Chapter – 11: 3-Dimensional Geometry
Exercise 1.1
Q1. Find the cosine directions, if a line makes angles 90°, 135°, 45° with x, y and z- axes respectively.
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
Let, the direction cosines of the line be p, q and r
p = cos 90° = 0
q = cos 135° = \(-\frac{1}{\sqrt{2}}\)
r = cos 45° = \(\frac{1}{\sqrt{2}}\)
Therefore, the direction cosines of the lines are 0, \(\frac{-1}{\sqrt{2}}\) and \(\frac{1}{\sqrt{2}}\).
Q2. Find the direction cosines of a line which makes equal angles with the coordinate axes.
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
Let the direction cosines of the line make an angle α with each of the coordinate axes.
p = cos α, q = cos α and r = cos α
Therefore, p2 + q2 + r2 = 1
cos2 α + cos2 α + cos2 α = 1
3cos2 α = 1
cos2 α = \(\frac{1}{3}\)
cos2 α = ± \(\frac{1}{\sqrt{3}}\)
Therefore, the direction cosines of the line, which is equally inclined to the coordinate axes, are \(\pm \frac{1}{\sqrt{3}}, \;\pm \frac{1}{\sqrt{3}}\; and \; \pm \frac{1}{\sqrt{3}}\).
Q3. If a line has the direction ratios 12, -18, -4 then what are its direction cosines?
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
If a line has direction ratios of 12, -18, -4 then its direction cosines are:
\(\frac{12}{\sqrt{\left ( 12 \right )^{2} + \left ( -18 \right )^{2} + \left ( -4 \right )^{2}}}\), \(\frac{-18}{\sqrt{\left ( 12 \right )^{2} + \left ( -18 \right )^{2} + \left ( -4 \right )^{2}}}\), \(\frac{ -4 }{\sqrt{\left ( 12 \right )^{2} + \left ( -18 \right )^{2} + \left ( -4 \right )^{2}}}\)
That is,
\(\frac{12}{22}, \frac{-18}{22}, \frac{-4}{22}\) \(\frac{6}{11}, \frac{-9}{11}, \frac{-2}{11}\\\)Thus, the direction cosines are \(\frac{6}{11}, \frac{-9}{11} \; and \; \frac{-2}{11}\)
Q4. Prove that the given points are collinear:
(-1, -2, 1), (2, 3, 4) and (8, 8, 7)
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
The points are as follows:
P (-1, -2, 1)
Q (2, 3, 4)
R (8, 13, -2)
It is known that the direction ratios of line joining the points (x1, y1, z1) and (x1, y2, z2) are given by:
(x2 – y1), (y2 – y1) and (z2 – z1)
The direction ratios are PQ are [2 – (-1)], [3 – (-2)] and (4 – 1) that is 3, 5 and -3.
The direction ratios are QR are (8 – 2), (13 – 3) and (-2 – 4) that is 6, 10 and -6.
It can be seen that the direction ratios of QR are 2 times that of PQ. That is they are proportional.
Therefore, PQ is parallel to QR and Q is the common point to both PQ and QR. P, Q and R are collinear.
Q5. Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (− 1, 1, 2) and (− 5, − 5, − 2).
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
The vertices of \(\Delta ABC\) are X (3, 5, −4), Y (−1, 1, 2) and Z (−5, −5, −2).
The direction ratios of side XY are (−1 − 3), (1 − 5), and [2 − (−4)], that is −4, −4, and 6.
i.e. \(\sqrt{\left ( -4 \right )^{2} + \left ( -4 \right )^{2} + \left ( 6 \right )^{2}}\) = \(\sqrt{ 16 + 16 + 36}\) = \(\sqrt{ 68 }\) = \(2\sqrt{ 17 }\)
Hence, the direction cosines of XY are:
= \(\frac{-4}{\sqrt{\left ( -4 \right )^{2} + \left ( -4 \right )^{2} + \left ( 6 \right )^{2}}}\), \(\frac{-4}{\sqrt{\left ( -4 \right )^{2} + \left ( -4 \right )^{2} + \left ( 6 \right )^{2}}}\), \(\frac{6}{\sqrt{\left ( -4 \right )^{2} + \left ( -4 \right )^{2} + \left ( 6 \right )^{2}}}\)
= \(\frac{-4}{2 \sqrt{17}}\), \(\frac{-4}{2 \sqrt{17}}\), \(\frac{6}{2 \sqrt{17}}\)
= \(\frac{-2}{ \sqrt{17}}\), \(\frac{-2}{ \sqrt{17}}\), \(\frac{3}{ \sqrt{17}}\)
The direction ratios of YZ are [−5 − (−1)], (−5 − 1), and (−2 − 2), that is −4, −6, and −4.
Hence, the direction cosines of XY are:
= \(\frac{-4}{\sqrt{\left ( -4 \right )^{2} + \left ( -6 \right )^{2} + \left ( -4 \right )^{2}}}\), \(\frac{-6}{\sqrt{\left ( -4 \right )^{2} + \left ( -6 \right )^{2} + \left ( -4 \right )^{2}}}\), \(\frac{-4}{\sqrt{\left ( -4 \right )^{2} + \left ( -6 \right )^{2} + \left ( -4 \right )^{2}}}\)
= \(\frac{-4}{2 \sqrt{17}}\), \(\frac{-6}{2 \sqrt{17}}\), \(\frac{-4}{2 \sqrt{17}}\)
The direction ratios of ZX are (−5 − 3), (−5 − 5), and (−2 − (−4), that is −8, −10, and 2.
Hence, the direction cosines of XZ are:
= \(\frac{-8}{\sqrt{\left ( -8 \right )^{2} + \left ( 10 \right )^{2} + \left ( 2 \right )^{2}}}\), \(\frac{-5}{\sqrt{\left ( -8 \right )^{2} + \left ( 10 \right )^{2} + \left ( 2 \right )^{2}}}\), \(\frac{2}{\sqrt{\left ( -8 \right )^{2} + \left ( 10 \right )^{2} + \left ( 2 \right )^{2}}}\)
= \(\frac{-8}{2 \sqrt{42}}\), \(\frac{-5}{2 \sqrt{42}}\), \(\frac{2}{2 \sqrt{42}}\)
Exercise 11.2
Q1. Prove that the three lines with direction cosines \(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}: \frac{4}{13}, \frac{12}{13}, \frac{3}{13}: \frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\) are mutually perpendicular.
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
Two lines with direction cosines: l1, m1, n1 and l2, m2, n2 are perpendicular to each other, if l1 l2 + m1 m2 + n1 n2 = 0
(i) For the lines with direction cosines, \(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\) and \(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\\\)
Therefore, l1 l2 + m1 m2 + n1 n2 = \(\frac{12}{13} \times \frac{4}{13} + \left ( \frac{-3}{13} \right ) \times \frac{12}{13} + \left ( \frac{-4}{13} \right ) \times \frac{3}{13}\\\)
i.e. l1 l2 + m1 m2 + n1 n2 = \(\frac{48}{169} – \frac{36}{169} – \frac{12}{13}\)
(or) l1 l2 + m1 m2 + n1 n2 = 0
Hence, the lines are perpendicular.
(ii) For the lines with direction cosines, \(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\) and \(\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\\\).
l1 l2 + m1 m2 + n1 n2 = \(\left [ \frac{3}{13} \times \frac{12}{13} \right ] + \left [ \left (\frac{-4}{13} \right ) \times \left (\frac{-3}{13} \right ) \right ] + \left [ \frac{12}{13} \times \left (\frac{-4}{13} \right ) \right ]\\\)
Therefore, l1 l2 + m1 m2 + n1 n2 = \(\frac{12}{169} – \frac{48}{169} + \frac{36}{169}\)
or, l1 l2 + m1 m2 + n1 n2 = 0
Hence, the lines are perpendicular.
(iii) For the lines with direction cosines, \(\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\) and \(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\\\)
l1 l2 + m1 m2 + n1 n2 =
Therefore, l1 l2 + m1 m2 + n1 n2 = \(\frac{36}{169} + \frac{12}{169} – \frac{48}{169}\)
i.e. l1 l2 + m1 m2 + n1 n2 = 0
Hence, the lines are perpendicular.
Therefore, all the lines are mutually perpendicular.
Q2. Show that the line through the points (1, −1, 2) (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
Let PQ be the line joining the points, (1, −1, 2) and (3, 4, − 2), and RS be the line joining the points, (0, 3, 2) and (3, 5, 6).
The direction ratios p1 + q1 + r1 of PQ are (3 − 1), [4 − (−1)], and (−2 − 2) that is 2, 5, and −4.
The direction ratios p1 + q1 + r1 of RS are (3 − 0), (5 − 3), and (6 −2) that is 3, 2, and 4.
PQ and RS will be perpendicular to each other, if l1 l2 + m1 m2 + n1 n2 = 0
l1 l2 + m1 m2 + n1 n2 = 2 × 3 + 5 ×2 + (-4) × 4
i.e. l1 l2 + m1 m2 + n1 n2 = 6 + 10 – 16 = 0
Hence, PQ and RS are perpendicular to each other.
Q3. Prove that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (−1, −2, 1) and (1, 2, 5).
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
Let PQ be the line through the points, (4, 7, 8) and (2, 3, 4), and RS be the line through the points, (−1, −2, 1) and (1, 2, 5).
The directions ratios p1, q1, r1 of PQ are (2 − 4), (3 − 7), and (4 − 8) that is −2, −4, and -4.
The direction ratios p2, q2, r2 of RS are [1 − (−1)], [2 − (−2)], and (5 − 1) that is 2, 4, and 4.
PQ will be parallel to RS, if \(\frac{p_{1}}{p_{2}} = \frac{q_{1}}{q_{2}} = \frac{r_{1}}{r_{2}}\)
\(\frac{p_{1}}{p_{2}} = \frac{-2}{2} = -1\\\)Also, \(\frac{q_{1}}{q_{2}} = \frac{-4}{4} = -1\\\)
And \(\frac{r_{1}}{r_{2}} = \frac{-4}{4} = -1\\\)
Therefore, \(\frac{p_{1}}{p_{2}} = \frac{q_{1}}{q_{2}} = \frac{r_{1}}{r_{2}}\)
Hence, PQ is parallel to RS.
Q4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector \(3\hat{i} + 2\hat{j} – 2\hat{k}\).
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
It is given that the line passes through the point A (1, 2, 3).
Therefore, the position vector through A is,
\(\bar{a} = \hat{i} + 2\hat{j} + 3\hat{k}\) \(\bar{b} = 3\hat{i} + 2\hat{j} – 2\hat{k}\)It is known that the line which passes through point A and parallel to \(\vec{b}\) is given by \(\vec{r} = \vec{a} + \lambda \vec{b}\) where \(\lambda\) is a constant.
\(\vec{r} = \hat{i} + 2 \hat{j} + 3\hat{k} + \lambda \left ( 3\hat{i} + 2\hat{j} – 2\hat{k} \right )\)This is required equation of the line.
Q5. Find the equation of the line in Cartesian and in vector form that passes through the point with position vector \(2\hat{i} – \hat{j} + 4\hat{k}\) and is in the direction \(\hat{i} + 2\hat{j} – \hat{k}\).
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
It is given that line passes through the point with position vector
\(\vec{a} = 2\hat{i} – \hat{j} + 4\hat{k}\) ——– (1)
\(\vec{b} = \hat{i} + 2\hat{j} – \hat{k}\) ———- (2)
It is known that a line through a point with position vector \(\vec{a}\) and parallel to \(\vec{b}\) is given by the equation,
\(\vec{r} = \vec{a} + \lambda \vec{b}\) \(\vec{r} = 2\hat{i} – \hat{j} + 4\hat{k} + \lambda \left (\hat{i} + 2\hat{j} – \hat{k} \right )\)Eliminating \(\lambda\), we obtain the Cartesian form of equation as,
\(\frac{x – 2}{1} = \frac{y + 1}{2} = \frac{z – 4}{-1}\)This is the required equation of the given line in Cartesian form.
Q6. Find the Cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by \(\frac{x + 3}{3} = \frac{y – 4}{5} = \frac{z + 8}{6}\).
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
It is given that line passes through the given point (−2, 4, −5) and it is parallel to \(\frac{x + 3}{3} = \frac{y – 4}{5} = \frac{z + 8}{6}\)
The direction ratios of the line \(\frac{x + 3}{3} = \frac{y – 4}{5} = \frac{z + 8}{6}\) are 3, 5, and 6.
The required line is parallel to \(\frac{x + 3}{3} = \frac{y – 4}{5} = \frac{z + 8}{6}\).
Hence, its direction ratios are 3k, 5k, and 6k, where k ≠ 0.
It is known that the equation of the line through the point \(\left ( x_{1}, y_{1}, z_{1} \right )\) and with direction ratios p, q, r is given by \(\frac{x – x_{1}}{p} = \frac{y – y_{1}}{q} = \frac{z – z_{1}}{r}\).
Hence, the equation of the required line is,
\(\frac{x + 2}{3k} = \frac{y – 4}{5k} = \frac{z + 5}{6k}\) \(\frac{x + 2}{3} = \frac{y – 4}{5} = \frac{z + 5}{6} = k\)
Q7. The Cartesian equation of a line is \(\frac{x – 5}{3} = \frac{y + 4}{7} = \frac{z – 6}{2}\). Give the vector form of the line.
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
The Cartesian equation of the line is,
\(\frac{x – 5}{3} = \frac{y + 4}{7} = \frac{z – 6}{2}\) ——— (1)
The given line passes through the point (5, -4, 6). The position vector of this point is \(\vec{a} = 5\hat{i} – 4\hat{j} + 6\hat{k}\).
Also, the direction ratios of the given line are 3, 7, and 2.
This means that the line is in the direction of vector, \(\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}\).
It is known that the line through position vector \(\vec{a}\) and in the direction of the vector \(\vec{b}\) is given by the equation,
\(\vec{r} = \vec{a} + \lambda \vec{b},\; \lambda \in R\) \(\vec{r} = \left (5\hat{i} – 4\hat{j} + 6\hat{k} \right ) + \lambda \left ( 3\hat{i} + 7\hat{j} + 2\hat{k} \right )\)This is the required equation of the given line in vector form.
Q8. Find the Cartesian and the vector equations of the lines that pass through the origin and (5, −2, 3).
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
The required line passes through the origin. Hence, its position vector is given by,
\(\vec{a} = \vec{0}\) ————– (1)
The direction ratios of the line through origin and (5, −2, 3) are,
(5 − 0) = 5
(−2 − 0) = −2
(3 − 0) = 3
The line is parallel to the vector given by the equation,
\(\vec{b} = 5\hat{i} – 2\hat{j} + 3\hat{k}\)The equation of the line in vector form through a point with position vector \(\vec{a}\) and parallel to \(\vec{b}\) is,
\(\vec{r} = \vec{a} + \lambda \vec{b},\; \lambda \in R\) \(\vec{r} = \vec{0} + \lambda \left ( 5\hat{i} – 2\hat{j} + 3\hat{k} \right )\) \(\vec{r} = \lambda \left ( 5\hat{i} – 2\hat{j} + 3\hat{k} \right )\)The equation of the line through the point \(\left ( x_{1}, y_{1}, z_{1} \right )\) and the direction ratios are p, q, r is given as,
\(\frac{x – x_{1}}{p} = \frac{y – y_{1}}{q} = \frac{z – z_{1}}{r}\)Hence, the equation of the required line in the Cartesian form is,
\(\frac{x – 0}{5} = \frac{y – 0}{-2} = \frac{z – 0}{3}\) \(\frac{x}{5} = \frac{y}{-2} = \frac{z}{3}\)
Q9. Find the Cartesian and the vector equations of the line that passes through the points (3, −2, −5) and (3, −2, 6).
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
Let the line passing through the points, M (3, −2, −5) and N (3, −2, 6), be MN.
Since MN passes through M (3, −2, −5), its position vector is given by, \(\vec{a} = 3\hat{i} – 2\hat{j} – 5\hat{k}\).
The direction ratios of MN are given by,
(3 − 3) = 0
(−2 + 2) = 0
(6 + 5) = 11
The equation of the vector in the direction of MN is,
\(\vec{b} = 0.\hat{i} – 0.\hat{j} + 11\hat{k}\) \(\vec{b} = 11\hat{k}\)The equation of MN in vector form is given by,
\(\vec{r} = \vec{a} + \lambda \vec{b},\; \lambda \in R\) \(\vec{r} = \left (3\hat{i} – 2\hat{j} – 5\hat{k} \right ) + 11 \lambda \hat{k}\)The equation of MN in Cartesian form is,
\(\frac{x – x_{1}}{p} = \frac{y – y_{1}}{q} = \frac{z – z_{1}}{r}\)i.e. \(\frac{x – 3}{0} = \frac{y + 2}{0} = \frac{z + 5}{11}\)
Q10. Calculate the angle between the given pairs of lines:
(i) \(\vec{r} = 2\hat{i} – 5\hat{j} + \hat{k} + \lambda \left ( 3\hat{i} – 2\hat{j} + 6\hat{k} \right )\) and \(\vec{r} = 7\hat{i} – 6\hat{k} + \mu \left ( \hat{i} + 2\hat{j} + 2\hat{k} \right )\)
(ii) \(\vec{r} = 3\hat{i} + \hat{j} – 2\hat{k} + \lambda \left ( \hat{i} – \hat{j} – 2\hat{k} \right ) \; and \; \vec{r} = 2\hat{i} – \hat{j} – 56\hat{k} + \mu \left ( 3\hat{i} – 5\hat{j} – 4\hat{k} \right )\)
Sol:
(i) Let \(\theta\) be the angle between the given lines.
The angle between the given pairs of lines is given by,
\(\cos \theta = \left | \frac{\vec{b_{1}}. \vec{b_{2}}}{\left | \vec{b_{1}} \right | \left | \vec{b_{2}} \right |} \right |\)The given lines are parallel to the vectors, \(\vec{b_{1}} = 3\hat{i} + 2\hat{j} + 6\hat{k}\) and \(\vec{b_{2}} = \hat{i} + 2\hat{j} + 2\hat{k}\) respectively.
Therefore,
\(\left |\vec{b_{1}} \right | = \sqrt{3^{2} + 2^{2} + 6^{2}}\)= 7
\(\\\boldsymbol{\Rightarrow }\) \(\left |\vec{b_{2}} \right | = \sqrt{\left ( 1 \right )^{2} + \left ( 2 \right )^{2} + \left ( 2 \right )^{2}}\) = 3
\(\\\boldsymbol{\Rightarrow }\) \(\vec{b_{1}}.\vec{b_{2}} = \left ( 3\hat{i} + 2\hat{j} + 6\hat{k} \right ).\left ( \hat{i} + 2\hat{j} + 2\hat{k} \right )\)
\(\\\boldsymbol{\Rightarrow }\) \(\vec{b_{1}}.\vec{b_{2}} = 3 \times 1 + 2 \times 2 + 6 \times 2\) = 3 + 4 + 12 = 19
\(\\\boldsymbol{\Rightarrow }\) \(\cos \theta = \frac{19}{7 \times 3}\)
\(\\\boldsymbol{\Rightarrow }\) \(\theta = \cos ^{-1} \left (\frac{19}{21} \right )\)
(ii) The given lines are parallel to the vectors, \(\vec{b_{1}} = \hat{i} – \hat{j} – 2\hat{k}\) and \(\vec{b_{2}} = 3\hat{i} – 5\hat{j} – 4\hat{k}\) respectively.
Therefore,
\(\left |\vec{b_{1}} \right | = \sqrt{\left (-1 \right )^{2} + \left (-1 \right )^{2} + \left (-2 \right )^{2}}= \sqrt{6}\)\(\\\boldsymbol{\Rightarrow }\) \(\left |\vec{b_{2}} \right | = \sqrt{\left (3 \right )^{2} + \left (-5 \right )^{2} + \left (-4 \right )^{2}}\)
\(\left |\vec{b_{2}} \right | = \sqrt{50}\) \(= 5\sqrt{2}\)
\(\\\boldsymbol{\Rightarrow }\) \(\vec{b_{1}}.\vec{b_{2}} = \left ( \hat{i} – \hat{j} – 2\hat{k} \right ).\left ( 3\hat{i} – 5\hat{j} – 4\hat{k}\right )\)
\(\\\boldsymbol{\Rightarrow }\) \(\vec{b_{1}}.\vec{b_{2}} = 1 \times 3 – 1 \left ( -5 \right ) – 2\left ( -4 \right )\) = 3 + 5 + 8 = 16
\(\\\boldsymbol{\Rightarrow }\) \(\cos \theta = \left | \frac{\vec{b_{1}}. \vec{b_{2}}}{\left | \vec{b_{1}} \right | \left | \vec{b_{2}} \right |} \right | = \frac{16}{\sqrt{6}.5\sqrt{2}}= \frac{16}{\sqrt{2}.\sqrt{3}.5\sqrt{2}} = \frac{16}{10\sqrt{3}}\)
\(\\\boldsymbol{\Rightarrow }\) \(\cos \theta = \frac{8}{5\sqrt{3}}\)
\(\\\boldsymbol{\Rightarrow }\) \(\theta = \cos ^{-1}\left (\frac{8}{5\sqrt{3}} \right )\)
Q11. Calculate the angle between the given pairs of lines:
(i) \(\frac{x – 2}{2} = \frac{y – 1}{5} = \frac{z + 3}{-3}\) and \(\frac{x + 2}{-1} = \frac{y – 4}{8} = \frac{z – 5}{4}\)
(ii) \(\frac{x }{2} = \frac{y}{2} = \frac{z}{1}\) and \(\frac{x – 5}{4} = \frac{y – 2}{1} = \frac{z – 3}{8}\)
Sol:
(i) Let \(\vec{b_{1}}\) and \(\vec{b_{2}}\) be the vectors parallel to the pair of lines, \(\frac{x – 2}{2} = \frac{y – 1}{5} = \frac{z + 3}{-3}\) and \(\frac{x + 2}{-1} = \frac{y – 4}{8} = \frac{z – 5}{4}\)
\(\vec{b_{1}} = 2\hat{i} + 5\hat{j} – 3\hat{k}\) and
\(\vec{b_{2}} = -\hat{i} + 8\hat{j} + 4\hat{k}\)\(\\\boldsymbol{\Rightarrow }\) \(\left |\vec{b_{1}} \right | = \sqrt{\left ( 2 \right )^{2} + \left ( 5 \right )^{2} + \left ( -3 \right )^{2}}= \sqrt{38}\)
\(\\\boldsymbol{\Rightarrow }\) \(\left |\vec{b_{2}} \right | = \sqrt{\left ( -1 \right )^{2} + \left ( 8 \right )^{2} + \left ( 4 \right )^{2}}= \sqrt{81}\)= 9
\(\\\boldsymbol{\Rightarrow }\) \(\vec{b_{1}}. \vec{b_{2}} = \left (2\hat{i} + 5\hat{j} – 3\hat{k} \right ).\left (-\hat{i} + 8\hat{j} + 4\hat{k} \right )\)
\(\\\boldsymbol{\Rightarrow }\) \(\vec{b_{1}}. \vec{b_{2}} = 2\left ( -1 \right ) + 5 \times 8 + \left ( -3 \right )4 \)= -2 + 40 – 12 = 26
The angle \(\theta\), between the given pair of lines is given by the relation,
\(\cos \theta = \left | \frac{\vec{b_{1}}. \vec{b_{2}}}{\left | \vec{b_{1}} \right | \left | \vec{b_{2}} \right |} \right | = \frac{26}{9\sqrt{38}}\)\(\\\boldsymbol{\Rightarrow }\) \(\theta = \cos ^{-1}\left (\frac{26}{9\sqrt{38}} \right )\)
(ii) Let \(\vec{b_{1}}\) and \(\vec{b_{2}}\) be the vectors parallel to the given pair of lines, \(\frac{x }{2} = \frac{y}{2} = \frac{z}{1}\) and \(\frac{x – 5}{4} = \frac{y – 2}{1} = \frac{z – 3}{8}\) respectively.
\(\\\boldsymbol{\Rightarrow }\) \(\vec{b_{1}} = 2\hat{i} + 2\hat{j} + \hat{k}\) \(\;\;and \;\;\vec{b_{2}} = 4\hat{i} + \hat{j} + 8\hat{k}\)
\(\\\boldsymbol{\Rightarrow }\) \(\left |\vec{b_{1}} \right | = \sqrt{\left ( 2 \right )^{2} + \left ( 2 \right )^{2} + \left ( 1 \right )^{2}}= \sqrt{9}\)= 3
\(\\\boldsymbol{\Rightarrow }\) \(\left |\vec{b_{2}} \right | = \sqrt{\left ( 4 \right )^{2} + \left ( 1 \right )^{2} + \left ( 8 \right )^{2}} = \sqrt{81}\) = 9
\(\\\boldsymbol{\Rightarrow }\) \(\vec{b_{1}}. \vec{b_{2}} = \left ( 2\hat{i} + 2\hat{j} + \hat{k} \right ).\left ( 4\hat{i} + \hat{j} + 8\hat{k} \right )\)
\(\\\boldsymbol{\Rightarrow }\) \(\vec{b_{1}}. \vec{b_{2}} = 2 \times 4 + 2 \times 1 + 1 \times 8\) = 8 + 2 + 8 = 18
The angle \(\theta\), between the given pair of lines is given by the relation,
\(\\\boldsymbol{\Rightarrow }\) \(\cos \theta = \left | \frac{\vec{b_{1}}. \vec{b_{2}}}{\left | \vec{b_{1}} \right | \left | \vec{b_{2}} \right |} \right |= \frac{18}{3 \times 9}= \frac{2}{3}\)
\(\\\boldsymbol{\Rightarrow }\) \(\theta = \cos ^{-1}\left (\frac{2}{3} \right )\)
Q12. Find the values of m so the line \(\frac{1 – x}{3} = \frac{7y – 14}{2m} = \frac{z – 3}{2}\) and \(\frac{7 – 7x}{3m} = \frac{y – 5}{1} = \frac{6 – z}{5}\) are at right angles.
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
The given equations can be written in the standard form as \(\frac{x – 1}{-3} = \frac{y – 2}{\frac{2m}{7}} = \frac{z – 3}{2}\) and \(\frac{x – 1}{\frac{-3m}{7}} = \frac{y – 5}{1} = \frac{z – 6}{-5}\)
The direction ratios of the lines are -3, \(\frac{2m}{7}\), 2 and \(\frac{-3m}{7}\),1 , -5 respectively.
Two lines with direction ratios, \(a_{1}, b_{1}, c_{1}\) and \(a_{2}, b_{2}, c_{2}\) are perpendicular to each other, if \(a_{1}a_{1} + b_{1}b_{2} + c_{1}c_{2} = 0\)
\(\left ( -3 \right ).\left ( \frac{-3m}{7} \right ) + \left ( \frac{2m}{7} \right ). \left ( 1 \right ) + 2.\left ( -5 \right ) = 0\) \(\frac{9m}{7} + \frac{2m}{7} = 10\) \(11m = 70\) \(m = \frac{70}{11}\\\)Therefore, the value of m is \(\frac{70}{11}\).
Q13. Show that the lines \(\frac{x – 5}{7} = \frac{y + 2}{-5} = \frac{z}{1}\) and \(\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\) are perpendicular to each other.
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
The equations of the given lines are lines \(\frac{x – 5}{7} = \frac{y + 2}{-5} = \frac{z}{1}\) and \(\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\).
The direction ratios of the given lines are 7, −5, 1 and 1, 2, 3 respectively.
Two lines with direction ratios, \(a_{1}, b_{1}, c_{1}\) and \(a_{2}, b_{2}, c_{2}\) are perpendicular to each other, if \(a_{1}a_{1} + b_{1}b_{2} + c_{1}c_{2} = 0\)
\(\left (7 \times 1 \right ) + \left (-5 \times 2 \right ) + \left (1 \times 3 \right )\\\)= 7 – 10 + 3 = 0
Hence, the given lines are perpendicular to each other.
Q14. Find the shortest distance between the given lines,
\(\vec{r} = \left ( \hat{i} + 2\hat{j} + \hat{k} \right ) + \lambda \left ( \hat{i} – \hat{j} + \hat{k} \right )\) and \(\vec{r} = \left ( 2\hat{i} – \hat{j} – \hat{k} \right ) + \mu \left ( 2\hat{i} + \hat{j} + 2\hat{k} \right )\)
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
The equations of the given lines are,
\(\vec{r} = \left ( \hat{i} + 2\hat{j} + \hat{k} \right ) + \lambda \left ( \hat{i} – \hat{j} + \hat{k} \right )\) \(\vec{r} = \left ( 2\hat{i} – \hat{j} – \hat{k} \right ) + \mu \left ( 2\hat{i} + \hat{j} + 2\hat{k} \right )\)It is known that the shortest distance between the lines, \(\vec{r} = \vec{a_{1}} + \lambda \vec{b_{1}}\) and \(\vec{r} = \vec{a_{2}} + \mu \vec{b_{2}}\), is given by:
\(\\d = \left | \frac{\left ( \vec{b_{1}} \times \vec{b_{2}} \right ). \left ( \vec{a_{2}} – \vec{a_{1}} \right ) }{\left | \vec{b_{1}} \times \vec{b_{2}} \right |} \right |\) . . . . . . . . . . . . . (1)
Comparing the given equations, we obtain:
\(\vec{a_{1}} = \hat{i} + 2\hat{j} + \hat{k}\) , \(\vec{b_{1}} = \hat{i} – \hat{j} + \hat{k}\)
And, \(\vec{a_{2}} = 2\hat{i} – \hat{j} – \hat{k}\), \(\vec{b_{2}} = 2\hat{i} + \hat{j} + 2\hat{k}\)
\(\boldsymbol{\Rightarrow }\) \(\vec{a_{2}} – \vec{a_{1}}= \left (2\hat{i} – \hat{j} – \hat{k} \right ) – \left ( \hat{i} + 2\hat{j} + \hat{k} \right )= \hat{i} – 3\hat{j} – 2\hat{k}\\\)
\(\\\vec{b_{1}} \times \vec{b_{2}}\) = \(\boldsymbol{\begin{vmatrix} \hat{i}&\hat{j} &\hat{k}\\ 1& -1&1\\ 2& 1& 2\end{vmatrix}}\\\)
\(\vec{b_{1}} \times \vec{b_{2}} = \left ( -2 – 1 \right )\hat{i} – \left ( 2 – 2 \right )\hat{j} + \left ( 1 + 2 \right )\hat{k} = -3\hat{i} + 3\hat{k}\)And, \(\left |\vec{b_{1}} \times \vec{b_{2}} \right | = \sqrt{\left ( -3 \right )^{2} + \left ( 3 \right )^{2}} = \sqrt{9 + 9}= 3\sqrt{2}\)
Substituting all the values in equation (1), we obtain:
\(d = \left | \frac{\left ( -3\hat{i} + 3\hat{k}\right ). \left ( \hat{i} – 3\hat{j} – 2\hat{k}\right )}{3\sqrt{2}} \right |= \left | \frac{-3.1 + 3.\left ( -2 \right )}{3\sqrt{2}} \right |\left | =\frac{-9}{3\sqrt{2}} \right |=\frac{3}{\sqrt{2}} =\frac{3 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}=\frac{3\sqrt{2}}{2}\)Hence, the shortest distance between the two lines is \(\frac{3\sqrt{2}}{2}\) units.
Q15. Find the shortest distance between the lines \(\frac{x + 1}{7} = \frac{y + 1}{-6} = \frac{z + 1}{1}\) and \(\frac{x – 3}{1} = \frac{y – 5}{-2} = \frac{z – 7}{1}\)
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
The given lines are as follows,
\(\frac{x + 1}{7} = \frac{y + 1}{-6} = \frac{z + 1}{1}\) and \(\frac{x – 3}{1} = \frac{y – 5}{-2} = \frac{z – 7}{1}\)
It is known that the shortest distance between the two lines, \(\frac{x – x_{1}}{a_{1}} = \frac{y – y_{1}}{b_{1}} = \frac{z – z_{1}}{c_{1}}\) and \(\frac{x – x_{2}}{a_{2}} = \frac{y – y_{2}}{b_{2}} = \frac{z – z_{2}}{c_{2}}\) is given by,
\(\boldsymbol{\Rightarrow \frac{\begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1}& b_{1} &c_{1} \\ a_{2}& b_{2} &c_{2} \end{vmatrix} }{\sqrt{(b_{1}c_{2}-b_{1}c_{1})^{2}-(c_{1}a_{2}-c_{2}a_{1})^{2}+(a_{1}b_{2}-a_{2}b_{1})^{2}}}}\\\). . . . . . . . . . . (1)
Comparing the given equations, we obtain:
\(x_{1} = -1, y_{1} = -1, z_{1} = -1\)\(a_{1} = 7, b_{1} = -6, c_{1} = 1\)
And, \(x_{2} = 3, y_{2} = 5, z_{2} = 7\)\(a_{2} = 1, b_{2} = -2, c_{2} = 1\)
Then,
\(\boldsymbol{\Rightarrow \begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1}& b_{1} &c_{1} \\ a_{2}& b_{2} &c_{2} \end{vmatrix}\;=\;\begin{vmatrix} 4 & 6 & 8\\ 7 & -6 & 8\\ 1 & -2 & 1 \end{vmatrix}}\\\)= \(\\4\left ( -6 + 2 \right ) – 6\left ( 7 – 1 \right ) + 8\left ( -14 + 6 \right )\)
= \(– 16 – 36 – 64\) = – 116
= \(\\\sqrt{\left ( b_{1}c_{2} – b_{2}c_{1} \right )^{2} + \left ( c_{1}a_{2} – c_{2}a_{1} \right )^{2} + \left (a_{1}b_{2} – a_{2}b_{1} \right )^{2} }\\\)
= \(\\\sqrt{\left ( -6 + 2 \right )^{2} + \left ( 1 + 7 \right )^{2} + \left ( – 14 + 6 \right )^{2}}=\sqrt{116}=2\sqrt{29}\\\)
Substituting all the values in equation (1), we obtain:
\(d = \frac{-116}{2\sqrt{29}}= \frac{-58}{\sqrt{29}}= \frac{-2 \times 29}{\sqrt{29}}= -2 \sqrt{29}\)Since distance is always non-negative, the distance between the given lines is \(2 \sqrt{29}\) units.
Q16. Find the shortest distance between the lines whose vector equations are \(\vec{r} = \left ( \hat{i} + 2\hat{j} + 3\hat{k} \right ) + \lambda \left ( \hat{i} – 3\hat{j} + 2\hat{k} \right )\) and \(\vec{r} = \left ( 4\hat{i} + 5\hat{j} + 6\hat{k} \right ) + \mu \left ( 2\hat{i} + 3\hat{j} + \hat{k} \right )\)
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
The given vectors are as follows:
\(\vec{r} = \left ( \hat{i} + 2\hat{j} + 3\hat{k} \right ) + \lambda \left ( \hat{i} – 3\hat{j} + 2\hat{k} \right )\\\) \(\vec{r} = \left ( 4\hat{i} + 5\hat{j} + 6\hat{k} \right ) + \mu \left ( 2\hat{i} + 3\hat{j} + \hat{k} \right )\)It is known that the shortest distance between the lines, \(\vec{r} = \vec{a_{1}} + \lambda \vec{b_{1}}\) and \(\vec{r} = \vec{a_{2}} + \mu \vec{b_{2}}\)
The lines are given by:
\(d = \left | \frac{\left ( \vec{b_{1}} \times \vec{b_{2}} \right ). \left ( \vec{a_{2}} – \vec{a_{1}} \right ) }{\left | \vec{b_{1}} \times \vec{b_{2}} \right |} \right |\) . . . . . . . . . . . (1)
Comparing the given equations with \(\vec{r} = \vec{a_{1}} + \lambda \vec{b_{1}}\) and \(\vec{r} = \vec{a_{2}} + \mu \vec{b_{2}}\)
\(\\\vec{a_{1}} = \hat{i} + 2\hat{j} + 3\hat{k}\), \(\vec{b_{1}} = \hat{i} – 3\hat{j} + 2\hat{k}\), \(\vec{a_{2}} = 4\hat{i} + 5\hat{j} + 6\hat{k}\), \(\vec{b_{2}} = 2\hat{i} + 3\hat{j} + \hat{k}\)
\(\vec{a_{2}} – \vec{a_{1}} = \left ( 4\hat{i} + 5\hat{j} + 6\hat{k} \right ) – \left ( \hat{i} + 2\hat{j} + 3\hat{k} \right )= 3\hat{i} + 3\hat{j} + 3\hat{k}\\\) \(\\\vec{b_{1}} \times \vec{b_{2}} = \begin{vmatrix} \hat{i} \;\;\;\;\;\; \hat{j} \;\;\;\;\;\; \hat{k}\\ 1 \;\;\; -3 \;\;\;\; 2\\ 2 \;\;\;\;\;\; 3 \;\;\;\;\;\; 1 \end{vmatrix}\)= \(\\\left ( -3 – 6 \right )\hat{i} – \left ( 1 – 4 \right )\hat{j} + \left ( 3 + 6 \right )\hat{k} = -9 \hat{i} + 3\hat{j} + 9\hat{k}\\\)
\(\\\left |\vec{b_{1}} \times \vec{b_{2}} \right | = \sqrt{\left ( -9 \right )^{2} + \left ( 3 \right )^{2} + \left ( 9 \right )^{2}}=\sqrt{81 + 9 + 81} =sqrt{171}=3\sqrt{19}\\\) \(\\\left (\vec{b_{1}} \times \vec{b_{2}} \right ). \left (\vec{a_{2}} – \vec{a_{1}} \right ) = \left (-9 \hat{i} + 3\hat{j} + 9\hat{k} \right ). \left ( 3\hat{i} + 3\hat{j} + 3\hat{k} \right )\)= \(\\\left (-9 \times 3 \right ) + \left ( 3 \times 3 \right ) + \left ( 9 \times 3 \right )\) = 9
Substituting all the values in equation (1), we obtain:
\(d = \left | \frac{9}{3\sqrt{19}} \right | = \left | \frac{3}{\sqrt{19}} \right |\)Hence, the shortest distance between the two given lines is \(\frac{3}{\sqrt{19}}\) units.
Q17. Find the shortest distance between the lines whose vector equations are \(\vec{r} = \left ( 1 – t \right )\hat{i} + \left ( t – 2 \right )\hat{j} + \left ( 3 – 2t \right )\hat{k}\) and \(\vec{r} = \left ( s + 1 \right )\hat{i} + \left ( 2s – 1 \right )\hat{j} + \left ( 2s + 1 \right )\hat{k}\)
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
The lines are as follows:
\(\vec{r} = \left ( 1 – t \right )\hat{i} + \left ( t – 2 \right )\hat{j} + \left ( 3 – 2t \right )\hat{k}\)\(\\\vec{r} = \left ( \hat{i} – 2\hat{j} + 3\hat{k} \right ) + t \left ( -\hat{i} + \hat{j} – 2\hat{k} \right )\) . . . . . . . . (1)
\(\vec{r} = \left ( s + 1 \right )\hat{i} + \left ( 2s – 1 \right )\hat{j} + \left ( 2s + 1 \right )\hat{k}\)\(\\\vec{r} = \left ( \hat{i} – \hat{j} + \hat{k} \right ) + s \left ( \hat{i} + 2\hat{j} – 2\hat{k} \right )\) . . . . . . . . . . (2)
It is known that the shortest distance between the lines, \(\vec{r} = \vec{a_{1}} + \lambda \vec{b_{1}}\) and \(\vec{r} = \vec{a_{2}} + \mu \vec{b_{2}}\) is given by:
\(\\d = \left | \frac{\left ( \vec{b_{1}} \times \vec{b_{2}} \right ). \left ( \vec{a_{2}} – \vec{a_{1}} \right ) }{\left | \vec{b_{1}} \times \vec{b_{2}} \right |} \right |\) . . . . . . . . (3)
Comparing the given equations with \(\vec{r} = \vec{a_{1}} + \lambda \vec{b_{1}}\) and \(\vec{r} = \vec{a_{2}} + \mu \vec{b_{2}}\),
\(\vec{a_{1}} = \hat{i} – 2\hat{j} + 3\hat{k}\), \(\vec{b_{1}} = -\hat{i} + \hat{j} – 2\hat{k}\)
And, \(\vec{a_{2}} = \hat{i} – \hat{j} – \hat{k}\), \(\vec{b_{2}} = \hat{i} + 2\hat{j} – 2\hat{k}\)
\(\vec{a_{2}} – \vec{a_{1}} = \left ( \hat{i} – \hat{j} – \hat{k} \right ) – \left ( \hat{i} – 2\hat{j} + 3\hat{k} \right ) = \hat{j} – 4\hat{k}\\\) \(\\\vec{b_{1}} \times \vec{b_{2}} = \begin{vmatrix} \hat{i} \;\;\;\;\;\; \hat{j} \;\;\;\;\;\; \hat{k}\\ -1 \;\;\; 1 \;\;\;\; -2\\ \;\;1 \;\;\;\;\;\; 2 \;\;\;\;-2 \end{vmatrix}\\\)= \(\\\left ( -2 + 4 \right )\hat{i} – \left ( 2 + 2 \right )\hat{j} + \left ( -2 – 1 \right )\hat{k}= 2\hat{i} – 4\hat{j} – 3\hat{k}\)
\(\left | \vec{b_{1}} \times \vec{b_{2}} \right | = \sqrt{\left ( 2 \right )^{2} + \left ( -4 \right )^{2} + \left ( -3 \right )^{2}}= \sqrt{4 + 16 + 9}= \sqrt{29}\\\)
\(\\\boldsymbol{\Rightarrow }\) \(\\\left (\vec{b_{1}} \times \vec{b_{2}} \right ). \left ( \vec{a_{2}} – \vec{a_{1}}\right ) = \left (2\hat{i} – 4\hat{j} – 3\hat{k} \right ). \left ( \hat{j} – 4\hat{k} \right )\) = -4 + 12 = 8
Substituting all the values in equation (3), we obtain:
\(d = \left | \frac{8}{\sqrt{29}} \right | = \frac{8}{\sqrt{29}}\)Hence, the shortest distance between the lines is \(\frac{8}{\sqrt{29}}\) units.
Exercise 11.3
Q1. In the following cases, find the cosine directions of the normal to the plane and the distance from the origin.
(a) z = 2
(b) x = y+ z = 1
(c) 2x + 3y – z = 5
(d) 5y + 8 = 0
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
(a) The equation of the plane is z = 2 OR 0x + 0y + z = 2
The direction ratios of normal are 0, 0, and 1.
\(\sqrt{0^{2} + 0^{2} + 1^{2}} = 1\)Dividing both the sides of equation (1) by 1, we get,
0.x + 0.y + 1.z = 2
This is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin. Hence, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.
(b) x + y+ z = 1 —— (1)
The direction ratios of normal are 1, 1, and 1.
\(\sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}\)
Dividing both the sides of equation (1) by \(\sqrt{3}\), we get
\(\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}y + \frac{1}{\sqrt{3}}z = \frac{1}{\sqrt{3}}\) —–(2)
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
Hence, the direction cosines of the normal are \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\) and \(\frac{1}{\sqrt{3}}\) the distance of normal from the origin is \(\frac{1}{\sqrt{3}}\) units.
(c) 2x + 3y − z = 5 —— (1)
The direction ratios of normal are 2, 3, and −1.
\(\sqrt{\left ( 2 \right )^{2} + \left ( 3 \right )^{2} + \left ( -1 \right )^{2}} = \sqrt{14}\)Dividing both the sides of equation (1) by \(\sqrt{14}\), we get,
\(\frac{2}{\sqrt{14}}x + \frac{3}{\sqrt{14}}y – \frac{1}{\sqrt{14}}z = \frac{5}{\sqrt{14}}\)This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
Hence, the direction cosines of the normal to the plane are \(\frac{2}{\sqrt{14}}\), \(\frac{3}{\sqrt{14}}\), and \(\frac{-1}{\sqrt{14}}\) and the distance of normal from the origin is \(\frac{5}{\sqrt{14}}\) units.
5y + 8 = 0
0.x – 5.y + 0.z = 8 —— (1)
The direction ratios of normal are 0, −5, and 0.
\(\sqrt{0 + \left ( -5 \right )^{2} + 0} = 5\)Dividing both sides of equation (1) by 5, we get,
\(-y = \frac{8}{5}\)This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin. Therefore, the direction cosines of the normal to the plane are 0, −1, and 0 and the distance of normal from the origin is \(\frac{8}{5}\) units.
Q2. Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector \(3\hat{i} + 5\hat{j} – 6\hat{k}\).
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
The normal vector is, \(\vec{n} = 3\hat{i} + 5\hat{j} – 6\hat{k}\)
\(\hat{n} = \frac{\vec{n}}{\left | \vec{n} \right |}\) = \(\frac{3\hat{i} + 5\hat{j} – 6\hat{k}}{\sqrt{\left ( 3 \right )^{2} + \left ( 5 \right )^{2} + \left ( 6 \right )^{2}} }\) = \(\frac{3\hat{i} + 5\hat{j} – 6\hat{k}}{\sqrt{70}}\\\)
It is known that the equation of the plane with position vector \(\vec{r}\) is given by, \(\vec{r}.\hat{n} = d\)
\(\vec{r}. \frac{3\hat{i} + 5\hat{j} – 6\hat{k}}{\sqrt{70}}= 7\)This is the required plane of the vector equation.
Q3. Find the Cartesian equation of the following planes:
(a) \(\vec{r}. \left (\hat{i} + \hat{j} – \hat{k} \right ) = 2\)
(b) \(\vec{r}. \left (2\hat{i} + 3\hat{j} – 4\hat{k} \right ) = 1\)
(c) \(\vec{r}. \left [ \left ( s – 2t \right )\hat{i} + \left ( 3 – t \right )\hat{j} + \left ( 2s + t \right )\hat{k} \right ] = 15\)
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
(a) It is given that equation of the plane is \(\vec{r}. \left (\hat{i} + \hat{j} – \hat{k} \right ) = 2\) ——– (1)
For any arbitrary point P (x, y, z) on the plane, position vector \(\vec{r}\) is given as,
\(\vec{r} = x\hat{i} + y\hat{j} – z\hat{k}\)Substituting the value of \(\vec{r}\) in equation (1), we get,
\(\left (x\hat{i} + y\hat{j} – z\hat{k} \right ). \left ( \hat{i} + \hat{j} – \hat{k}\right ) = 2\) \(x + y – z = 2\\\)This is the Cartesian equation of the plane.
(b) \(\vec{r}. \left (2\hat{i} + 3\hat{j} – 4\hat{k} \right ) = 1\) ——– (1)
For any arbitrary point P (x, y, z) on the plane, position vector \(\vec{r}\) is given by,
\(\vec{r} = x\hat{i} + y\hat{j} – z\hat{k}\)Substituting the value of \(\vec{r}\) in equation (1), we get,
\(\left (x\hat{i} + y\hat{j} + z\hat{k} \right ). \left ( 2\hat{i} + 3\hat{j} – 4\hat{k}\right ) = 1\)2x + 3y – 4z = 1
This is the Cartesian equation of the plane.
(c) \(\vec{r}. \left [ \left ( s – 2t \right )\hat{i} + \left ( 3 – t \right )\hat{j} + \left ( 2s + t \right )\hat{k} \right ] = 15\) ———- (1)
For any arbitrary point P (x, y, z) on the plane, position vector \(\vec{r}\) is given by,
\(\vec{r} = x\hat{i} + y\hat{j} – z\hat{k}\)
Substituting the value of \(\vec{r}\) in equation (1), we get,
\(\left (x\hat{i} + y\hat{j} – z\hat{k} \right ). \left [ \left ( s – 2t \right )\hat{i} + \left ( 3 – t \right )\hat{j} + \left ( 2s + t \right )\hat{k} \right ] = 15\)\(\left ( s – 2t \right )x + \left ( 3 – t \right )y + \left ( 2s + t \right )z = 15\)
This is the Cartesian equation of the given plane.
Q4. In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
(a) 2x + 3y + 4z – 12 = 0
(b) 3y + 4z – 6 = 0
(c) x + y + z = 1
(d) 5y + 8 = 0
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
(a) Let the coordinates of the foot of perpendicular P from the origin to the plane be \(\left ( x_{1} + y_{1} + z_{1}\right )\).
2x + 3y + 4z – 12 = 0
2x + 3y + 4z = 12 ———- (1)
The direction ratios of normal are 2, 3, and 4.
\(\sqrt{\left ( 2 \right )^{2} + \left ( 3 \right )^{2} + \left ( 4 \right )^{2}} = \sqrt{29}\)Dividing both sides of equation (1) by \(\sqrt{29}\), we get,
\(\frac{2}{\sqrt{29}}x + \frac{3}{\sqrt{29}}y + \frac{4}{\sqrt{29}}z = \frac{12}{\sqrt{29}}\)This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld, md, nd).
Hence, the coordinates of the foot of the perpendicular are,
\(\left ( \frac{2}{\sqrt{29}}, \frac{12}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{12}{\sqrt{29}}, \frac{4}{\sqrt{29}}, \frac{12}{\sqrt{29}} \right )\)That is:
\(\left ( \frac{24}{29}, \frac{36}{49}, \frac{48}{29} \right )\)
(b) Let the coordinates of the foot of perpendicular P from the origin to the plane be \(\left ( x_{1} + y_{1} + z_{1}\right )\).
3y + 4z – 6 = 0
0x + 3y + 4z = 6 ———– (1)
The direction ratios of the normal are 0, 3, and 4.
\(\sqrt{0 + 3^{2} + 4^{2}} = 5\)
Dividing both sides of equation (1) by 5, we get,
\(0x + \frac{3}{5}y + \frac{4}{5}z = \frac{6}{5}\)This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld, md, nd).
Hence, the coordinates of the foot of the perpendicular are,
\(\left ( 0, \frac{3}{5}, \frac{6}{5}, \frac{4}{5}, \frac{6}{5} \right )\)That is,
\(\left ( 0, \frac{18}{25}, \frac{24}{25} \right )\)
(c) Let the coordinates of the foot of perpendicular P from the origin to the plane be \(\left ( x_{1} + y_{1} + z_{1}\right )\).
x + y + z = 1 ————- (1)
The direction ratios of the normal are 1, 1, and 1.
\(\sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}\)Dividing both sides of equation (1) by \(\sqrt{3}\), we get,
\(\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}y + \frac{1}{\sqrt{3}}z = \frac{1}{\sqrt{3}}\)This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld, md, nd).
Hence, the coordinates of the foot of the perpendicular are,
\(\left ( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right )\\\)That is:
\(\left ( \frac{1}{3}, \frac{1}{3}, \frac{1}{3} \right )\)
(d) Let the coordinates of the foot of perpendicular P from the origin to the plane be \(\left ( x_{1} + y_{1} + z_{1}\right )\).
5y + 8 = 0
0.x – 5y + 0.z = 8 ———– (1)
The direction ratios of the normal are 0, −5, and 0.
\(\sqrt{0 + \left ( -5 \right )^{2} + 0} = 5\)Dividing both sides of equation (1) by 5, we get,
\(-y = \frac{8}{5}\)This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld, md, nd).
Hence, the coordinates of the foot of the perpendicular are,
\(\left ( 0, -1\left ( \frac{8}{5} \right ), 0 \right )\\\)That is:
\(\left [ 0, -1\left ( \frac{8}{5} \right ), 0 \right ]\)
Q5. Find the vector and Cartesian equation of the planes
(a) that passes through the point (1, 0, −2) and the normal to the plane is \(\hat{i} + \hat{j} – \hat{k}\)
(b) that passes through the point (1, 4, 6) and the normal vector to the plane is \(\hat{i} – 2\hat{j} + \hat{k}\)
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
The position vector of point (1, 0, −2) is \(\vec{a} = \hat{i} – 2\hat{k}\)
The normal vector \(\vec{N}\) perpendicular to the plane is \(\vec{N} = \hat{i} + \hat{j} – \hat{k}\)
The vector equation of the plane is given by:
\(\left ( \vec{r} – \vec{a} \right ). \vec{N} = 0\)\(\left [ \vec{r} – \left ( \hat{i} – 2\hat{k} \right ) \right ]. \left ( \hat{i} + \hat{j} – \hat{k} \right ) = 0\) . . . . . . . . . (1)
\(\vec{r}\) is the position vector of any point P (x, y, z) in the plane.\(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\)
Therefore, equation (1) becomes:
\(\left [ x\hat{i} + y\hat{j} + z\hat{k} – \left ( \hat{i} – 2\hat{k} \right ) \right ]. \left ( \hat{i} + \hat{j} – \hat{k} \right ) = 0\\\) \(\\\left [ \left ( x – 1 \right )\hat{i} + y\hat{j} + \left ( z + 2\right )\hat{k} \right ]. \left ( \hat{i} + \hat{j} – \hat{k} \right ) = 0\\\)(x – 1) + y – (z + 2) = 0
x + y – z – 3 = 0
x + y – z = 3
This is the Cartesian equation of the required plane.
(a) The position vector of the point (1, 4, 6) is \(\vec{a} = \hat{i} + 4\hat{j} + 6\hat{k}\)
The normal vector \(\vec{N}\) perpendicular to the plane is \(\vec{N} = \hat{i} – 2\hat{j} + \hat{k}\).
The vector equation of the plane is given by:
\(\left (\vec{r} – \vec{a} \right ). \vec{N} = 0\\\)\(\\\left [\vec{r} – \left (\hat{i} + 4\hat{j} + 6\hat{k} \right ) \right ]. \left ( \hat{i} – 2\hat{j} + \hat{k} \right ) = 0\) ——— (1)
\(\vec{r}\) is the position vector of any point P (x, y, z) in the plane.
\(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\)Hence, equation (1) becomes:
\(\left [ \left ( x\hat{i} + y\hat{j} + z\hat{k} \right ) – \left ( \hat{i} + 4\hat{j} + 6\hat{k} \right ) \right ]. \left ( \hat{i} – 2\hat{j} + \hat{k} \right ) = 0\) \(\left [ \left (x – 1 \right )\hat{i} + \left (y – 4 \right )\hat{j} + \left (z – 6 \right )\hat{k} \right ]. \left ( \hat{i} – 2\hat{j} + \hat{k} \right ) = 0\) \(\left (x – 1 \right ) – 2\left (y – 4 \right ) + \left (z – 6 \right ) = 0\) \(x – 2y + z + 1 = 0\)This is the Cartesian equation of the required plane.
Q6. Find the equations of the planes that pass through three points.
(a) (1, 1, −1), (6, 4, −5), (−4, −2, 3)
(b) (1, 1, 0), (1, 2, 1), (−2, 2, −1)
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
(a) (1, 1, −1), (6, 4, −5), (−4, −2, 3) are the given points.
\(\\\boldsymbol{\begin{vmatrix} 1 & 1 & -1\\ 6 & 4 & -5\\ -4 & -2 & 3 \end{vmatrix}}\\\)= \(\left ( 12 – 10 \right ) – \left ( 18 – 20 \right ) – \left ( -12 + 16 \right )\) = 2 + 2 – 4 = 0
Since A, B, C are collinear points, there will be infinite number of planes passing through the given points.
(b) (1, 1, 0), (1, 2, 1), (−2, 2, −1) are the given points.
\(\boldsymbol{\begin{vmatrix} 1 & 1 & 0\\ 1 & 2 & 1\\ -2 & 2 & -1 \end{vmatrix}}\)= \(\left ( -2 – 2 \right ) – \left ( 2 + 2 \right )\) = -8 \(\neq 0\)
Hence, a plane will pass through the points A, B, and C.
It is known that the equation of the plane through the points, \(\left (x_{1}, y_{1}, z_{1} \right ), \left (x_{2}, y_{2}, z_{2} \right )\) and \(\left (x_{3}, y_{3}, z_{3} \right )\) is,
\(\begin{vmatrix} \ x – x_{1} \;\;\;\;\;\; y – y_{1} \;\;\;\;\;\; z – z_{1}\\ x_{2} – x_{1} \;\;\;\;\;\; y_{2} – y_{1} \;\;\;\;\;\; z_{2} – z_{1}\\ x_{3} – x_{1} \;\;\;\;\;\; y_{3} – y_{1} \;\;\;\;\;\; z_{3} – x_{1} \end{vmatrix} = 0\)\(\begin{vmatrix} \ x – 1 \;\;\;\;\;\; y – 1 \;\;\;\;\;\; z\\ 0 \;\;\;\;\;\;\;\;\;\;\;\;\; 1 \;\;\;\;\;\;\;\;\;\; 1\\ -3 \;\;\;\;\;\;\;\;\;\; 1 \;\;\;\;\;\;\;\; -1 \end{vmatrix} = 0\)
\(\left (-2 \right )\left ( x – 1 \right ) – 3\left ( y – 1 \right ) + 3z = 0\) = 0
\(-2x – 3y + 3z + 2 + 3 = 0\) = 0
-2x – 3y + 3z = -5
2x + 3y – 3z = 5
This is the Cartesian equation of the required plane.
Q7. Find the intercepts cut off by the plane 2x + y – z = 5.
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
2x + y – z = 5 ———- (1)
Dividing both sides of equation (1) by 5, we get
\(\frac{2}{5}x + \frac{y}{5} – \frac{z}{5} = 1\)
\(\frac{x}{\frac{5}{2}} + \frac{y}{5} + \frac{z}{-5} = 1\) ———— (2)
It is known that the equation of a plane in intercept form is \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\), where a, b, c are the intercepts cut off by the plane at x, y, and z axes respectively.
Hence, for the given equation,
\(a = \frac{5}{2}\), b = 5 and c = -5
Therefore, the intercepts cut off by the plane are \(\frac{5}{2}\), 5 and -5.
Q8. Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.
Sol:
The equation of the plane ZOX is y = 0.
Any plane parallel to it is of the form,
y = a
Since the y-intercept of the plane is 3,
a = 3
Therefore, the equation of the required plane is y = 3
Q9. Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
The equation of any plane through the intersection of the planes, 3x − y + 2z − 4 = 0 and x + y + z − 2 = 0, is,
\((3x – y + 2z – 4) + \alpha (x + y + z – 2) = 0\), where \(\alpha \in R\) ———– (1)
The plane passes through the point (2, 2, 1).
Hence, this point will satisfy equation (1).
(3 × 2 – 2 + 2 × 1 – 4) + \(\alpha\) (2 + 2+ 1 – 2) = 0
2 + 3\(\alpha\) = 0
\(\alpha = -\frac{2}{3}\)Substituting \(\alpha = -\frac{2}{3}\) in equation (1), we get:
\(\\\left ( 3x – y + 2z – 4 \right ) – \frac{2}{3}\left ( x + y + z – 2 \right ) = 0\)\(3\left ( 3x – y + 2z – 4 \right ) – 2 \left ( x + y + z – 2 \right ) = 0\)
\(\left ( 9x – 3y + 6z – 12 \right ) – 2 \left ( x + y + z – 2 \right ) = 0\)
\(7x – 5y + 4z – 8 = 0\\\)
This is the required equation of the plane.
Q10. Find the vector equation of the plane passing through the intersection of the planes \(\vec{r}. \left ( 2\hat{i} + 2\hat{j} – 3\hat{k} \right ) = 7\) , \(\vec{r}. \left ( 2\hat{i} + 5\hat{j} + 3\hat{k} \right ) = 9\) and through the point (2, 1, 3).
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
\(\vec{r}. \left ( 2\hat{i} + 2\hat{j} – 3\hat{k} \right ) = 7\) \(\vec{r}. \left ( 2\hat{i} + 5\hat{j} + 3\hat{k} \right ) = 9\)
The equations of plane are as follows:
\(\vec{r}. \left ( 2\hat{i} + 2\hat{j} – 3\hat{k} \right ) – 7 = 0\) ——– (1)
\(\vec{r}. \left ( 2\hat{i} + 5\hat{j} + 3\hat{k} \right ) – 9 = 0\) ——— (2)
The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by,
\(\left [\vec{r}. \left ( 2\hat{i} + 2\hat{j} – 3\hat{k} \right ) – 7 \right ] + \lambda \left [\vec{r}. \left ( 2\hat{i} + 5\hat{j} + 3\hat{k} \right ) – 9 \right ] = 0\), where \(\lambda \in R\)
\(\vec{r}. \left [\left (2\hat{i} + 2\hat{j} – 3\hat{k} \right ) + \lambda \left (2\hat{i} + 5\hat{j} + 3\hat{k} \right ) \right ] = 9\lambda + 7\)
\(\vec{r}. [\left (2 + 2\lambda \right )\hat{i} + \left (2 + 5\lambda \right )\hat{j} + \left (3\lambda – 3 \right )\hat{k}] = 9\lambda + 7\) ——– (3)
The plane passes through the point (2, 1, 3). Therefore, its position vector is given by,
\(\vec{r} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}\)
Substituting in equation (3), we get
\(\left ( 2\hat{i} + \hat{j} – 3\hat{k}\right ). \left [ \left (2 + 2\lambda \right )\hat{i} + \left (2 + 5\lambda \right )\hat{j} + \left (3\lambda – 3 \right )\hat{k} \right ] = 9\lambda + 7\)\(\left (2 + 2\lambda \right ) + \left (2 + 5\lambda \right ) + \left (3\lambda – 3 \right ) = 9\lambda + 7\)
\(18\lambda – 3 = 9\lambda + 7\)
\(9\lambda = 10\)
\(\lambda = \frac{10}{9}\)
Substituting \(\lambda = \frac{10}{9}\) in equation (3), we get
\(\vec{r}. \left ( \frac{38}{9} \hat{i} + \frac{68}{9} \hat{j} + \frac{3}{9} \hat{k}\right ) = 17\)\(\vec{r}. \left ( 38 \hat{i} + 68 \hat{j} + 3 \hat{k}\right ) = 153\)
This is the vector equation of the required plane.
Q11. Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
The equation of the plane through the intersection of the planes, x + y + z = 1 and 2x + 3y + 4z = 5 is,
\(\left ( x + y + z – 1 \right ) + \lambda \left ( 2x + 3y + 4z – 5 \right ) = 0\)\(\left ( 2\lambda + 1 \right )x + \left ( 3\lambda + 1 \right )y + \left ( 4\lambda + 1 \right )z – \left ( 5\lambda + 1 \right ) = 0\) ——— (1)
The direction ratios, \(a_{1}, b_{1}, c_{1}\) of this plane are \(\left ( 2\lambda + 1 \right )\), \(\left ( 3\lambda + 1 \right )\), and \(\left ( 4\lambda + 1 \right )\).
The plane in equation (1) is perpendicular to x – y + z = 0.
Its direction ratios, \(a_{2}, b_{2}, c_{2}\) are 1, -1 and 1.
Since the planes are perpendicular,
\(a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0\) \(\left ( 2\lambda + 1 \right ) – \left ( 3\lambda + 1 \right ) + \left ( 4\lambda + 1 \right ) = 0\) \(3\lambda + 1 = 0\)\(\lambda = -\frac{1}{3}\)
Substituting \(\lambda = -\frac{1}{3}\) in equation (1), we get
\(\frac{1}{3}x – \frac{1}{3}z + \frac{2}{3} = 0\)
x – z + 2 = 0
This is the required equation of the plane.
Q12. Find the angle between the planes whose vector equations are \(\vec{r}. \left ( 2\hat{i} + 2\hat{j} – 3\hat{k} \right ) = 5\) and \(\vec{r}. \left ( 3\hat{i} – 3\hat{j} + 5\hat{k} \right ) = 3\)
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
The equations of the given planes are \(\vec{r}. \left ( 2\hat{i} + 2\hat{j} – 3\hat{k} \right ) = 5\) and \(\vec{r}. \left ( 3\hat{i} – 3\hat{j} + 5\hat{k} \right ) = 3\).
It is known that if \(\vec{n_{1}}\) and \(\vec{n_{2}}\) are \(\vec{r}. \vec{n_{1}} = d_{1}\) and \(\vec{r}. \vec{n_{2}} = d_{2}\) normal to the planes, , then the angle between them, \(\theta\), is given by,
\(\cos \theta = \left | \frac{\vec{n_{1}}. \vec{n_{2}}}{\left | \vec{n_{1}} \right |\left | \vec{n_{2}} \right |} \right |\) ———- (1)
Here, \(\vec{n_{1}} = 2\hat{i} + 2\hat{j} – 3\hat{k}\) and \(\vec{n_{2}} = 3\hat{i} – 3\hat{j} + 5\hat{k}\)
Therefore,
\(\vec{n_{1}}. \vec{n_{2}} = \left ( 2\hat{i} + 2\hat{j} – 3\hat{k} \right ). \left ( 3\hat{i} – 3\hat{j} + 5\hat{k} \right )\)= (2 × 3) + (2 × -3 ) + (-3 × 5) = -15
\(\left |\vec{n_{1}} \right | = \sqrt{\left ( 2 \right )^{2} + \left ( 2 \right )^{2} + \left ( -3 \right )^{2}} = \sqrt{17}\)
\(\left |\vec{n_{2}} \right | = \sqrt{\left ( 3 \right )^{2} + \left ( -3 \right )^{2} + \left ( 5 \right )^{2}} = \sqrt{43}\)
Substituting the value of \(\vec{n_{1}}, \vec{n_{2}}, \left |\vec{n_{1}} \right | \; and \; \left |\vec{n_{2}} \right |\)
\(\cos \theta = \left | \frac{-15}{\sqrt{17}. \sqrt{43}} \right |\)
\(\cos \theta = \frac{15}{\sqrt{731}}\)
\(\theta = \cos ^{-1} \left (\frac{15}{\sqrt{731}} \right )\)
Q13. In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0
(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
(d) 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0
(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0
Sol:
The direction ratios of normal to the plane, \(L_{1}: a_{1}x + b_{1}y + c_{1}z = 0\), are \(a_{1}, b_{1}, c_{1}\) and \(L_{2}: a_{2}x + b_{2}y + c_{2}z = 0\) are \(a_{2}, b_{2}, c_{2}\)
\(L_{1}\parallel L_{2}, \;if \;\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}\) \(L_{1}\perp L_{2}, \;if \;a_{1}a_{2} = b_{1}b_{2} = c_{1}c_{2} = 0\)
The angle between \(L_{1}\) and \(L_{2}\) is given by,
\(\theta = \cos ^{-1}\left | \frac{a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} }{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}. \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}} } } \right |\)
(a) The equations of the planes are 7x + 5y + 6z + 30 = 0 and 3x − y − 10z + 4 = 0
Here,
\(a_{1} = 7\), \(b_{1} = 5\), \(c_{1} = 6\)
\(a_{2} = 3\), \(b_{2} = -1\), \(c_{2} = -10\)
\(a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = (7 \times 3) + (5 \times -1) + \left ( 6 \times -10 \right )\)
= – 44
\(\neq 0\)Hence, the given planes are not perpendicular.
\(\frac{a_{1}}{a_{2}} = \frac{7}{3}, \frac{b_{1}}{b_{2}} = \frac{5}{-1} = -5, \frac{c_{1}}{c_{2}} = \frac{6}{-10} = \frac{-3}{5}\)
It can be seen that, \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, the given planes are not parallel.
The angle between them is given by,
\(\theta = \cos ^{-1}\left | \frac{(7 \times 3) + \left ( 5 \times -1 \right ) + \left ( 6 \times -10 \right )}{\sqrt{\left ( 7 \right )^{2} + \left ( 5 \right )^{2} + \left ( 6 \right )^{2} \times \sqrt{\left ( 3 \right )^{2} + \left ( -1 \right )^{2} + \left ( -10 \right )^{2}}}} \right |\)\(\theta = \cos ^{-1} \left | \frac{21 – 5 – 60}{\sqrt{110} \times \sqrt{110}} \right |\)
\(\theta = \cos ^{-1} \frac{44}{110}\)
\(\theta = \cos ^{-1} \frac{2}{5}\)
(b) The equations of the planes are 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0.
Here,
\(a_{1} = 2\), \(b_{1} = 1\), \(c_{1} = 3\)
\(a_{2} = 1\), \(b_{2} = -2\), \(c_{2} = 0\)
\(a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = (2 \times 1) + (1 \times -2) + \left ( 3 \times 0 \right )\) = 0
Thus, the given planes are perpendicular to each other.
(c) The equations of the given planes are 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
Here,
\(a_{1} = 2\), \(b_{1} = -2\), \(c_{1} = 4\)
\(a_{2} = 3\), \(b_{2} = -3\), \(c_{2} = 6\)
\(a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = (2 \times 3) + (-2 \times -3) + \left ( 4 \times 6 \right )\)
= 6 + 6 + 24
= 36
\(\neq 0\)Therefore, the given planes are not perpendicular to each other.
\(\frac{a_{1}}{a_{2}} = \frac{2}{3}, \frac{b_{1}}{b_{2}} = \frac{-2}{-3} = \frac{2}{3}, \frac{c_{1}}{c_{2}} = \frac{4}{6} = \frac{2}{3}\)
\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}}\)
Therefore, the given planes are parallel to each other.
(d) The equations of the planes are 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0
Here,
\(a_{1} = 2\), \(b_{1} = -1\), \(c_{1} = 3\)
\(a_{2} = 2\), \(b_{2} = -1\), \(c_{2} = 3\)
\(\frac{a_{1}}{a_{2}} = \frac{2}{2} = 1, \frac{b_{1}}{b_{2}} = \frac{-1}{-1} = 1, \frac{c_{1}}{c_{2}} = \frac{3}{3} = 1\)
\(\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}}\)
Therefore, the given planes are parallel to each other.
(e) The equations of the given planes are 4x + 8y + z – 8 = 0 and y + z – 4 = 0
Here,
\(a_{1} = 4\), \(b_{1} = 8\), \(c_{1} = 1\)
\(a_{2} = 0\), \(b_{2} = 1\), \(c_{2} = 1\)
\(a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = (4 \times 0) + (8 \times 1) + \left ( 1 \times 1 \right )\)
= 9
\(\neq 0\)Thus, the given lines are not perpendicular to each other.
\(\frac{a_{1}}{a_{2}} = \frac{4}{0} = 1, \frac{b_{1}}{b_{2}} = \frac{8}{1} = 8, \frac{c_{1}}{c_{2}} = \frac{1}{1} = 1\)
\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Thus, the given lines are not parallel to each other.
The angle between the planes is given by,
\(\theta = \cos ^{-1}\left | \frac{\left ( 4 \times 0 \right ) + \left ( 8 \times 1 \right ) + \left ( 1 \times 1 \right )}{\sqrt{\left ( 4 \right )^{2} + \left ( 8 \right )^{2} + \left ( 1 \right )^{2} \times \sqrt{\left ( 0 \right )^{2} + \left ( 1 \right )^{2} + \left ( 1 \right )^{2}}}} \right |\)\(\theta = \cos ^{-1}\left | \frac{9}{9 \times \sqrt{2}} \right |\)
\(\theta = \cos ^{-1}\left (\frac{1}{\sqrt{2}} \right )\)
\(\theta = 45^{\circ}\)
Q14. In the following cases, find the distance of each of the given points from the corresponding given plane.
Point Plane
(a) (0, 0, 0) 3x – 4y + 12z = 3
(b) (3, −2, 1) 2x – y + 2z + 3 = 0
(c) (2, 3, −5) x + 2y – 2z = 9
(d) (−6, 0, 0) 2x – 3y + 6z – 2 = 0
Sol: Based on the formula given in THREE DIMENSIONAL GEOMETRY
It is known that the distance between a point, p \(\left ( x_{1}, y_{1}, z_{1} \right )\), and a plane, Ax + By + Cz = D, is given by,
\(d = \left | \frac{Ax_{1} + By_{1} + Cz_{1} – D}{\sqrt{A^{2} + B^{2} + C^{2}}} \right |\) ——– (1)
(a) The given point is (0, 0, 0) and the plane is 3x – 4y + 12z = 3.
\(d = \left | \frac{3 \times 0 – 4 \times 0 + 12 \times 0 – 3}{\sqrt{3^{2} + \left ( -4 \right )^{2} + 12^{2}}} \right |\) \(d = \frac{3}{\sqrt{169}}\) \(d = \frac{3}{13}\)
(b) The given point is (3, − 2, 1) and the plane is 2x – y + 2z + 3 = 0
\(d = \left | \frac{2 \times 3 – (-2) + 2 \times 1 + 3}{\sqrt{2^{2} + \left ( -2 \right )^{2} + 2^{2}}} \right |\) \(d = \left |\frac{13}{3} \right |\) \(d = \frac{13}{3}\)
(c) The given point is (2, 3, −5) and the plane is x + 2y – 2z = 9
\(d = \left | \frac{2 + 2 \times 3 – 2 }{\sqrt{2^{2} + \left ( -2 \right )^{2} + 2^{2}}} \right |\) \(d = \left | \frac{2 + 2 \times 3 – 2 \left ( -5 \right ) – 9}{\sqrt{\left ( 1 \right )^{2} + \left ( 2 \right )^{2} + \left ( -2 \right )^{2}}} \right |\) \(d = \frac{9}{3}\) \(d = 3\)
(d) The given point is (−6, 0, 0) and the plane is 2x – 3y + 6z – 2 = 0.
\(d = \left | \frac{2 \times -6 – 2 \times 0 + 6 \times 0 – 2}{\sqrt{\left ( 2 \right )^{2} + \left ( -3 \right )^{2} + \left ( 6 \right )^{2}}} \right |\) \(d = \left | \frac{-14}{\sqrt{49}} \right |\) \(d = \frac{14}{7}\) \(d = 2\)