CBSE Class 12 Mathematics Three Dimensional Geometry Notes Set 02

Important Notes and Formulas

Introduction

In earlier classes we have learnt about points, lines, circles and conic sections in two dimensional geometry. In two dimensions a point is represented by an ordered pair (x, y) (where x and y are both real numbers).
In this chapter, we shall study the direction cosines and direction ratios of a line joining two points and also discuss the equations of lines in space under different conditions, the angle between two lines and the shortest distance between two skew lines.

Rectangular Co-ordinate System

In the cartesian system, three lines which are mutually perpendicular form what is called the rectangular cartesian co-ordinate system.
Co-ordinate axes and co-ordinate planes:
When three mutually perpendicular planes intersect at a point, then mutually perpendicular lines are obtained and these lines also pass through that point. If we take the point of intersection as the origin, then the three planes are known as co-ordinate planes and the three lines are known as co-ordinate axes.

Octants:
Every plane bisects the space. Hence the three co-ordinate planes divide the space into eight parts. These parts are known as octants.

Direction Cosines and Direction Ratios of a Line

(i) Direction cosines of a line:
If α, β, γ are the angles made by a line with the x-axis, y-axis and z-axis respectively, then cos α, cos β and cos γ are called the direction cosines of the line, denoted by l, m and n respectively, and the relation between l, m, n is given by
\( l^2 + m^2 + n^2 = 1 \)
The direction cosines of the x-axis, y-axis and z-axis are respectively 1, 0, 0; 0, 1, 0; 0, 0, 1.

(ii) Direction ratios:
Any three numbers a, b, c proportional to the direction cosines l, m, n are called the direction ratios of the line,
i.e. \( \frac{l}{a} = \frac{m}{b} = \frac{n}{c} \).

(iii) Relation between direction cosines and direction ratios:
\( \frac{l}{a} = \frac{m}{b} = \frac{n}{c} \)
∴ \( \frac{l^2}{a^2} = \frac{m^2}{b^2} = \frac{n^2}{c^2} = \frac{l^2 + m^2 + n^2}{a^2 + b^2 + c^2} \)
⇒ \( l = \frac{\pm a}{\sqrt{a^2 + b^2 + c^2}}; \quad m = \frac{\pm b}{\sqrt{a^2 + b^2 + c^2}}; \quad n = \frac{\pm c}{\sqrt{a^2 + b^2 + c^2}} \)

(iv) Relation between the direction cosines of a line:
Consider a line RS with direction cosines l, m, n. Through the origin draw a line parallel to the given line and take a point P(x, y, z) on the line. From P draw a perpendicular PA on the x-axis (figure).
Let OP = r. Then \( \cos\alpha = \frac{OA}{OP} = \frac{x}{r} \). This gives x = lr.
Similarly, y = mr and z = nr.
Thus \( x^2 + y^2 + z^2 = r^2(l^2 + m^2 + n^2) \)
But \( x^2 + y^2 + z^2 = r^2 \)
Hence \( l^2 + m^2 + n^2 = 1 \)

(v) Direction cosines of a line passing through two points:
Let the two points be \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \).
Let l, m, n be the direction cosines of the line PQ and let it make angles α, β and γ with the X, Y and Z-axis, respectively.
Draw perpendiculars from P and Q to the XY-plane to meet at R and S. Draw a perpendicular from P to QS to meet at N. Now, in the right angled triangle PNQ, ∠PQN = γ.
Therefore, \( \cos\gamma = \frac{NQ}{PQ} = \frac{z_2 - z_1}{PQ} \)
Similarly \( \cos\alpha = \frac{x_2 - x_1}{PQ} \) and \( \cos\beta = \frac{y_2 - y_1}{PQ} \)
Hence, the direction cosines of the line segment joining the points \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \) are
\( \frac{x_2 - x_1}{PQ}, \frac{y_2 - y_1}{PQ}, \frac{z_2 - z_1}{PQ} \)
where \( PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \)
Note: The direction ratios of the line segment joining \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \) may be taken as \( x_2 - x_1, y_2 - y_1, z_2 - z_1 \) or \( x_1 - x_2, y_1 - y_2, z_1 - z_2 \).

Equation of a Line in Space

A line is uniquely determined if
(i) it passes through a given point and has a given direction, or
(ii) it passes through two given points.

Equation of a line through a given point and parallel to a given vector \( \vec{b} \):
Let \( \vec{a} \) be the position vector of the given point A with respect to the origin O of the rectangular coordinate system. Let l be the line which passes through the point A and is parallel to a given vector \( \vec{b} \). Let \( \vec{r} \) be the position vector of an arbitrary point P on the line (figure).
Then \( \overrightarrow{AP} \) is parallel to the vector \( \vec{b} \), i.e., \( \overrightarrow{AP} = \lambda\vec{b} \), where λ is some real number.
But \( \overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA} \)
i.e. \( \lambda\vec{b} = \vec{r} - \vec{a} \)
Conversely, for each value of the parameter λ, this equation gives the position vector of a point P on the line. Hence, the vector equation of the line is given by
\( \vec{r} = \vec{a} + \lambda\vec{b} \) ...(i)

Derivation of cartesian form from vector form
Let the coordinates of the given point A be \( (x_1, y_1, z_1) \) and the direction ratios of the line be a, b, c. Consider the coordinates of any point P to be (x, y, z). Then
\( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}; \quad \vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k} \) and \( \vec{b} = a\hat{i} + b\hat{j} + c\hat{k} \)
Substituting these values in (i) and equating the coefficients of \( \hat{i}, \hat{j} \) and \( \hat{k} \),
\( x = x_1 + \lambda a; \quad y = y_1 + \lambda b; \quad z = z_1 + \lambda c \) ...(ii)
These are the parametric equations of the line. Eliminating the parameter λ from (ii),
\( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \) ...(iii)
This is the cartesian equation of the line.
Note: If l, m, n are the direction cosines of the line, the equation of the line is
\( \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} \)

Equation of a line passing through two given points:
Let \( \vec{a} \) and \( \vec{b} \) be the position vectors of two points \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \), respectively, that are lying on a line (figure).
Let \( \vec{r} \) be the position vector of an arbitrary point P(x, y, z). Then P is a point on the line if and only if \( \overrightarrow{AP} = \vec{r} - \vec{a} \) and \( \overrightarrow{AB} = \vec{b} - \vec{a} \) are collinear vectors. Therefore, P is on the line if and only if
\( \vec{r} - \vec{a} = \lambda(\vec{b} - \vec{a}) \)
or \( \vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}), \ \lambda \in R \) ...(i)
This is the vector equation of the line.

Derivation of cartesian form from vector form
We have
\( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}, \ \vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k} \) and \( \vec{b} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k} \).
Substituting these values in (i),
\( x\hat{i} + y\hat{j} + z\hat{k} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k} + \lambda[(x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}] \)
Equating the coefficients of \( \hat{i}, \hat{j}, \hat{k} \),
\( x = x_1 + \lambda(x_2 - x_1); \ y = y_1 + \lambda(y_2 - y_1); \ z = z_1 + \lambda(z_2 - z_1) \)
On eliminating λ,
\( \frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1} \)
which is the equation of the line in Cartesian form.

Perpendicular and Parallel Lines

Let the two lines have their d.c.'s given by \( l_1, m_1, n_1 \) and \( l_2, m_2, n_2 \) respectively. Then they are perpendicular if θ = 90°, i.e. cos θ = 0, i.e. \( l_1 l_2 + m_1 m_2 + n_1 n_2 = 0 \).
Also the two lines are parallel if θ = 0, i.e. sin θ = 0, i.e. \( \frac{l_1}{l_2} = \frac{m_1}{m_2} = \frac{n_1}{n_2} \).
Note: If instead of d.c.'s, the d.r.'s \( a_1, b_1, c_1 \) and \( a_2, b_2, c_2 \) are given, then the lines are perpendicular if \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \) and parallel if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).

Shortest Distance between Two Lines

If two lines in space intersect at a point, then obviously the shortest distance between them is zero. Lines which do not intersect and are also not parallel are called skew lines. In other words, the lines which are not coplanar are skew lines. For skew lines, the direction of the shortest distance vector would be perpendicular to both the lines. The magnitude of the shortest distance vector would be equal to that of the projection of \( \overrightarrow{AB} \) along the direction of the line of shortest distance; \( \overrightarrow{LM} \) is parallel to \( \vec{p} \times \vec{q} \),
i.e. \( |\overrightarrow{LM}| = |\text{Projection of } \overrightarrow{AB} \text{ on } \overrightarrow{LM}| = |\text{Projection of } \overrightarrow{AB} \text{ on } \vec{p} \times \vec{q}| = \left| \frac{\overrightarrow{AB} \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right| = \left| \frac{(\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right| \)

(a) The two lines directed along \( \vec{p} \) and \( \vec{q} \) will intersect only if the shortest distance = 0, i.e. \( (\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q}) = 0 \).
(b) If two lines are given by \( \vec{r}_1 = \vec{a}_1 + k_1\vec{b} \) and \( \vec{r}_2 = \vec{a}_2 + k_2\vec{b} \), i.e. they are parallel, then
\( d = \left| \frac{\vec{b} \times (\vec{a}_2 - \vec{a}_1)}{|\vec{b}|} \right| \)

Solved Examples

Illustration 1. If a line makes angles 120°, 60° and 45° with the X, Y and Z axis respectively, find its direction cosines. [1 Mark]
Answer:
Required direction cosines are either
<cos 120°, cos 60°, cos 45°> or <−cos 120°, −cos 60°, −cos 45°>
\( \left\langle -\frac{1}{2}, \frac{1}{2}, \frac{1}{\sqrt{2}} \right\rangle \) or \( \left\langle \frac{1}{2}, -\frac{1}{2}, -\frac{1}{\sqrt{2}} \right\rangle \) [1]

Illustration 2. Find the direction cosines of the ray from P to Q, where P is the point (1, −2, 2) and Q is the point (3, −5, −4). [1 Mark]
Answer:
\( \overrightarrow{PQ} \) = P.V. of Q − P.V. of P = \( (3\hat{i} - 5\hat{j} - 4\hat{k}) - (\hat{i} - 2\hat{j} + 2\hat{k}) = 2\hat{i} - 3\hat{j} - 6\hat{k} \)
⇒ \( |\overrightarrow{PQ}| = \sqrt{(2)^2 + (-3)^2 + (-6)^2} = \sqrt{4 + 9 + 36} = 7 \)
Hence, unit vector in the direction of \( \overrightarrow{PQ} = \frac{1}{7}(2\hat{i} - 3\hat{j} - 6\hat{k}) = \frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k} \)
∴ Direction cosines of the ray from P to Q are \( \left(\frac{2}{7}, \frac{-3}{7}, \frac{-6}{7}\right) \). [1]

Illustration 3. Find the vector equation for the line passing through the points (6, 4, −2) and (2, 3, 5). [1 Mark]
Answer:
Let \( \vec{a} \) and \( \vec{b} \) be the position vectors of the points A(6, 4, −2) and B(2, 3, 5).
Then \( \vec{a} = 6\hat{i} + 4\hat{j} - 2\hat{k} \)
and \( \vec{b} = 2\hat{i} + 3\hat{j} + 5\hat{k} \)
Therefore \( \vec{b} - \vec{a} = -4\hat{i} - \hat{j} + 7\hat{k} \)
Let \( \vec{r} \) be the position vector of any point on the line, then the vector equation of the line is
\( \vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}) \)
\( \vec{r} = 6\hat{i} + 4\hat{j} - 2\hat{k} + \lambda(-4\hat{i} - \hat{j} + 7\hat{k}) \) [1]

Illustration 4. The cartesian equation of a line is \( \frac{x - 5}{2} = \frac{y + 3}{4} = \frac{z + 6}{2} \). Find the vector equation for the line. [1 Mark]
Answer:
Comparing the given equation with the standard form
\( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \)
⇒ \( x_1 = 5, y_1 = -3, z_1 = -6; \ a = 2, b = 4, c = 2 \)
Thus the required line passes through the point (5, −3, −6) and is parallel to the vector \( 2\hat{i} + 4\hat{j} + 2\hat{k} \). If \( \vec{r} \) is the position vector of any point on the line, then the vector equation is
\( \vec{r} = (5\hat{i} - 3\hat{j} - 6\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 2\hat{k}) \) [1]

Illustration 5. Find the equations of the two lines through the origin which intersect the line \( \frac{x - 3}{2} = \frac{y - 3}{1} = \frac{z}{1} \) at angles of \( \frac{\pi}{3} \). [Exemplar] [4 Marks]
Answer:
Given line is
\( \frac{x - 3}{2} = \frac{y - 3}{1} = \frac{z - 0}{1} = t \) (say) ...(i) [1]
Any point on the line (i) is P(3 + 2t, 3 + t, t).
Direction numbers of OP are
<3 + 2t − 0, 3 + t − 0, t − 0>, i.e., <3 + 2t, 3 + t, t> [1]
Since OP makes an angle of \( \frac{\pi}{3} \) with the given line (i), therefore
\( \cos\frac{\pi}{3} = \frac{|2 \times (3 + 2t) + 1 \times (3 + t) + 1t|}{\sqrt{6}\sqrt{(3 + 2t)^2 + (3 + t)^2 + t^2}} \)
⇒ \( \frac{1}{2} = \frac{|6t + 9|}{\sqrt{6}\sqrt{6t^2 + 18t + 18}} \)
⇒ \( 6(6t^2 + 18t + 18) = 4(36t^2 + 108t + 81) \)
⇒ \( t^2 + 3t + 2 = 0 \) ⇒ t = −1 or t = −2 [1]
Substituting the values of t, P is (1, 2, −1) and (−1, 1, −2).
Required lines are \( \frac{x - 0}{1 - 0} = \frac{y - 0}{2 - 0} = \frac{z - 0}{-1 - 0} \) and \( \frac{x - 0}{-1 - 0} = \frac{y - 0}{1 - 0} = \frac{z - 0}{-2 - 0} \)
⇒ \( \frac{x}{1} = \frac{y}{2} = \frac{z}{-1} \) and \( \frac{x}{-1} = \frac{y}{1} = \frac{z}{-2} \) [1]

Illustration 6. Find the shortest distance between the lines \( \vec{r} = (4\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k}) \) and \( \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(2\hat{i} + 4\hat{j} - 5\hat{k}) \). [6 Marks]
Answer:
We know that the shortest distance between the lines \( \vec{r} = \vec{a}_1 + \lambda\vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \lambda\vec{b}_2 \) is given by
\( d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| \) [1]
Comparing the given equations with the equations \( \vec{r} = \vec{a}_1 + \lambda\vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \mu\vec{b}_2 \) respectively,
\( \vec{a}_1 = 4\hat{i} - \hat{j}, \ \vec{a}_2 = \hat{i} - \hat{j} + 2\hat{k}, \ \vec{b}_1 = \hat{i} + 2\hat{j} - 3\hat{k} \) and \( \vec{b}_2 = 2\hat{i} + 4\hat{j} - 5\hat{k} \) [1]
Now, \( \vec{a}_2 - \vec{a}_1 = -3\hat{i} + 0\hat{j} + 2\hat{k} \) and \( \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} = 2\hat{i} - \hat{j} + 0\hat{k} \) [1]
∴ \( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-3\hat{i} + 0\hat{j} + 2\hat{k}) \cdot (2\hat{i} - \hat{j} + 0\hat{k}) = -6 \) and \( |\vec{b}_1 \times \vec{b}_2| = \sqrt{4 + 1 + 0} = \sqrt{5} \) [2]
∴ Shortest distance \( d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \left| \frac{-6}{\sqrt{5}} \right| = \frac{6}{\sqrt{5}} \) units. [1]

Illustration 7. Find the shortest distance between the lines \( \frac{x - 8}{3} = \frac{y + 9}{-16} = \frac{z - 10}{7} \) and \( \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5} \). Also find the equations of the line of shortest distance. [6 Marks]
Answer:
Given lines are \( \frac{x - 8}{3} = \frac{y + 9}{-16} = \frac{z - 10}{7} = t \) (say) ...(i)
and \( \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5} = s \) (say) ...(ii)
Any point on (i) is M(8 + 3t, −9 − 16t, 10 + 7t) and any point on (ii) is N(15 + 3s, 29 + 8s, 5 − 5s). [1]
Direction numbers of MN are <15 + 3s − 8 − 3t, 29 + 8s + 9 + 16t, 5 − 5s − 10 − 7t>,
i.e., <7 + 3s − 3t, 38 + 8s + 16t, −5 − 5s − 7t>. [1]
Now |MN| will be the shortest distance between (i) and (ii) iff MN is perpendicular to both (i) and (ii),
i.e., iff 3(7 + 3s − 3t) + (−16)(38 + 8s + 16t) + 7(−5 − 5s − 7t) = 0
and 3(7 + 3s − 3t) + 8(38 + 8s + 16t) + (−5)(−5 − 5s − 7t) = 0
i.e., iff −154s − 314t − 622 = 0, i.e., 77s + 157t + 311 = 0 ...(iii)
and 98s + 154t + 350 = 0, i.e., 7s + 11t + 25 = 0 ...(iv) [1]
On solving (iii) and (iv) simultaneously, we get
t = −1 and s = −2.
t = −1 gives M(5, 7, 3) and s = −2 gives N(9, 13, 15). [1]
∴ The shortest distance between the given lines \( = \sqrt{(9 - 5)^2 + (13 - 7)^2 + (15 - 3)^2} \)
\( = \sqrt{16 + 36 + 144} = \sqrt{196} = 14 \) units [1]
Also, the equations of the line MN are \( \frac{x - 5}{9 - 5} = \frac{y - 7}{13 - 7} = \frac{z - 3}{15 - 3} \), i.e., \( \frac{x - 5}{2} = \frac{y - 7}{3} = \frac{z - 3}{6} \) [1]

Illustration 8. Find the distance of the point (2, 4, −1) from the line \( \frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9} \). [6 Marks]
Answer:
Vector equation of the given line is \( \vec{r} = \vec{a}_1 + t\vec{v} \), where \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \),
\( \vec{a}_1 = -5\hat{i} - 3\hat{j} + 6\hat{k} \) and \( \vec{v} = \hat{i} + 4\hat{j} - 9\hat{k} \). [1]
Position vector of the given point is \( \vec{a}_2 = 2\hat{i} + 4\hat{j} - \hat{k} \). [1]
Hence, the required distance \( = \left| \frac{\vec{v} \times (\vec{a}_2 - \vec{a}_1)}{|\vec{v}|} \right| = \frac{|(\hat{i} + 4\hat{j} - 9\hat{k}) \times (7\hat{i} + 7\hat{j} - 7\hat{k})|}{|\hat{i} + 4\hat{j} - 9\hat{k}|} \) [2]
⇒ \( d = \frac{7}{\sqrt{1^2 + 4^2 + 9^2}}|(\hat{i} + 4\hat{j} - 9\hat{k}) \times (\hat{i} + \hat{j} - \hat{k})| \)
Now, \( (\hat{i} + 4\hat{j} - 9\hat{k}) \times (\hat{i} + \hat{j} - \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & -9 \\ 1 & 1 & -1 \end{vmatrix} = 5\hat{i} - 8\hat{j} - 3\hat{k} \)
⇒ \( d = \frac{7}{\sqrt{98}}|5\hat{i} - 8\hat{j} - 3\hat{k}| \) [1]
\( = \frac{7\sqrt{5^2 + 8^2 + 3^2}}{\sqrt{98}} = 7 \) units [1]

Illustration 9. Determine the equation of the line passing through the point (−1, 3, −2) and perpendicular to the lines \( \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \) and \( \frac{x + 2}{-3} = \frac{y - 1}{2} = \frac{z + 1}{5} \). [6 Marks]
Answer:
We know that the equations of any line through a given point \( (x_1, y_1, z_1) \) are \( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \) [1]
∴ Equation of any line through (−1, 3, −2) can be taken as
\( \frac{x - (-1)}{a} = \frac{y - 3}{b} = \frac{z - (-2)}{c} \) ...(1) [1]
If it is perpendicular to the lines \( \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \) ...(2)
and \( \frac{x + 2}{-3} = \frac{y - 1}{2} = \frac{z + 1}{5} \) ...(3)
then 1a + 2b + 3c = 0 ⇒ a + 2b + 3c = 0 ...(4)
and (−3)a + 2b + 5c = 0 ⇒ −3a + 2b + 5c = 0 ...(5) [1]
Subtracting (5) from (4), we get
4a − 2c = 0 ⇒ c = 2a ...(6)
and hence, from (4),
a + 2b + 3 × 2a = 0 ⇒ 2b = −7a ⇒ \( b = -\frac{7}{2}a \) ...(7) [1]
From (6) and (7), we find
\( a : b : c = 1 : -\frac{7}{2} : 2 \) ⇒ a : b : c = 2 : −7 : 4. [1]
Putting these values in (1), we get the required line as \( \frac{x + 1}{2} = \frac{y - 3}{-7} = \frac{z + 2}{4} \). [1]

Do Yourself

Do Yourself 1
(i) If a line makes angles α, β and γ with the axes, then prove that \( \sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2 \).
(ii) Find the direction cosines of the line which is perpendicular to the lines with direction cosines proportional to (1, −2, −2) and (0, 2, 1).

Do Yourself 2
Find the equation of the line passing through the point (4, 2, 3) and having direction ratios 1, −1, 2.

Do Yourself 3
(i) Prove that the lines 2x = y = −z and x − 1 = −y − 2 = −2z + 1 are skew.
(ii) Show that the lines \( \vec{r} = \hat{i} + \hat{j} - \hat{k} + \lambda(3\hat{i} - \hat{j}) \) and \( \vec{r} = 4\hat{i} - \hat{k} + \mu(2\hat{i} + \hat{k}) \) intersect. Also, find their point of intersection.

Answers for Do Yourself

Do Yourself 1
(ii) \( \frac{2}{3}, -\frac{1}{3}, \frac{2}{3} \)

Do Yourself 2
\( \frac{x - 4}{1} = \frac{y - 2}{-1} = \frac{z - 3}{2} \)

Do Yourself 3
(ii) \( 4\hat{i} - \hat{k} \)

Practice Assignment

Question 1. Find the distance of the point (4, 3, 5) from the y-axis.
Answer:
\( \sqrt{41} \)

Question 2. Write the vector equation of the straight line \( \frac{x - 3}{3} = \frac{y - 2}{1} = \frac{z - 1}{0} \).
Answer:
\( \vec{r} = (3\hat{i} + 2\hat{j} + \hat{k}) + \lambda(3\hat{i} + \hat{j}) \)

Question 3. Find the distance of the point (0, 0, 0) from the line \( \frac{x}{1} = \frac{y - 3}{2} = \frac{z - 6}{2} \).
Answer:
3