{ 7 } \right ) + \frac{ 6 }{ 7 } \times \frac{2}{7}$$

$$\vec{ i }.\vec{ j } = \frac{6}{49} – \frac{18}{49} + \frac{12}{49}$$ $$\vec{ i }.\vec{ j } = 0$$

$$\boldsymbol{\Rightarrow }$$ $$\vec{ j }.\vec{ k } = \frac{3}{ 7 }\times \frac{ 6 }{ 7 } + \frac{ -6 }{ 7 } \times \left ( \frac{ 2 }{ 7 } \right ) + \frac{ 2 }{ 7 } \times \frac{ -3}{7}$$

$$\vec{ j }.\vec{ k } = \frac{18}{49} – \frac{12}{49} – \frac{6}{49}$$ $$\vec{ j }.\vec{ k } = 0$$

$$\boldsymbol{\Rightarrow }$$ $$\vec{ k }.\vec{ i } = \frac{6}{ 7 }\times \frac{ 2 }{ 7 } + \frac{ 2 }{ 7 } \times \left ( \frac{ 3 }{ 7 } \right ) + \frac{ -3 }{ 7 } \times \frac{ 6}{7}$$

$$\vec{ k }.\vec{ i } = \frac{12}{49} – \frac{6}{49} – \frac{18}{49}$$ $$\vec{ k }.\vec{ i } = 0$$

Hence, the given three vectors are mutually perpendicular to each other.

Q 6: Find:

$$\left | \vec{m} \right | and \left | \vec{n} \right |$$, if $$\left ( \vec{m} + \vec{n} \right ).\left ( \vec{m} + \vec{n} \right ) = 8$$ and $$\left | \vec{m} \right | = 8\left | \vec{n} \right |$$

Solution 6: Based on the formula given in Vector Algebra

$$\left ( \vec{m} + \vec{n} \right ).\left ( \vec{m} – \vec{n} \right ) = 8$$

$$\boldsymbol{\Rightarrow }$$ $$\vec{ m}.\vec{ m} – \vec{ m}\vec{n} + \vec{ n}\vec{ m} – \vec{n}.\vec{ n} = 8$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{ m} \right |^{ 2} – \left | \vec{ n} \right |^{ 2} = 8$$

$$\boldsymbol{\Rightarrow }$$ $$8 \left | \vec{ n} \right |^{2} – \left | n \right |^{ 2} = 8$$ . . . . . . . . . $$\left [ \left | \vec{ m } = 8\left | \vec{ n} \right | \right | \right ]$$

$$\boldsymbol{\Rightarrow }$$ $$64 \left | \vec{n} \right |^{2} – \left | \vec{n} \right |^{2} = 8$$

$$\boldsymbol{\Rightarrow }$$ $$63 \left | \vec{ n } \right |^{2} = 8$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{n} \right |^{2} = \frac{ 8}{ 63}$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{n} \right |^{2} = \sqrt{\frac{ 8 }{ 63 }}$$ [ magnitude of a vector is non-negative]

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{n} \right |^{2} = \frac{ 2 \sqrt{2}}{ 3 \sqrt{7}}$$

$$\left | \vec{m} \right | = 8 \left | \vec{n} \right |$$ $$\left | \vec{m} \right | = \frac{ 8 \times 2\sqrt{2}}{ 3\sqrt{7}}$$ $$\left | \vec{m} \right | = \frac{ 16 \sqrt{2}}{ 3 \sqrt{7}}$$

Q 7: Find the product of the following : $$\left ( 3 \vec{m} – 5 \vec{n}\right ).\left ( 2\vec{m} + 7\vec{n}\right )$$

Solution 7: Based on the formula given in Vector Algebra

$$\left ( 3 \vec{m} – 5 \vec{n}\right ).\left ( 2\vec{m} + 7\vec{n}\right )$$ $$= 3 \vec{m}.2 \vec{m} + 3 \vec{m}.7 \vec{n} – 5 \vec{n}.2 \vec{m} – 5 \vec{n}.7 \vec{n}$$ $$= 6 \vec{m}. \vec{m} + 21 \vec{m}. \vec{n} – 10 \vec{n}. \vec{m} – 35 \vec{n}. \vec{n}$$ $$= 6 \left | \vec{m} \right |^{2} + 11 \vec{m}.\vec{n} – 35 \left | \vec{n} \right |^{2}$$

Q 8: Find the magnitude of two vectors $$\vec{m} and \vec{n}$$, having the same magnitude and such that the angle between them is 60° and their scalar product is $$\frac{ 1 }{ 2 }$$

Solution 8: Based on the formula given in Vector Algebra

Let θ be the angle between the vectors $$\vec{m} and \vec{n}$$.

As given in the question we have:

$$\left | \vec{m} \right | = \left | \vec{n} \right |, \vec{m}.\vec{n} = \frac{1}{2} and \theta = 60 ^{\circ}$$. . . . . . . . . . . . . . . . . . . . . . . ( 1 )

We know that:

$$\vec{ m }.\vec{ n } = \left | \vec{ m } \right |\left | \vec{ n } \right | \cos \theta$$

Therefore,

$$\frac{1}{2} = \left | \vec{m} \right |\left | \vec{m} \right | cos 60 ^{\circ}$$ . . . . . . . . . . . . . . . . . . . [using ( 1 )]

$$\boldsymbol{\Rightarrow }$$ $$\frac{1}{2} = \left | \vec{m} \right |^{2} \times \frac{1}{2}$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{m} \right |^{2} = 1$$

$$\boldsymbol{\Rightarrow }$$ $$\left | \vec{m} \right |^{2} = \left | \vec{n} \right |^{2} = 1$$

Q 9: Find:

$$\left | \vec{y} \right |$$ , if for a unit vector $$\vec{b}$$ , $$\left (\vec{y } – \vec{b} \right ).\left (\vec{y } + \vec{b} \right ) = 12$$

Solution 9: Based on the formula given in Vector Algebra

$$\left (\vec{y } – \vec{b} \right ).\left (\vec{y } + \vec{b} \right ) = 12$$

$$\boldsymbol{\Rightarrow }$$ $$\vec{y}.\vec{y} + \vec{y}.\vec{b} – \vec{b}.\vec{y} – \vec{b}.\vec{b} = 12$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{y} \right |^{ 2 } – \left |\vec{b} \right |^{ 2 } = 12$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{y} \right |^{ 2 } – 1 = 12 \left [ \left |\vec{b} \right | = 1 as \vec{b} is a unit vector \right ]$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{y} \right |^{ 2 } = 13$$

Therefore,

$$\left | \vec{ y } \right | = \sqrt{ 13 }$$

Q 10: If $$\vec{i} = 2\hat{a} + 2\hat{b} + 3\hat{c}$$ ,

$$\vec{j} = -\hat{a} + 2\hat{b} + \hat{c}$$

and $$\vec{k} = 3\hat{a} + \hat{b}$$ are such that $$\vec{i} + \lambda \vec{j}$$ is perpendicular to $$\vec{k}$$ then find the value of λ

Solution 10: Based on the formula given in Vector Algebra

The given vectors are $$\vec{i} = 2\hat{a} + 2\hat{b} + 3\hat{c}$$ , $$\vec{j} = -\hat{a} + 2\hat{b} + \hat{c}$$ and $$\vec{k} = 3\hat{a} + \hat{b}$$

Now,

$$\vec{i} + \lambda \vec{j} = \left ( 2 \hat{a} + 2 \hat{b} + 3 \hat{c} \right ) + \lambda \left ( – \hat{a} + 2 \hat{b} + \hat{c} \right ) = ( 2 – \lambda ) \hat{a} + ( 2 + 2 \lambda )\hat{b} + \left ( 3 + \lambda \right )\hat{c}$$

If $$\left (\vec{i} + \lambda \vec{j} \right )$$ is perpendicular to $$\vec{k}$$ ,then

$$\left (\vec{i} + \lambda \vec{j} \right ).\vec{k} = 0$$

$$\boldsymbol{\Rightarrow }$$ $$\left [\left ( 2 – \lambda \right )\hat{a} + \left ( 2 + 2\lambda \right )\hat{b} + \left ( 3 + \lambda \right )\hat{c} \right ]\left ( 3\hat{a} + \hat{b} \right ) = 0$$

$$\boldsymbol{\Rightarrow }$$ $$\left [\left ( 2 – \lambda \right )3 + \left ( 2 + 2\lambda \right )1 + \left ( 3 + \lambda \right )0 \right ] = 0$$

$$\boldsymbol{\Rightarrow }$$ $$6 – 3\lambda + 2 + 2\lambda = 0$$

$$\boldsymbol{\Rightarrow }$$ $$– \lambda + 8 = 0$$

$$\boldsymbol{\Rightarrow }$$ $$\lambda = 8$$

Hence, the required value of λ is 8.

Q 11: Show that: Based on the formula given in Vector Algebra

$$\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a}$$ is perpendicular to $$\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a}$$ , for any two non zero vectors $$\vec{a} and \vec{b}$$

Solution 11: Based on the formula given in Vector Algebra

$$\boldsymbol{\Rightarrow }$$ $$\left (\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a} \right ).\left (\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a} \right )$$

$$\left (\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a} \right ).\left (\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a} \right )$$ = $$\left |\vec{a} \right |^{2} \vec{b}.\vec{b} – \left |\vec{a} \right |\left |\vec{b} \right |\vec{b}.\vec{a} + \left |\vec{b} \right |\left |\vec{a} \right |\vec{a}.\vec{b} – \left |\vec{b} \right |^{2} \vec{a}.\vec{a}$$

$$= \left | \vec{a} \right |^{ 2 } \left |\vec{b} \right |^{ 2 }$$

$$\left (\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a} \right ).\left (\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a} \right )$$ = 0

Hence, $$\left |\vec{a} \right |\vec{b} + \left |\vec{b} \right |\vec{a}$$ is perpendicular to $$\left |\vec{a} \right |\vec{b} – \left |\vec{b} \right |\vec{a}$$

Q 12: If $$\vec{a}.\vec{a} = 0 and \vec{a}.\vec{b} = 0$$, then what can be concluded about the vector $$\vec{b}$$ ?

Solution: Based on the formula given in Vector Algebra

It is given that $$\vec{a}.\vec{a} = 0 and \vec{a}.\vec{b} = 0$$

Now we know that,

that $$\vec{a}.\vec{a} = 0$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{a} \right |^{ 2 } = 0$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{a} \right |= 0$$

Therefore, $$\vec{a}$$ is a zero vector

Hence, vector $$\vec{b}$$ satisfying $$\vec{a}.\vec{b} = 0$$ can be any vector

Q 13:

If $$\vec{a} , \vec{b} , \vec{c}$$ are unit vectors such that $$\vec{a} + \vec{b} + \vec{c}$$ = 0, find the value of $$\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a}$$

Solution 13: Based on the formula given in Vector Algebra

$$\left |\vec{a} + \vec{b }+ \vec{c} \right |^{ 2 } = \left (\vec{a} + \vec{b} + \vec{c} \right )\left (\vec{a} + \vec{b} + \vec{c} \right )$$

$$\left |\vec{a} + \vec{b }+ \vec{c} \right |^{ 2 }$$ = $$\left |\vec{a} \right | ^{2}+ \left |\vec{b} \right |^{ 2 } +\left | \vec{c} \right |^{ 2 } + 2 \left ( \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right )$$

$$\boldsymbol{\Rightarrow }$$ $$0 = 1 + 1 + 1 + 2 \left ( \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right )$$

$$\boldsymbol{\Rightarrow }$$ $$\left ( \vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} \right ) = \frac{ – 3 }{ 2 }$$

Q 14: If either vector $$\vec{a} = \vec{0} or \vec{b} = \vec{0 }$$ , then $$\vec{a}.\vec{b} = 0$$. But the converse need not be true. Justify your answer with an example.

Solution 14:

Consider $$\vec{a} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k} and \vec{b} = 3 \hat{i} + 3 \hat{j } – 6 \hat{k}$$

Then,

$$\vec{a}.\vec{b} = 2.3 + 4.3 + 3 \left ( – 6 \right ) = 6 + 12 – 18 = 0$$

We now observe that :

$$\left |\vec{a} \right | = \sqrt{ 2 ^{2} + 4 ^{ 2 } + 3 ^{2}} = \sqrt{ 29 }$$

Therefore,

$$\vec{a} \neq \vec{0}$$ $$\left |\vec{ b } \right | = \sqrt{ 3 ^{2} + 3 ^{ 2 } + ( -6 ) ^{2}} = \sqrt{ 54 }$$

Therefore,

$$\vec{b} \neq \vec{0}$$

Hence, the converse of the given statement need not be true

Q 15: If the vertices A, B, C of a triangle ABC are ( 1 , 2 , 3 ) , ( – 1 , 0 , 0 ) , ( 0 , 1 , 2 ) , respectively, then find ABC. [ ABC is the angle between the vectors $$\vec{BA} and \vec{BC}$$ ]

Solution 15: Based on the formula given in Vector Algebra

The vertices of ∆ABC are given as A ( 1 , 2 , 3 ) , B ( – 1 , 0 , 0 ) , and C ( 0 , 1 , 2 ). Also, it is given that ∠ ABC is the angle between the vectors $$\vec{BA} and \vec{BC}$$

$$\vec{BA} = \left \{ 1 – \left ( – 1 \right ) \right \}\hat{ i} + \left ( 2 – 0 \right ) \hat{j} + \left ( 3 – 0 \right ) \hat{k} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$$ $$\vec{BC} = \left \{ 0 – \left ( – 1 \right ) \right \}\hat{ i} + \left ( 1 – 0 \right ) \hat{j} + \left ( 2 – 0 \right ) \hat{k} = \hat{i} + \hat{j} + 2 \hat{k}$$

Therefore,

$$\vec{BA }. \vec{BC} = \left ( 2 \hat{i} + 2 \hat{j} + 3 \hat{k} \right ).\left ( \hat{i} + \hat{j} + 2 \hat{k} \right ) = 2 \times 1 + 2 \times 1 + 3 \times 2 = 2 + 2 + 6 = 10$$ $$\left |\vec{BA } \right | = \sqrt{ 2 ^{ 2 } + 2 ^{2} + 3^{ 2 }} = \sqrt{ 4 + 4 + 9 } = \sqrt{ 17 }$$ $$\left |\vec{BC } \right | = \sqrt{ 1 + 1 + 2 ^{ 2 }} = \sqrt{ 6 }$$

Now, it is known that :

$$\vec{BA} . \vec{BC} = \left |\vec{BA} \right |\left |\vec{BC} \right | cos \left ( \angle ABC \right )$$

Therefore,

$$10 = \sqrt{ 17 } \times \sqrt{ 6 } cos \left ( \angle ABC \right )$$

$$\boldsymbol{\Rightarrow }$$ $$cos \left ( \angle ABC \right ) = \frac{ 10 }{ \sqrt{ 17 } \times \sqrt{6}}$$

$$\boldsymbol{\Rightarrow }$$ $$\angle ABC = \cos^{-1}\left ( \frac{ 10 }{ \sqrt{ 102 }} \right )$$

Q 16: Show that the points A ( 1 , 2 , 7 ) , B ( 2 , 6 , 3 ) and C ( 3 , 10 , – 1 ) are collinear.

Solution 16: Based on the formula given in Vector Algebra

The given points are A ( 1 , 2 , 7 ) , B ( 2 , 6 , 3 ) , and C ( 3 , 10 , – 1 ).

Therefore,

$$\vec{AB} = \left ( 2 – 1 \right ) \hat{i} + \left ( 6 – 2 \right ) \hat{j} + \left ( 3 – 7 \right ) \hat{k} = \hat{i} + 4 \hat{j} – 4 \hat{k}$$ $$\vec{BC} = \left ( 3 – 2 \right ) \hat{i} + \left ( 10 – 6 \right ) \hat{j} + \left ( – 1 – 3 \right ) \hat{k} = \hat{i} + 4 \hat{j} – 4 \hat{k}$$ $$\vec{AC} = \left ( 3 – 1 \right ) \hat{i} + \left ( 10 – 2 \right ) \hat{j} + \left ( – 1 – 7 \right ) \hat{k} = 2 \hat{i} + 8 \hat{j} – 8 \hat{k}$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{AB} \right | = \sqrt{ 1 ^{2} + 4 ^{2} + \left ( – 4 \right )^{ 2 }} = \sqrt{ 1 + 16 + 16 } = \sqrt{ 33 }$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{BC} \right | = \sqrt{ 1 ^{2} + 4 ^{2} + \left ( – 4 \right )^{ 2 }} = \sqrt{ 1 + 16 + 16 } = \sqrt{ 33 }$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{AC} \right | = \sqrt{ 2 ^{2} + 8 ^{2} + 8 ^{2} } = \sqrt{ 4 + 64 + 64 } = \sqrt{ 132 }$$ = $$2 \sqrt{33}$$

Therefore,

$$\left |\vec{AC} \right | = \left |\vec{AB} \right | + \left |\vec{BC} \right |$$

Hence, the given points A, B, and C are collinear.

Q 17: Show that the vectors $$2 \hat{i} + \hat{j} + \hat{k} , \hat{i} – 3 \hat{j } – 5 \hat{k} and 3 \hat{i} – 4 \hat{j} – 4 \hat{k}$$ form the vertices of a right angled triangle.

Solution 17: Based on the formula given in Vector Algebra

Let vectors $$2 \hat{i} + \hat{j} + \hat{k} , \hat{i} – 3 \hat{j } – 5 \hat{k} and 3 \hat{i} – 4 \hat{j} – 4 \hat{k}$$ be position vectors of points A, B, and C respectively.

i.e , $$\vec{OA} = 2 \hat{i} – \hat{j} + \hat{k} , OB = \hat{i} – 3 \hat{j} – 5 \hat{k} and OC = 3 \hat{i} – 4 \hat{j } – 4 \hat{k}$$

Therefore,

$$\vec{AB} = \left ( 1 – 2 \right ) \hat{i} + \left ( – 3 + 1 \right ) \hat{j} + \left ( – 5 – 1 \right ) \hat{k} = – \hat{i} – 2 \hat{j} – 6 \hat{k}$$ $$\vec{BC} = \left ( 3 – 1 \right ) \hat{i} + \left ( – 4 + 3 \right ) \hat{j} + \left ( – 4 + 5 \right ) \hat{k} = – 2\hat{i} – \hat{j} – \hat{k}$$ $$\vec{AC} = \left ( 2 – 3 \right ) \hat{i} + \left ( – 1 + 4 \right ) \hat{j} + \left ( 1 + 4 \right ) \hat{k} = – \hat{i} + 3 \hat{j} + 5 \hat{k}$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{AB} \right | = \sqrt{ \left ( – 1 \right )^{ 2 } + \left ( – 2 \right )^{ 2 } + \left ( – 6 \right ) ^{ 2 }} = \sqrt{ 1 + 4 + 36 } = \sqrt{ 41 }$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{BC} \right | = \sqrt{ \left ( 2 \right )^{ 2 } + \left ( – 1 \right )^{ 2 } + \left ( 1 \right ) ^{ 2 }} = \sqrt{ 1 + 4 + 1 } = \sqrt{ 6 }$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{AC} \right | = \sqrt{ \left ( – 1 \right )^{ 2 } + \left ( 3 \right )^{ 2 } + \left ( 5 \right ) ^{ 2 }} = \sqrt{ 1 + 9 + 25 } = \sqrt{ 35 }$$

Therefore,

$$\left |\vec{BC} \right |^{ 2 } + \left |\vec{AC} \right | ^{2} = 6 + 35 = 41 = \left |\vec{AB} \right | ^{ 2 }$$

Hence, ∆ ABC is a right – angled triangle.

Q 18:

If $$\vec{a}$$ is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λ $$\vec{a}$$ is unit vector if

(A) λ = 1

(B) λ = – 1

(C) $$a = \left |\lambda \right |$$

(D) $$a = \frac{ 1 }{ \left | \lambda \right |}$$

Solution 18:

Vector $$\lambda \vec{a}$$ is a unit vector if $$\left |\lambda \vec{a} \right | = 1$$

Now we know that,

$$\left |\lambda \vec{a} \right | = 1$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\lambda \right | \left |\vec{a} \right | = 1$$

$$\boldsymbol{\Rightarrow }$$ $$\left |\vec{a} \right | = \frac{ 1 }{ \left |\lambda \right | }$$ . . . . . . . . . . . . [ λ ≠ 0 ]

$$\boldsymbol{\Rightarrow }$$ $$a = \frac{ 1 }{ \left |\lambda \right | }$$ . . . . . . . . . . . . . . . . . . [$$\left |\vec{a} \right | = a$$ ]

Therefore, vector $$\lambda \vec{a}$$ is a unit vector if $$a = \frac{ 1 }{ \left | \lambda \right |}$$

#### Exercise 10.4

Q 1: Find $$\left | \vec{a }\times \vec{b} \right |$$ , if $$\vec{a} = \hat{i} – 7 \hat{j} + 7 \hat{k}$$ and $$\vec{ b } = 3 \hat{i} – 2 \hat{j} + 2 \hat{k}$$

Solution 1: Based on the formula given in Vector Algebra

We have,

$$\vec{a} = \hat{i} – 7 \hat{j} + 7 \hat{k}$$ and $$\vec{ b } = 3 \hat{i} – 2 \hat{j} + 2 \hat{k}$$

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & – 7 & 7 \\ 3 & – 2 & 2 \end{vmatrix}$$

$$\vec{a} \times \vec{b}$$ = $$\hat{i} \left ( – 14 + 14 \right ) – \hat{j} \left ( 2 – 21 \right ) + \hat{k} \left ( – 2 + 21 \right ) = 19 \hat{j} + 19 \hat{k}$$

Therefore,

$$\left |\vec{a} \times \vec{b} \right | = \sqrt{\left ( 19 \right ) ^{ 2} + \left ( 19 \right ) ^{2}} = \sqrt{ 2 \times \left ( 19 \right ) ^{ 2 }} = 19 \sqrt{2}$$

Q 2: Find a unit vector perpendicular to each of the vector $$\vec{a} + \vec{b} and \vec{a} – \vec{b}$$, where $$\vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} – 2 \hat{k}$$

Solution 2: Based on the formula given in Vector Algebra

We have,

$$\vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k} and \vec{b} = \hat{i} + 2 \hat{j} – 2 \hat{k}$$

Therefore,

$$\vec{a} + \vec{b} = 4 \hat{i} + 4 \hat{j} , \vec{a} – \vec{b} = 2 \hat{i} + 4 \hat{k}$$ $$\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix}$$ $$\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) = \hat{i} \left ( 16 \right ) – \hat{j} \left ( 16 \right ) + \hat{k} \left ( – 8 \right ) = 16 \hat{i} – 16 \hat{j} – 8 \hat{k}$$ $$\left |\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) \right | = \sqrt{16 ^{2} + \left ( – 16 \right )^{2} + \left ( – 8 \right ) ^{ 2 }}$$ $$\left |\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) \right | = \sqrt{ 2 ^{2} \times 8 ^{ 2} + 2 ^{2} \times 8 ^{2 } + 8 ^{2}}$$ $$\left |\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) \right | = 8 \sqrt{ 2^{2} + 2 ^{2} + 1 } = 8 \sqrt{9} = 8 \times 3 = 24$$

Therefore, the unit vector perpendicular to each of the vectors $$\vec{a} + \vec{b} and \vec{a} – \vec{b}$$ is given by the relation,

= $$\pm \frac{ \left ( \vec{a} + \vec{b} \right ) \times \left ( \vec{a} – \vec{b} \right )}{\left |\left (\vec{a} + \vec{b} \right ) \times \left (\vec{a} – \vec{b} \right ) \right | } = \pm \frac{ 16 \hat{i} – 16 \hat{j} – 8 \hat{k}}{ 24 }$$

= $$\pm \frac{ 2 \hat{i} – 2 \hat{j} – \hat{k} }{ 3 } = \pm \frac{ 2 }{ 3 } \hat{i} \mp \frac{2}{3} \hat{j} \mp \frac{1}{ 3} \hat{k}$$

Q 3: If a unit vector $$\vec{a}$$ makes angle $$\frac{ \pi }{3} with \hat{i} , \frac{ \pi }{ 4} angle with \hat{j}$$ and an acute angle θ with $$\vec{k}$$ , then find θ and hence, the compounds of $$\vec{a}$$

Solution 3: Based on the formula given in Vector Algebra

Let unit vector $$\vec{a}$$ have ( a 1 , a 2 , a 3 ) components as follows

$$\vec{a} = a _{1} \hat{i} + a _{2} \hat{ j} + a _{ 3 } \hat{ k }$$

Since, $$\vec{ a }$$ is a unit vector , $$\left |\vec{a} \right | = 1$$

Also, it is given that $$\vec{a}$$ makes angles $$\frac{ \pi }{3} with \hat{i} , \frac{ \pi }{ 4} angle with \hat{j}$$ and an acute angle θ with $$\vec{k}$$

Then, we have :

$$\cos \frac{ \pi }{ 3} = \frac{ a _{1}}{ \left |\vec{a} \right | }$$

$$\boldsymbol{\Rightarrow }$$ $$\frac{ 1 }{ 2 } = a _{1} . . . . . . . . . . \left [ \left |\vec{a} \right | = 1 \right ]$$

$$\cos \frac{ \pi }{ 4 } = \frac{ a _{2 } }{ \left |\vec{a} \right | }$$

$$\boldsymbol{\Rightarrow }$$ $$\frac{ 1 }{ \sqrt{ 2 }} = a _{ 2 } . . . . . . . . . . \left [ \left |\vec{a} \right | = 1 \right ]$$

Also , $$\cos \theta = \frac{ a _{ 3 }}{ \left |\vec{a} \right | }$$

$$\boldsymbol{\Rightarrow }$$ $$a _{ 3 } = \cos \theta$$

$$\left |a \right | = 1$$

$$\boldsymbol{\Rightarrow }$$ $$\sqrt{ {a_{ 1 }}^{ 2 } + {a_{ 2 }}^{ 2 } + {a_{ 3 }}^{ 2 } } = 1$$

$$\boldsymbol{\Rightarrow }$$ $$\frac{ 1 }{ 2 }^{ 2 } + \frac{ 1 }{ \sqrt{ 2 }}^{ 2 } + \cos ^{ 2 } \theta = 1$$

$$\boldsymbol{\Rightarrow }$$ $$\frac{ 1 }{ 4 } + \frac{ 1 }{ 2 } + \cos ^{ 2 } \theta = 1$$

$$\boldsymbol{\Rightarrow }$$ $$\frac{ 3 }{ 4 } + \cos ^{ 2 } \theta = 1$$

$$\boldsymbol{\Rightarrow }$$ $$\cos ^{ 2 } \theta = 1 – \frac{ 3 }{ 4 } = \frac{ 1 }{ 4 }$$

$$\boldsymbol{\Rightarrow }$$ $$\cos \theta = \frac{ 1 }{ 2 } \\ \theta = \cos^{-1} \left ( \frac{ 1 }{ 2 } \right ) \\ \theta = \frac{ \pi }{ 3 }$$

Therefore,

$$a_{ 3 } = \cos \frac{ \pi }{ 3 } = \frac{1}{ 2 }$$

Therefore, $$\theta = \frac{ \pi }{ 3} and the components of \vec{a} are \left ( \frac{ 1 }{ 2} , \frac{ 1 }{ \sqrt{ 2}} , \frac{ 1 }{2} \right )$$

Q 4: Show that:

$$\left (\vec{a} – \vec{b} \right ) \times \left (\vec{a} + \vec{b} \right ) = 2 \left ( \vec{a} \times \vec{b} \right )$$

Solution: Based on the formula given in Vector Algebra

To prove:

$$\left (\vec{a} – \vec{b} \right ) \times \left (\vec{a} + \vec{b} \right ) = 2 \left ( \vec{a} \times \vec{b} \right )$$

= $$\left (\vec{a} – \vec{b} \right ) \times \left (\vec{a} + \vec{b} \right ) = \left ( \vec{a} -\vec{ b} \right ) \times \vec{a} + \left ( \vec{a} – \vec{b} \right ) \times \vec{b}$$ . . . . . . . . . . . . [ By distributivity of vector product over addition ]

= $$\vec{a} \times \vec{a} – \vec{b} \times \vec{a} + \vec{a} \times \vec{b} – \vec{b} \times \vec{b}$$ . . . . . . . . . . . . . . . [ again, by distributivity of vector product over addition ]

= $$\vec{0} + \vec{a} \times \vec{b} + \vec{a} \times \vec{b} – \vec{0}$$

= $$2 \vec{a} \times \vec{b}$$

Q 5: Find $$\lambda and \mu if \left (2 \hat{i} + 6 \hat{j }+ 27 \hat{k } \right ) \times \left ( \hat{i} + \lambda \hat{j} + \mu \hat{k} \right ) = \vec{0}$$

Solution:

$$\left (2 \hat{i} + 6 \hat{j }+ 27 \hat{k } \right ) \times \left ( \hat{i} + \lambda \hat{j} + \mu \hat{k} \right ) = \vec{0}$$

$$\boldsymbol{\Rightarrow }$$ $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda \mu & \end{vmatrix}$$ = $$0 \hat{i} + 0 \hat{j} + 0 \hat{k}$$

$$\boldsymbol{\Rightarrow }$$ $$\hat{i} \left ( 6 \mu – 27 \mu \right ) – \hat{j} \left ( 2 \mu – 27 \right ) + \hat{k} \left ( 2 \lambda – 6 \right ) = 0 \hat{i} + 0 \hat{j} + 0 \hat{k}$$

On comparing the corresponding components, we have :

$$6 \mu – 27 \lambda = 0 \\ 2 \mu – 27 = 0 \\ 2 \lambda – 6 = 0$$

Now,

$$2 \lambda – 6 = 0 \Rightarrow \lambda = 3$$

$$\boldsymbol{\Rightarrow }$$ $$2 \mu – 27 = 0 \\ \Rightarrow \mu = \frac{ 27 }{ 2 }$$

Therefore, $$\lambda = 3 and \mu = \frac{ 27 }{ 2 }$$

Q 6: Given that:

$$\vec{a}.\vec{b} = 0 and \vec{a} \times \vec{b} = \vec{0}$$

What can you conclude about the vectors $$\vec{a} and \vec{b}$$ ?

Solution: Based on the formula given in Vector Algebra

$$\vec{a}.\vec{b} = 0$$

Then ,

1. i) either $$\vec{a } = 0 or \vec{b } = 0 , or \vec{a} \perp \vec{b} \left ( in case \vec{a} and \vec{b} are non – zero \right ) \\ \vec{a} \times \vec{b} = 0$$
2. ii) Either $$\vec{a } = 0 or \vec{b } = 0 , or \vec{a} \parallel \vec{b} \left ( in case \vec{a} and \vec{b} are non – zero \right ) \\ \vec{a} \times \vec{b} = 0$$

But, $$\vec{a} and \vec{b}$$ cannot be perpendicular and parallel simultaneously.

Therefore, $$\left |\vec{a} \right | = 0 or \left |\vec{b} \right | = 0$$

Q 7: Let the vectors $$\vec{a }, \vec{b} ,\vec{ c} given as a_{ 1 } \hat{i} + a_{ 2 } \hat{j} + a_{3} \hat{k} , b_{1} \hat{i} + b_{2} \hat{j} + b_{ 3 } \hat{k} , c_{ 1 } \hat{i} + c_{ 2} \hat{j} + c_{ 3 } \hat{k}$$

Then show that = $$\vec{a} \times \left ( \vec{ b } + \vec{ c } \right ) = \vec{a} \times \vec{b} + \vec{a }\times \vec{ c }$$

Solution: Based on the formula given in Vector Algebra

We have,

$$a_{ 1 } \hat{i} + a_{ 2 } \hat{j} + a_{3} \hat{k} , b_{1} \hat{i} + b_{2} \hat{j} + b_{ 3 } \hat{k} , c_{ 1 } \hat{i} + c_{ 2} \hat{j} + c_{ 3 } \hat{k}$$ $$\left ( \vec{b} + \vec{c} \right ) = \left ( b_{ 1 } + c _{1} \right ) \hat{i} + \left ( b_{ 2 } + c_{ 2 } \right ) \hat{j} + \left ( b_{ 3 } + c_{ 3 } \right ) \hat{k}$$

Now, $$\vec{a} \times \left ( \vec{b} + \vec{c} \right )\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_{ 1 } & a_{ 2} & a_{ 3 } \\ b_{1} + c_{1} & b_{2} + c_{ 2 } & b_{ 3 } + c_{ 3 } \end{vmatrix}$$

$$= \hat{i} \left [ a_{ 2} \left ( b_{ 3 } + c_{3 } \right ) – a_{ 3 } \left ( b_{ 2 } + c_{ 2 } \right ) \right ] – \hat{j} \left [ a_{ 1 } \left ( b_{ 3 } + c_{ 3 } \right ) – a_{ 3 } \left ( b_{ 1 } + c_{ 1 } \right ) \right ] + \hat{k} \left [ a_{ 1 } \left ( b_{ 2 } + c _{ 2 }\right ) – a_{ 2 } \left ( b_{ 1 } + c_{ 1 } \right ) \right ]$$

$$= \hat{i} \left [ a_{ 2 } b_{3} + a_{2} c_{3} – a_{3} b_{2} – a_{3} c_{2} \right ] + \hat{j} \left [ – a_{1} b_{3} – a_{1} c_{3} + a_{3} b_{1} + a_{3} c_{1} \right ] + \hat{k} \left [ a_{1} b_{2} + a_{1} c_{2} – a_{2}b _{1} – a_{2} c_{1} \right ]$$ . . . . . . . . . . . . . . . . . ( 1 )

$$\boldsymbol{\Rightarrow }$$ $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1 } & a_{2} & a_{3} \\ b_{ 1} & b_{ 2} & b_{3} \end{vmatrix}$$

= $$\hat{i} \left [ a_{ 2 } b_{ 3 } – a_{ 3 } b_{2} \right ] + \hat{j} \left [ a_{3} b_{1} – a_{1} b_{3} \right ] + \hat{k} \left [ a_{1} b_{2} – a_{2} b_{1} \right ]$$ . . . . . . . . . . . . . . . . . . . . . . . . . . ( 2 )

$$\boldsymbol{\Rightarrow }$$ $$\vec{a} \times \vec{ c } = \begin{vmatrix} \hat{ I } & \hat{ j } & \hat{ k } \\ a_{1 } & a_{2} & a_{3} \\ c_{ 1} & c_{ 2 } & c_{3 } \end{vmatrix}$$

$$\hat{i} \left [ a_{ 2 } c_{ 3 } – a_{ 3 } c_{2} \right ] + \hat{j} \left [ a_{3} c_{1} – a_{1} c_{3} \right ] + \hat{k} \left [ a_{1} c_{2} – a_{2} c_{1} \right ]$$ . . . . . . . . . . . . . . . . . . . . . . . . . . ( 3 )

On adding (2) and (3), we get:

$$\left ( \vec{a} \times \vec{b} \right ) + \left ( \vec{a} \times \vec{ c} \right ) = \hat{i} \left [ a_{ 2 } b_{ 3 } + a_{ 2 } c_{ 3 } + – a _{ 3 } b_{ 2 } – a_{ 3 } c_{ 2 } \right ] + \hat{j} \left [ a_{ 3 } b_{ 1 } + a_{ 3 } c_{ 1 } + – a_{ 1 } b_{ 3 } – a_{ 1 } c_{ 3 } \right ] + \hat{ k } \left [ a_{ 1 } b_{ 2 } + a_{ 1 } c_{ 2 } + – a_{ 2 } b_{ 1 } – a_{ 2 } c_{ 1 } \right ]$$ . . . . . . . . . . . . . . . . . . . . . . . . . . ( 4 )

Now, from (1) and (4), we have:

$$\vec{a} \times \left ( \vec{b} + \vec{c} \right ) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$$

Hence, the given result is proved.

Question 8: Suppose, any of the vector $$\overrightarrow{m} \;or\; \overrightarrow{n} = \overrightarrow{0}$$, then $$\overrightarrow{m} \times \overrightarrow{n} = \overrightarrow{0}$$.

Is the above given statement true?

Answer 8: Based on the formula given in Vector Algebra

By taking any two non – zero vectors, for the condition $$\overrightarrow{m} \times \overrightarrow{n} = \overrightarrow{0}$$.

Suppose, $$\overrightarrow{m} = \hat{i} + 2 \hat{j} + 3 \hat{k} \;and\; \overrightarrow{n} = 2 \hat{i} + 4 \hat{j} + 6 \hat{k} \\ Then,$$

$$\overrightarrow{PQ} \times \overrightarrow{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 4 & 6 \end{vmatrix}$$ $$\hat{i} (4 – 4) – \hat{j} (6 – 6) + \hat{k} (4 – 4) = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} = \hat{0} \\ Now,\; we\; find\; that, \\ \left [ \overrightarrow{m} \right ] = \sqrt{(1) ^{2} + (2) ^{2} + (3) ^{2}} = \sqrt{14} \\ So,\; \overrightarrow{m} \neq \overrightarrow{0} \\ \left [ \overrightarrow{n} \right ] = \sqrt{(2) ^{2} + (4) ^{2} + (6) ^{2}} = \sqrt{56} \\ So,\; \overrightarrow{n} \neq \overrightarrow{0}$$

Thus, the above given statement is not true.

Question 9: P (1, 1, 2), Q (2, 3, 5) and R (1, 5, 5) are the vertices of a triangle. Obtain the area.

Answer 9: Based on the formula given in Vector Algebra

Given:

Vertices of a triangle P (1, 1, 2), Q (2, 3, 5) and R (1, 5, 5) are as follows

The contiguous sides $$\overrightarrow{PQ} \;and\; \overrightarrow{QR}$$ of the triangle is given as,

$$\overrightarrow{PQ} = (2 – 1) \hat{i} + (3 – 1) \hat{j} + (5 – 2) \hat{k} = \hat{i} + 2 \hat{j} + 3 \hat{k} \;and\; \overrightarrow{QR} = (1 – 2) \hat{i} + (5 – 3) \hat{j} + (5 – 5) \hat{k} = – \hat{i} + 2 \hat{j} \\ Area\; of\; \Delta \;triangle = \frac{1}{2} \left | \overrightarrow{PQ} \times \overrightarrow{QR} \right |$$ $$\overrightarrow{PQ} \times \overrightarrow{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ – 1 & 2 & 0 \end{vmatrix} \\ \hat{i} (- 6) – \hat{j} (3) + \hat{k} (2 + 2) = – 6 \hat{i} -3 \hat{j} + 4 \hat{k} \\ \left | \overrightarrow{PQ} \times \overrightarrow{QR} \right | = \sqrt{(- 6) ^{2} + (- 3) ^{2} + 4 ^{2}} = \sqrt{36 + 9 + 16} = \sqrt{61}$$

Thus, $$\frac{\sqrt{61}}{2}$$ is the area of triangle ABC.

Question 10: $$\overrightarrow{m} = \hat{i} – \hat{j} + 3 \hat{k} \;and\; \overrightarrow{n} = 2 \hat{i} – 7 \hat{j} + \hat{k}$$ are the adjacent sides of a vector. Obtain the area of the parallelogram.

The area of the parallelogram whose contiguous sides are as follows$$\overrightarrow{m} \;and\; \overrightarrow{n} is \left | \overrightarrow{m} \times \overrightarrow{n} \right |$$

Given to us,

$$\overrightarrow{m} = \hat{i} – \hat{j} + 3 \hat{k} \;and\; \overrightarrow{n} = 2 \hat{i} – 7 \hat{j} + \hat{k}$$ are the adjacent sides of a vector.

$$\overrightarrow{m} \times \overrightarrow{n} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & – 1 & 3 \\ 2 & – 7 & 1 \end{vmatrix} \\ \hat{i} (- 1 + 21) – \hat{j} (1 – 6) + (- 7 + 2) \hat{k} = 20 \hat{i} + 5 \hat{j} – 5 \hat{k} \\ \left | \overrightarrow{m} \times \overrightarrow{n} \right | = \sqrt{20 ^{2} + 5 ^{2} + 5 ^{2}} = \sqrt{400 + 25 + 25} = 15 \sqrt{2}$$

$$15 \sqrt{2}$$ is the area of the given parallelogram.

Question 11: Suppose, vectors $$\overrightarrow{m} \;and\; \overrightarrow{n} \; in\; such\; a\; way\; that \left | \overrightarrow{m} \right | = 3 \;and \left | \overrightarrow{n} \right | = \frac{\sqrt{2}}{3},\; then\; \overrightarrow{m} \times \overrightarrow{n}$$ is a unit vector, suppose the angle between the two vectors is

$$(i)\; \frac{\pi }{6}, \;(ii)\; \frac{\pi }{4},\; (iii)\; \frac{\pi }{3} \;(iv) \frac{\pi }{2}$$

Answer 11: Based on the formula given in Vector Algebra

Given, $$\left | \overrightarrow{m} \right | = 3 \;and \left | \overrightarrow{n} \right | = \frac{\sqrt{2}}{3},\; then\; as\; we\; know\; \overrightarrow{m} \times \overrightarrow{n} = 1 \\ \overrightarrow{m} \times \overrightarrow{n} = \left | \overrightarrow{m} \right | \left | \overrightarrow{n} \right | \;sin \theta \hat{s}$$, where s is a unit vector perpendicular to both $$\overrightarrow{m} \;and\; \overrightarrow{n}$$ and $$\Theta$$ is the angle between $$\overrightarrow{m} \;and\; \overrightarrow{n}$$

Now, $$\overrightarrow{m} \;and\; \overrightarrow{n} \;is\; a \;unit \; vector \;if\; \left | \overrightarrow{m} \times \overrightarrow{n} \right | = 1 \\ \left | \overrightarrow{m} \times \overrightarrow{n} \right | = 1 \\ \left | \left | \overrightarrow{m} \right | \left | \overrightarrow{n} \right | sin\; \Theta \hat{s} \right | = 1 \\ \left | \left | \overrightarrow{m} \right | \left | \overrightarrow{n} \right | sin\; \Theta \right | = 1 \\ 3 \times \frac{\sqrt{2}}{3} \times sin\; \Theta = 1 \\ sin\; \Theta = \frac{1}{\sqrt{2}} \\ \Theta = \frac{\pi }{4}$$

The correct answer is option (ii)

Question 12: The area of the rectangle with P, Q, R and S as the vertices with positive vectors $$– \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \hat{i} – \frac{1}{2} \hat{j} + 4 \hat{k} \;and\; – \hat{i} – \frac{1}{2} \hat{j} + 4 \hat{k} \; respectively\; is \\ (i) \frac{1}{2}, \;\;\;\; (ii) 1 \\ (iii) 2, \;\;\;\; (iv) 4$$

Answer 12: Based on the formula given in Vector Algebra

Given,

The area of the rectangle PQRS with P, Q, R and S as the vertices with positive vectors such as follows:

$$\overrightarrow{OP} = – \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k},\; \overrightarrow{OQ} = \hat{i} + \frac{1}{2} \hat{j} + 4 \hat{k}, \;\overrightarrow{OR} = \hat{i} – \frac{1}{2} \hat{j} + 4 \hat{k} \;and\; \overrightarrow{OS} – \hat{i} – \frac{1}{2} \hat{j} + 4 \hat{k}$$

The contiguous sides $$\overrightarrow{PQ} \;and\; \overrightarrow{QR}$$ of the rectangle is given as follows,

$$\overrightarrow{PQ} = (1 + 1) \hat{i} + (\frac{1}{2} – \frac{1}{2}) \hat{j} + (4 – 4) \hat{k} = 2 \hat{i} \;and\; \overrightarrow{QR} = (1 – 1) \hat{i} + \left (- \frac{1}{2} – \frac{1}{2} \right ) \hat{j} + (4 – 4) \hat{k} = – \hat{j}$$ $$\overrightarrow{PQ} \times \overrightarrow{QR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 0 & – 1 & 0 \end{vmatrix} \\ \hat{k} (- 2) = – 2 \hat{k} \\ \left | \overrightarrow{PQ} \times \overrightarrow{QR} \right | = \sqrt{(- 2) ^{2}} = 2$$

Area of a rectangle = 2 square units.

Option (iii) is the correct .