Chapter 11 Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 Three Dimensional Geometry

Class 12 Maths Chapter 11 Three Dimensional Geometry explains how points, lines, and planes are represented in three-dimensional space and how distances and angles are calculated. The chapter helps students understand geometry using coordinates and develop better visualisation skills by connecting algebra with geometry. These concepts are useful for CBSE board exams as well as competitive examinations.

NCERT Solutions for Class 12 Maths Chapter 11 Fby ALLEN follow the latest CBSE syllabus and cover the questions from the NCERT textbook. Step-by-step solutions help students understand Three Dimensional Geometry, use the required formulas, and solve questions accurately. Regular practice with these solutions can improve calculation accuracy and help students prepare for CBSE board exams and competitive examinations.

1.0Key Concepts of Class 12 Maths Chapter 11 Three Dimensional Geometry

Class 12 Maths Chapter 11, Three Dimensional Geometry, focuses on studying points and lines in space using coordinates. The main concepts covered in NCERT Solutions for Class 12 Maths Chapter 11 include:

Coordinate System in Space: Understand the three mutually perpendicular axes and locate points in three-dimensional space using ordered triples.
Distance Between Two Points: Learn how to calculate the distance between two points in three-dimensional space.
Section Formula: Find the coordinates of a point that divides a line segment internally or externally in a given ratio.
Direction Ratios and Direction Cosines: Understand how the direction of a line is represented using direction ratios and direction cosines.
Equation of a Line: Study different forms of the equation of a line in three-dimensional geometry.
Angle Between Two Lines: Learn how to calculate the angle between two lines using their direction ratios or direction cosines.

2.0NCERT Class 12 Maths Chapter 11 Three Dimensional Geometry : Detailed Solutions

3.0EXERCISE - 11.1

If a line makes angles 90°, 135°, 45° with x, y and z-axes respectively, find its direction cosines. Sol. Let direction cosines of the line be $l, m$ , and n . $l = cos 9 0^{\circ} = 0$ ;

$m = cos (18 0^{\circ} - 4 5^{\circ}) = - cos 4 5^{\circ} = - \frac{1}{2}; n = cos 4 5^{\circ} = \frac{1}{2}$

Therefore, the direction cosines of the line are $0, - \frac{1}{2}$ and $\frac{1}{2}$

Find the direction cosines of a line which makes equal angles with the coordinate axes. Sol. Let the line makes an angle ' $α$ ' with each of the co-ordinate axes. $∴ l = cos α, m = cos α, n = cos α$ $l^{2} + m^{2} + n^{2} = 1$ $\Rightarrow cos^{2} α + cos^{2} α + cos^{2} α = 1$ $\Rightarrow 3 cos^{2} α = 1$ $\Rightarrow cos^{2} α = \frac{1}{3} \Rightarrow cos α = \pm \frac{1}{3}$

Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are $\pm \frac{1}{3}, \pm \frac{1}{3}$ , and $\pm \frac{1}{3}$

If a line has the direction ratios -18, 12, - 4, then what are its direction cosines? Sol. If a line has direction ratios of -18, 12 and -4, then its direction cosines are

$\frac{- 18}{( - 18 ) ^{2} + ( 12 ) ^{2} + ( - 4 ) ^{2}}, \frac{12}{( - 18 ) ^{2} + ( 12 ) ^{2} + ( - 4 ) ^{2}}, \frac{- 4}{( - 18 ) ^{2} + ( 12 ) ^{2} + ( - 4 ) ^{2}} i.e., \frac{- 18}{22}, \frac{12}{22}, \frac{- 4}{22} \Rightarrow \frac{- 9}{11}, \frac{6}{11}, \frac{- 2}{11}$

Thus, the direction cosines are $\frac{- 9}{11}, \frac{6}{11}$ and $\frac{- 2}{11}$ .

Show that the points (2, 3, 4), (-1, -2, 1), (5, 8, 7) are collinear. Sol. The given points are $A (2, 3, 4), B (- 1, - 2, 1)$ , and C (5, 8, 7). It is known that the direction ratios of line joining the points, $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ , are given by, $(x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1})$ . The direction ratios of AB are

$a_{1} = (- 1 - 2), b_{1} = (- 2 - 3) and c_{1} = (1 - 4)$

i.e., $a_{1} = - 3, b_{1} = - 5$ , and $c_{1} = - 3$ .

The direction ratios of BC are

$a_{2} = (5 - (- 1)), b_{2} = (8 - (- 2)) and c_{2} = (7 - 1)$

i.e., $a_{2} = 6, b_{2} = 10$ and $c_{2} = 6$ .

$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}} = - \frac{1}{2} i.e. they are proportional.$

Therefore, AB is parallel to BC. Since point B is common to both AB and BC , points $A, B$ , and C are collinear.

Find the direction cosines of the sides of the triangle whose vertices are (3, 5, -4), (-1,1,2) and (-5, -5, -2) Sol. The vertices of $△ ABC$ are $A (3, 5, - 4), B (- 1, 1, 2)$ and C (-5, -5, -2) The direction ratios of side AB are (-1-3), (1-5), and (2-(-4)) i.e., (-4, -4, 6).

Therefore, the direction cosines of AB are

$\frac{- 4}{( - 4 ) ^{2} + ( - 4 ) ^{2} + ( 6 ) ^{2}}, \frac{- 4}{( - 4 ) ^{2} + ( - 4 ) ^{2} + ( 6 ) ^{2}}, \frac{6}{( - 4 ) ^{2} + ( - 4 ) ^{2} + ( 6 ) ^{2}} \frac{- 4}{2 17}, \frac{- 4}{2 17}, \frac{6}{2 17} = \frac{- 2}{17}, \frac{- 2}{17}, \frac{3}{17}$

The direction ratios of BC are (-5 - (-1)), (-5-1), and (-2 - 2) i.e., (- 4, - 6,- 4). Therefore, the direction cosines of BC are

$\frac{- 4}{( - 4 ) ^{2} + ( - 6 ) ^{2} + ( - 4 ) ^{2}}, \frac{- 6}{( - 4 ) ^{2} + ( - 6 ) ^{2} + ( - 4 ) ^{2}}, \frac{- 4}{( - 4 ) ^{2} + ( - 6 ) ^{2} + ( - 4 ) ^{2}} = \frac{- 4}{2 17}, \frac{- 6}{2 17}, \frac{- 4}{2 17} = \frac{- 2}{17}, \frac{- 3}{17}, \frac{- 2}{17}$

The direction ratios of AC are (-5 - 3), (-5 - 5), and (-2 - (-4)) i.e., (- 8, -10 and 2). Therefore, the direction cosines of AC are

$\frac{- 8}{( - 8 ) ^{2} + ( - 10 ) ^{2} + ( 2 ) ^{2}} \frac{- 10}{( - 8 ) ^{2} + ( - 10 ) ^{2} + ( 2 ) ^{2}} \frac{2}{( - 8 ) ^{2} + ( - 10 ) ^{2} + ( 2 ) ^{2}} = \frac{- 8}{2 42}, \frac{- 10}{2 42}, \frac{2}{2 42} = \frac{- 4}{42}, \frac{- 5}{42}, \frac{1}{42}$

EXERCISE 11.2

Show 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. Two lines with direction cosines, $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ , are perpendicular to each other, if $l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0$ (i) For the lines with direction cosines, $\frac{12}{13}, \frac{- 3}{13}, \frac{- 4}{13}$ and $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$ , we obtain

$l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = (\frac{12}{13}) \times (\frac{4}{13}) + (\frac{- 3}{13}) \times (\frac{12}{13}) + (\frac{- 4}{13}) \times (\frac{3}{13}) = \frac{48}{169} - \frac{36}{169} - \frac{12}{169} = 0$

Therefore, the lines are perpendicular (ii) For the lines with direction cosines, $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$ and $\frac{3}{13}, \frac{- 4}{13}, \frac{12}{13}$ , we obtain

$l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = (\frac{4}{13}) \times (\frac{3}{13}) + (\frac{12}{13}) \times (\frac{- 4}{13}) + (\frac{3}{13}) \times (\frac{12}{13}) = \frac{12}{169} - \frac{48}{169} + \frac{36}{169} = 0$

Therefore, the lines are perpendicular, (iii) For the lines with direction cosines, $\frac{3}{13}, \frac{- 4}{13}, \frac{12}{13}$ and $\frac{12}{13}, \frac{- 3}{13}, \frac{- 4}{13}$ , we obtain

$l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = (\frac{3}{13}) \times (\frac{12}{13}) + (\frac{- 4}{13}) \times (\frac{- 3}{13}) + (\frac{12}{13}) \times (\frac{- 4}{13}) = \frac{36}{169} + \frac{12}{169} - \frac{48}{169} = 0$

Therefore, the lines are perpendicular. Thus, all lines are mutually perpendicular.

Show that the line through the points (1, -1, 2) and ( $3, 4, - 2$ ) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6). Sol. Let AB be the line joining the points, (1, -1, 2) and (3, 4, -2), and CD be the line joining the points, (0, 3, 2) and (3, 5, 6). The direction ratios, $a_{1}, b_{1}, c_{1}$ , of AB are ( $3 - 1$ ), (4-(-1)), and (-2-2) i.e., (2, 5, -4). The direction ratios, $a_{2}, b_{2}, c_{2}$ , of CD are (3-0), (5-3), and (6-2) i.e., (3, 2, 4). AB and CD will be perpendicular to each other, if

$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

$\Rightarrow a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}$

$= 2 \times 3 + 5 \times 2 + (- 4) \times 4 = 6 + 10 - 16 = 0$

Therefore, AB and CD are perpendicular to each other.

Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (-1, -2, 1), (1, 2, 5). Sol. Let AB be the line through the points, $(4, 7, 8)$ and (2, 3, 4), and CD be the line through the points, (-1, -2, 1) and (1, 2, 5). The directions ratios, $a_{1}, b_{1}, c_{1}$ of AB are (2-4), (3-7), and (4-8) i.e., (-2, -4, -4). The direction ratios, $a_{2}, b_{2}, c_{2}$ of CD are (1-(-1)), (2-(-2)), and (5-1) i.e., (2, 4, 4). AB will be parallel to CD, if $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$

$\frac{a _{1}}{a _{2}} = \frac{- 2}{2} = - 1, \frac{b _{1}}{b _{2}} = \frac{- 4}{4} = - 1 and \frac{c _{1}}{c _{2}} = \frac{- 4}{4} = - 1$

$∴ \frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}} = - 1$

Thus, AB is parallel to CD.

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. It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector point A is $a = \hat{i} + 2 \hat{j} + 3 \hat{k}$ . Let, $b = 3 \hat{i} + 2 \hat{j} - 2 \hat{k}$ It is known that the line which passes through point A and parallel to $b$ is given by $r = a + λ b,$ where $λ$ is a constant $r = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ (3 \hat{i} + 2 \hat{j} - 2 \hat{k})$ . This is the required equation of the line.
Find the equation of the line in vector and in Cartesian 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. $a = 2 \hat{i} - \hat{j} + 4 \hat{k}$ and $b = \hat{i} + 2 \hat{j} - \hat{k}$ It is known that a line through a point with position vector $a$ and parallel to $b$ is given by the equation

$r = a + λ b \Rightarrow r = 2 \hat{i} - \hat{j} + 4 \hat{k} + λ (\hat{i} + 2 \hat{j} - \hat{k})$

This is the required equation of the line in vector form.

$r = x \hat{i} + y \hat{j} + z \hat{k}$

$\Rightarrow x \hat{i} + y \hat{j} + z \hat{k} = (λ + 2) \hat{i} + (2 λ - 1) \hat{j} + (- λ + 4) \hat{k}$ Eliminating $λ$ , we obtain the Cartesian form 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.

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. It is given that the line passes through the point (-2, 4, -5) and is parallel to

$\frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6}$

The direction rations of the line,

$\frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6} are (3, 5, 6) .$

The required line is parallel to

$\frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6}$

Therefore, its direction ratios are $3 k, 5 k$ , and 6 k , where $k \neq = 0$ It is known that the equation of the line through the point $(x_{1}, y_{1}, z_{1})$ and with direction ratios,

$a, b, c, is given by \frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}$

Therefore the equation of the required line is

$\frac{x + 2}{3 k} = \frac{y - 4}{5 k} = \frac{z + 5}{6 k}$

$\Rightarrow \frac{x + 2}{3} = \frac{y - 4}{5} = \frac{z + 5}{6} = k$

The Cartesian equation of a line is $\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2}$ . Write its Vector form. Sol. The Cartesian equation of the line is $\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2}$ The given line passes through the point (5, - 4, 6). The position vector of this point is $a = 5 \hat{i} - 4 \hat{j} + 6 \hat{k}$ Also, the direction ratios of the given line are 3, 7, and 2. This means that the line is in the direction of vector, $b = 3 \hat{i} + 7 \hat{j} + 2 \hat{k}$ It is known that the line through position vector $a$ and in the direction of the vector $b$ is given by the equation,

$r = a + λ b, λ \in R$

$\Rightarrow r = (5 \hat{i} - 4 \hat{j} + 6 \hat{k}) + λ (3 \hat{i} + 7 \hat{j} + 2 \hat{k})$

This is the required equation of the given line in vector form.

Find the angle between the following pairs of lines: (i) $r = 2 \hat{i} - 5 \hat{j} + \hat{k} + λ (3 \hat{i} + 2 \hat{j} + 6 \hat{k})$ and $r = 7 \hat{i} - 6 \hat{k} + μ (\hat{i} + 2 \hat{j} + 2 \hat{k})$ (ii) $r = 3 \hat{i} + \hat{j} - 2 \hat{k} + λ (\hat{i} - \hat{j} - 2 \hat{k})$ and $r = 2 \hat{i} - \hat{j} - 56 \hat{k} + μ (3 \hat{i} - 5 \hat{j} - 4 \hat{k})$

Sol.

(i) Let $θ$ be the angle between the given lines. The angle between the given pairs of lines is given by, $cos θ = \frac{b _{1} \cdot b _{2}}{∣ b _{1} ∣ ∣ b _{2} ∣}$ The given lines are parallel to the vectors, $b_{1} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$ and $b_{2} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ respectively. $∴ b_{1} = 3^{2} + 2^{2} + 6^{2} = 7$

$b_{2} = (1)^{2} + (2)^{2} + (2)^{2} = 3$

$\Rightarrow b_{1} \cdot b_{2} = (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) (\hat{i} + 2 \hat{j} + 2 \hat{k})$

$= 3 \times 1 + 2 \times 2 + 6 \times 2 = 3 + 4 + 12 = 19$

$\Rightarrow cos θ = \frac{19}{7 \times 3}$ $\Rightarrow θ = cos^{- 1} (\frac{19}{21})$ (ii) The given lines are parallel to the vectors, $b_{1} = \hat{i} - \hat{j} - 2 \hat{k}$ and $b_{2} = 3 \hat{i} - 5 \hat{j} - 4 \hat{k}$ respectively

$b_{1} = (1)^{2} + (- 1)^{2} + (- 2)^{2} = 6 b_{2} = (3)^{2} + (- 5)^{2} + (- 4)^{2} = 50 = 52$

$b_{1} \cdot b_{2} = (\hat{i} - \hat{j} - 2 \hat{k}) (3 \hat{i} - 5 \hat{j} - 4 \hat{k}) = 1.3 - 1 (- 5) - 2 (- 4) = 3 + 5 + 8 = 16$

$cos θ = \frac{b _{1} \cdot b _{2}}{b _{1} b _{2}}$

$\Rightarrow cos θ = \frac{16}{6 .5 2} = \frac{16}{2 . 3 .5 2} = \frac{16}{10 3}$

$\Rightarrow cos θ = \frac{8}{5 3}$

$\Rightarrow θ = cos^{- 1} (\frac{8}{5 3})$

Find the angle between the following 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 $b_{1}$ and $b_{2}$ be the vectors parallel to the pair of lines,

$\frac{x - 2}{2} = \frac{y - 1}{5} = \frac{z + 3}{- 3} and \frac{x + 2}{- 1} = \frac{y - 4}{8} = \frac{z - 5}{4},$

respectively $∴ b_{1} = 2 \hat{i} + 5 \hat{j} - 3 \hat{k}$ and $b_{2} = - \hat{i} + 8 \hat{j} + 4 \hat{k}$

$b_{1} = b_{2} = b_{1} \cdot b_{2} (2)^{2} + (5)^{2} + (- 3)^{2} = 38 (- 1)^{2} + (8)^{2} + (4)^{2} = 81 = 9 = (2 \hat{i} + 5 \hat{j} - 3 \hat{k}) \cdot (- \hat{i} + 8 \hat{j} + 4 \hat{k}) = 2 (- 1) + 5 \times 8 + (- 3) \cdot 4 = - 2 + 40 - 12 = 26$

The angle $θ$ between the given pair of lines is given by the relation,

$cos θ = \frac{b _{1} \cdot b _{2}}{b _{1} b _{2}}$

$\Rightarrow cos θ = \frac{26}{9 38} \Rightarrow θ = cos^{- 1} (\frac{26}{9 38})$ (ii) Let $b_{1}$ and $b_{2}$ be the vectors parallel to the given pair of lines, $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and

$\frac{x - 5}{4} = \frac{y - 2}{1} = \frac{z - 3}{8}, respectively$

$\Rightarrow b_{1} = 2 \hat{i} + 2 \hat{j} + \hat{k}$ and $b_{2} = 4 \hat{i} + \hat{j} + 8 \hat{k}$ $∴ b_{1} = (2)^{2} + (2)^{2} + (1)^{2} = 9 = 3$

$b_{2} b_{1} \cdot b_{2} = 4^{2} + 1^{2} + 8^{2} = 81 = 9 = (2 \hat{i} + 2 \hat{j} + \hat{k}) \cdot (4 \hat{i} + \hat{j} + 8 \hat{k}) = 2 \times 4 + 2 \times 1 + 1 \times 8 = 8 + 2 + 8 = 18$

If $θ$ is the angle between the given pair of lines, then

$cos θ = \frac{b _{1} \cdot b _{2}}{∣ b _{1} ∣ ∣ b _{2} ∣}$

$\Rightarrow cos θ = \frac{18}{3 \times 9} = \frac{2}{3} \Rightarrow θ = cos^{- 1} (\frac{2}{3})$

Find the values of p so that the lines

$\frac{1 - x}{3} = \frac{7 y - 14}{2 p} = \frac{z - 3}{2} and \frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5} are at right angles.$

Sol. The given equations can be written in the standard form as

$\frac{x - 1}{- 3} = \frac{y - 2}{\frac{2 p}{7}} = \frac{z - 3}{2} and \frac{x - 1}{\frac{- 3 p}{7}} = \frac{y - 5}{1} = \frac{z - 6}{- 5}$

The direction ratios of the lines are $- 3, \frac{2 p}{7}$ ,

$2 and \frac{- 3 p}{7}, 1, - 5 respectively.$

Two lines with direction ratios, $a_{1}, b_{1}, c_{1}$ and $a_{2}$ , $b_{2},, c_{2}$ , are perpendicular to each other, if $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ $∴ (- 3) . (\frac{- 3 p}{7}) + (\frac{2 p}{7}) . (1) + 2 (- 5) = 0$ $\Rightarrow \frac{9 p}{7} + \frac{2 p}{7} = 10$ $\Rightarrow 11 p = 70$ $\Rightarrow p = \frac{70}{11}$

Thus, the value of p is $\frac{70}{11}$

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. The equations of the given lines are

$\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} and \frac{x}{1} = \frac{y}{2} = \frac{z}{3}$

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

$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

$∴ 7 \times 1 + (- 5) \times 2 + 1 \times 3 = 7 - 10 + 3 = 0$

Therefore, the given lines are perpendicular to each other.

Find the shortest distance between the lines:

$r = (\hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$

and

$r = (2 \hat{i} - \hat{j} - \hat{k}) + μ (2 \hat{i} + \hat{j} + 2 \hat{k})$

Sol. The equations of the given lines are

$r = (\hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$

and

$r = (2 \hat{i} - \hat{j} - \hat{k}) + μ (2 \hat{i} + \hat{j} + 2 \hat{k})$

It is known that the shortest distance between the lines, $r = a_{1} + λ b_{1}$ and $r = a_{2} + μ b_{2}$ is given

$by d = \frac{( b _{1} \times b _{2} ) \cdot ( a _{2} - a _{1} )}{b _{1} \times b _{2}}$

Comparing the given equations, we obtain

$a_{1} = \hat{i} + 2 \hat{j} + \hat{k}, b_{1} = \hat{i} - \hat{j} + \hat{k}, a_{2} = 2 \hat{i} - \hat{j} - \hat{k}, b_{2} = 2 \hat{i} + \hat{j} + 2 \hat{k} a_{2} - a_{1} = (2 \hat{i} - \hat{j} - \hat{k}) - (\hat{i} + 2 \hat{j} + \hat{k}) = \hat{i} - 3 \hat{j} - 2 \hat{k} b_{1} \times b_{2} = \hat{i} 12 \hat{j} - 1 1 \hat{k} 12$

$\Rightarrow b_{1} \times b_{2} = (- 2 - 1) \hat{i} - (2 - 2) \hat{j} + (1 + 2) \hat{k} = - 3 \hat{i} + 3 \hat{k}$

Substituting all the values in equation (1), we obtain

$d = \frac{( - 3 i ^ + 3 k ^ ) \cdot ( i ^ - 3 j ^ - 2 k ^ )}{3 2} = \frac{- 3 \cdot 1 + 3 ( - 2 )}{3 2}$

$\Rightarrow d = \frac{- 9}{3 2}$

Therefore, the shortest distance between the two lines is $\frac{3 2}{2}$ units.

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. The given lines are

$\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} and \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}$

It is known that the shortest distance between

$the two lines, \frac{x - x _{1}}{a _{1}} = \frac{y - y _{1}}{b _{1}} = \frac{z - z _{1}}{c _{1}} and$

$\frac{x - x _{2}}{a _{2}} = \frac{y - y _{2}}{b _{2}} = \frac{z - z _{2}}{c _{2}} is given as : d = \frac{x _{2} - x _{1} a _{1} a _{2} y _{2} - y _{1} b _{1} b _{2} z _{2} - z _{1} c _{1} c _{2}}{( b _{1} c _{2} - b _{2} c _{1} ) ^{2} + ( c _{1} a _{2} - c _{2} a _{1} ) ^{2} + ( a _{1} b _{2} - a _{2} b _{1} ) ^{2}}$

Comparing the given equations, we obtain

$x_{1} = - 1, y_{1} = - 1, z_{1} = - 1; x_{2} = 3, y_{2} = 5, z_{2} = 7 a_{1} = 7, b_{1} = - 6, c_{1} = 1; a_{2} = 1, b_{2} = - 2, c_{2} = 1$

Then, $x_{2} - x_{1} a_{1} a_{2} y_{2} - y_{1} b_{1} b_{2} z_{2} - z_{1} c_{1} c_{2}$

$= = 471 6 - 6 - 2 811 4 (- 6 + 2) - 6 (7 - 1) + 8 (- 14 + 6) - 16 - 36 - 64 = - 116$

$\Rightarrow (b_{1} c_{2} - b_{2} c_{1})^{2} + (c_{1} a_{2} - c_{2} a_{1})^{2} + (a_{1} b_{2} - a_{2} b_{1})^{2} = (- 6 + 2)^{2} + (1 - 7)^{2} + (- 14 + 6)^{2} = 16 + 36 + 64 = 116 = 229$

Substituting all the values in equation (1), we obtain

$d = \frac{- 116}{116} = \frac{116}{116} = 229$

Since distance is always non-negative, the distance between the given lines is $229$ units.

Find the shortest distance between the lines whose vector equations are:

$r = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ (\hat{i} - 3 \hat{j} + 2 \hat{k}) and r = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + μ (2 \hat{i} + 3 \hat{j} + \hat{k})$

Sol. The given lines are

$r = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ (\hat{i} - 3 \hat{j} + 2 \hat{k}) and r = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + μ (2 \hat{i} + 3 \hat{j} + \hat{k})$

It is known that the shortest distance between the lines, $r = a_{1} + λ b_{1}$ and $r = a_{2} + μ b_{2}$ is given by,

$d = \frac{( b _{1} \times b _{2} ) \cdot ( a _{2} - a _{1} )}{∣ b _{1} \times b _{2} ∣}$

Comparing the given equations with

$r = a_{1} + λ b_{1} and r = a_{2} + μ b_{2}; we have a_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}; a_{2} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} b_{1} = \hat{i} - 3 \hat{j} + 2 \hat{k}; b_{2} = 2 \hat{i} + 3 \hat{j} + \hat{k} a_{2} - a_{1} = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) - (\hat{i} + 2 \hat{j} + 3 \hat{k}) = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} b_{1} \times b_{2} = \hat{i} 12 \hat{j} - 3 3 \hat{k} 21 = (- 3 - 6) \hat{i} - (1 - 4) \hat{j} + (3 + 6) \hat{k} = - 9 \hat{i} + 3 \hat{j} + 9 \hat{k}$

$\Rightarrow b_{1} \times b_{2} = (- 9)^{2} + (3)^{2} + (9)^{2} = 81 + 9 + 81 = 171 = 319$

$\Rightarrow (b_{1} \times b_{2}) \cdot (a_{2} - a_{1}) = (- 9 \hat{i} + 3 \hat{j} + 9 \hat{k}) \cdot (3 \hat{i} + 3 \hat{j} + 3 \hat{k}) = - 9 \times 3 + 3 \times 3 + 9 \times 3 = 9$

Substituting all the values in equation (1), we obtain

$d = \frac{9}{3 19} = \frac{3}{19}$

Therefore, the shortest distance between the two given lines is $\frac{3}{19}$ units.

Find the shortest distance between the lines whose vector equations are

$r = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - 2 t) \hat{k} and r = (s + 1) \hat{i} + (2 s - 1) \hat{j} - (2 s + 1) \hat{k}$

Sol. The given lines are

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow r = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - 2 t) \hat{k} r = (\hat{i} - 2 \hat{j} + 3 \hat{k}) + t (- \hat{i} + \hat{j} - 2 \hat{k}) r = (s + 1) \hat{i} + (2 s - 1) \hat{j} - (2 s + 1) \hat{k} r = (\hat{i} - \hat{j} - \hat{k}) + s (\hat{i} + 2 \hat{j} - 2 \hat{k})$

It is known that the shortest distance between the lines, $r = a_{1} + λ b_{1}$ and $r = a_{2} + μ b_{2}$ is given by

$d = \frac{( b _{1} \times b _{2} ) \cdot ( a _{2} - a _{1} )}{b _{1} \times b _{2}}$

For the given equations,

$a_{1} = \hat{i} - 2 \hat{j} + 3 \hat{k}; a_{2} = \hat{i} - \hat{j} - \hat{k} b_{1} = - \hat{i} + \hat{j} - 2 \hat{k}; b_{2} = \hat{i} + 2 \hat{j} - 2 \hat{k} a_{2} - a_{1} = (\hat{i} - \hat{j} - \hat{k}) - (\hat{i} - 2 \hat{j} + 3 \hat{k}) = \hat{j} - 4 \hat{k} b_{1} \times b_{2} = \hat{i} - 1 1 \hat{j} 12 \hat{k} - 2 - 2 = (- 2 + 4) \hat{i} - (2 + 2) \hat{j} + (- 2 - 1) \hat{k} = 2 \hat{i} - 4 \hat{j} - 3 \hat{k}$

$\Rightarrow b_{1} \times b_{2} = (2)^{2} + (- 4)^{2} + (- 3)^{2} = 4 + 16 + 9 = 29$

$∴ (b_{1} \times b_{2}) \cdot (a_{2} - a_{1}) = (2 \hat{i} - 4 \hat{j} - 3 \hat{k}) \cdot (\hat{j} - 4 \hat{k}) = - 4 + 12 = 8$

Substituting all the values in equation (3), we obtain $d = \frac{8}{29} = \frac{8}{29}$ Therefore, the shortest distance between the lines is $\frac{8}{29}$ units.

MISCELLANEOUS EXERCISE

Find the angle between the lines whose direction ratios are (a, b, c) and (b - c, $c - a, a - b)$ . Sol. The angle $θ$ between the lines with direction cosines, (a, b, c) and $(b - c, c - a, a - b)$ , is given by,

$cos θ = \frac{a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2}}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} + a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}$

$\Rightarrow cos θ$

$= \frac{a ( b - c ) + b ( c - a ) + c ( a - b )}{a ^{2} + b ^{2} + c ^{2} + ( b - c ) ^{2} + ( c - a ) ^{2} + ( a - b ) ^{2}}$

$\Rightarrow cos θ = 0$ $\Rightarrow θ = cos^{- 1} (0)$ $\Rightarrow θ = 9 0^{\circ}$ Thus, the angle between the lines is 90°.

Find the equation of a line parallel to x-axis and passing through the origin. Sol. The line parallel to x-axis and passing through the origin is x-axis itself. Let A be a point on x-axis. Therefore, the coordinates of A are given by $(a, 0, 0)$ , where $a \in R$ . Direction ratios of OA are (a -0), (0-0), (0-0) i.e. a, 0, 0 The equation of OA is given by

$\frac{x - 0}{a} = \frac{y - 0}{0} = \frac{z - 0}{0}$

$\Rightarrow \frac{x}{1} = \frac{y}{0} = \frac{z}{0} = a$

Thus, the equation of line parallel to x-axis and passing through origin is $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$

If the lines $\frac{x - 1}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3 k} = \frac{y - 1}{1} = \frac{z - 6}{- 5}$ are perpendicular, find the value of k. Sol. The direction ratios of the lines,

$\frac{x - 1}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2} and \frac{x - 1}{3 k} = \frac{y - 1}{1} = \frac{z - 6}{- 5},$

are - 3, 2k, 2 and 3k, 1, -5 respectively. It is known that two lines with direction ratios, $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$ , are perpendicular if

$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

$∴ - 3 (3 k) + 2 k \times 1 + 2 (- 5) = 0$ $\Rightarrow - 9 k + 2 k - 10 = 0$ $\Rightarrow 7 k = - 10$ $\Rightarrow k = \frac{- 10}{7}$ Therefore, for $k = - \frac{10}{7}$ , the given lines are perpendicular to each other.

Find the shortest distance between lines

$r = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} + λ (\hat{i} - 2 \hat{j} + 2 \hat{k}) and r = - 4 \hat{i} - \hat{k} + μ (3 \hat{i} - 2 \hat{j} - 2 \hat{k})$

Sol. The given lines are

$r = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} + λ (\hat{i} - 2 \hat{j} + 2 \hat{k}) r = - 4 \hat{i} - \hat{k} + μ (3 \hat{i} - 2 \hat{j} - 2 \hat{k})$

It is known that the shortest distance between two lines, $r = a_{1} + λ b_{1}$ and $r = a_{2} + μ b_{2}$ is given

$by d = \frac{( b _{1} \times b _{2} ) \cdot ( a _{2} - a _{1} )}{∣ b _{1} \times b _{2} ∣}$

Comparing $r = a_{1} + λ b_{1}$ and $r = a_{2} + μ b_{2}$ to equations (1) and (2), we obtain

$a_{1} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k}, b_{1} = \hat{i} - 2 \hat{j} + 2 \hat{k}, a_{2} = - 4 \hat{i} - \hat{k}, b_{2} = 3 \hat{i} - 2 \hat{j} - 2 \hat{k}$

$\Rightarrow a_{2} - a_{1} = (- 4 \hat{i} - \hat{k}) - (6 \hat{i} + 2 \hat{j} + 2 \hat{k}) = - 10 \hat{i} - 2 \hat{j} - 3 \hat{k}$

$\Rightarrow b_{1} \times b_{2} = i 13 \hat{j} - 2 - 2 \hat{k} 2 - 2 = (4 + 4) \hat{i} - (- 2 - 6) \hat{j} + (- 2 + 6) \hat{k} = 8 \hat{i} + 8 \hat{j} + 4 \hat{k}$

$\Rightarrow b_{1} \times b_{2} = (8)^{2} + (8)^{2} + (4)^{2} = 12 (b_{1} \times b_{2}) \cdot (a_{2} - a_{1}) = (8 \hat{i} + 8 \hat{j} + 4 \hat{k}) \cdot (- 10 \hat{i} - 2 \hat{j} - 3 \hat{k}) = - 80 - 16 - 12 = - 108$

Substituting all the values in equation (3), we obtain $d = \frac{- 108}{12} = 9$ units Therefore, the shortest distance between the two given lines is 9 units.

Find the vector equation of the line passing through the point (1, 2, -4) and perpendicular to the two lines: $\frac{x - 8}{3} = \frac{y + 19}{- 16} = \frac{z - 10}{7}$ and $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}$

Sol. Let the required line be parallel to the vector $b$ given by, $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ The position vector of the point (1, 2, -4) is $a = \hat{i} + 2 \hat{j} - 4 \hat{k}$ The equation of the line passing through (1, 2, -4) and parallel to vector $b$ is $r = a + λ b$

$\Rightarrow r = (\hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) .$

The equations of the lines are

$\frac{x - 8}{3} = \frac{y + 19}{- 16} = \frac{z - 10}{7}$

and $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}$ Lines (1) and line (2) are perpendicular to each other

$3 b_{1} - 16 b_{2} + 7 b_{3} = 0$

Also, lines (1) and line (3) are perpendicular to each other.

$3 b_{1} + 8 b_{2} - 5 b_{3} = 0$

From equations (4) and (5), we obtain

$\Rightarrow \Rightarrow \frac{b _{1}}{( - 16 ) ( - 5 ) - 8 \times 7} = \frac{b _{2}}{7 \times 3 - 3 ( - 5 )} = \frac{b _{3}}{3 \times 8 - 3 ( - 16 )} \frac{b _{1}}{24} = \frac{b _{2}}{36} = \frac{b _{3}}{72} \frac{b _{1}}{2} = \frac{b _{2}}{3} = \frac{b _{3}}{6}$

∴ Direction ratios of $b$ are 2, 3, and 6.

$∴ b = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$

Substituting $b = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ in equation (1), we obtain $r = (\hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ This is the equation of the required line.

4.0Class 12 Maths NCERT Solutions – Chapter-wise Links

Get NCERT Solutions for Class 12 Maths chapter-wise, with answers to NCERT exercises, key formulas, and detailed solutions to make question practice easier and more effective.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 12	Linear Programming
Chapter 13	Probability

5.0Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise-wise Solutions

Exercise	Number of Questions	Important Topics
Exercise 11.1	5 Questions & Solutions	Direction cosines, direction ratios and their relationship
Exercise 11.2	15 Questions & Solutions	Equations of lines in space, angle between two lines and shortest distance between two lines
Miscellaneous Exercise	5 Questions & Solutions	Mixed questions on direction cosines, equations of lines, angles and distances

6.0Key Features and Benefits of Class 12 Maths Chapter 11 Three Dimensional Geometry

Clear Understanding of Spatial Concepts: The solutions explain points, lines, directions, and other concepts in space clearly, helping students visualise three-dimensional geometry.
Step-by-Step Methods: Detailed solutions make it easier to apply formulas correctly and reduce confusion while solving lengthy calculations.
NCERT-Based Exercises: The exercises follow the NCERT syllabus and cover question types that are useful for CBSE board exam preparation.
Improved Accuracy: Regular practice helps students improve accuracy while solving coordinate geometry and angle-based problems.
Useful for Competitive Exams: A strong foundation in three-dimensional geometry can support preparation for Mathematics Olympiads and competitive entrance examinations.
Application in Further Studies: Understanding three-dimensional geometry helps students connect mathematical concepts with topics used in physics, engineering, and other technical studies.

Frequently Asked Questions

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Can NCERT Solutions for Class 12 Maths Chapter 11 help with questions on angles between two lines?

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Frequently Asked Questions

Where can I find NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry?

What do NCERT Solutions for Class 12 Maths Chapter 11 explain about direction cosines and direction ratios?

How are equations of lines solved using NCERT Solutions in Class 12 Maths Chapter 11?

Which topics should I revise from NCERT Solutions for Class 12 Maths Chapter 11 before exams?

Can NCERT Solutions for Class 12 Maths Chapter 11 help with questions on angles between two lines?

How does the Miscellaneous Exercise in NCERT Class 12 Maths Chapter 11 help with practice?

Can NCERT Solutions for Class 12 Maths Chapter 11 help with competitive exams like JEE?

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NCERT Solutions Class 12 Maths Chapter 11 Three Dimensional Geometry

1.0Key Concepts of Class 12 Maths Chapter 11 Three Dimensional Geometry

2.0NCERT Class 12 Maths Chapter 11 Three Dimensional Geometry : Detailed Solutions

3.0EXERCISE - 11.1

EXERCISE 11.2

MISCELLANEOUS EXERCISE

4.0Class 12 Maths NCERT Solutions – Chapter-wise Links

5.0Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise-wise Solutions

6.0Key Features and Benefits of Class 12 Maths Chapter 11 Three Dimensional Geometry

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