Three Dimensional Geometry — NCERT Solutions

Given: A line makes angles $\alpha = 90^\circ$, $\beta = 135^\circ$, $\gamma = 45^\circ$ with the $x$-, $y$- and $z$-axes respectively.

Formula: Direction cosines are $l = \cos\alpha,\ m = \cos\beta,\ n = \cos\gamma$.

Working:
$$l = \cos 90^\circ = 0$$
$$m = \cos 135^\circ = -\frac{1}{\sqrt{2}}$$
$$n = \cos 45^\circ = \frac{1}{\sqrt{2}}$$

Verification: $l^2 + m^2 + n^2 = 0 + \dfrac{1}{2} + \dfrac{1}{2} = 1$ ✓

Answer: The direction cosines are $0,\ -\dfrac{1}{\sqrt{2}},\ \dfrac{1}{\sqrt{2}}$.

Given: The line makes equal angles with all three coordinate axes, i.e., $\alpha = \beta = \gamma$.

Concept: If $l, m, n$ are direction cosines, then $l^2 + m^2 + n^2 = 1$.

Working:
Since $\alpha = \beta = \gamma$, we have $l = m = n = \cos\alpha$.

Substituting in the identity:
$$l^2 + l^2 + l^2 = 1 \implies 3l^2 = 1 \implies l = \pm\frac{1}{\sqrt{3}}$$

Answer: The direction cosines are $\dfrac{1}{\sqrt{3}},\ \dfrac{1}{\sqrt{3}},\ \dfrac{1}{\sqrt{3}}$ or $-\dfrac{1}{\sqrt{3}},\ -\dfrac{1}{\sqrt{3}},\ -\dfrac{1}{\sqrt{3}}$.

Given: Direction ratios are $a = -18,\ b = 12,\ c = -4$.

Formula:
$$l = \frac{a}{\sqrt{a^2+b^2+c^2}},\quad m = \frac{b}{\sqrt{a^2+b^2+c^2}},\quad n = \frac{c}{\sqrt{a^2+b^2+c^2}}$$

Working:
$$\sqrt{a^2+b^2+c^2} = \sqrt{(-18)^2+(12)^2+(-4)^2} = \sqrt{324+144+16} = \sqrt{484} = 22$$

$$l = \frac{-18}{22} = -\frac{9}{11},\quad m = \frac{12}{22} = \frac{6}{11},\quad n = \frac{-4}{22} = -\frac{2}{11}$$

Answer: The direction cosines are $-\dfrac{9}{11},\ \dfrac{6}{11},\ -\dfrac{2}{11}$.

Given: Points $A(2,3,4)$, $B(-1,-2,1)$, $C(5,8,7)$.

Concept: Three points are collinear if the direction ratios of $AB$ and $BC$ are proportional.

Direction ratios of AB:
$$(-1-2,\ -2-3,\ 1-4) = (-3,\ -5,\ -3)$$

Direction ratios of BC:
$$(5-(-1),\ 8-(-2),\ 7-1) = (6,\ 10,\ 6)$$

Check proportionality:
$$\frac{-3}{6} = \frac{-5}{10} = \frac{-3}{6} = -\frac{1}{2}$$

The direction ratios of $AB$ and $BC$ are proportional, so $AB \parallel BC$. Since point $B$ is common to both, the points $A$, $B$, $C$ are collinear. $\hfill\blacksquare$

Given: Vertices $A(3,5,-4)$, $B(-1,1,2)$, $C(-5,-5,-2)$.

Side AB:
Direction ratios: $(-1-3,\ 1-5,\ 2-(-4)) = (-4,\ -4,\ 6)$
$$|AB| = \sqrt{16+16+36} = \sqrt{68} = 2\sqrt{17}$$
Direction cosines of $AB$:
$$\frac{-4}{2\sqrt{17}},\ \frac{-4}{2\sqrt{17}},\ \frac{6}{2\sqrt{17}} = -\frac{2}{\sqrt{17}},\ -\frac{2}{\sqrt{17}},\ \frac{3}{\sqrt{17}}$$

Side BC:
Direction ratios: $(-5-(-1),\ -5-1,\ -2-2) = (-4,\ -6,\ -4)$
$$|BC| = \sqrt{16+36+16} = \sqrt{68} = 2\sqrt{17}$$
Direction cosines of $BC$:
$$\frac{-4}{2\sqrt{17}},\ \frac{-6}{2\sqrt{17}},\ \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}},\ -\frac{3}{\sqrt{17}},\ -\frac{2}{\sqrt{17}}$$

Side CA:
Direction ratios: $(3-(-5),\ 5-(-5),\ -4-(-2)) = (8,\ 10,\ -2)$
$$|CA| = \sqrt{64+100+4} = \sqrt{168} = 2\sqrt{42}$$
Direction cosines of $CA$:
$$\frac{8}{2\sqrt{42}},\ \frac{10}{2\sqrt{42}},\ \frac{-2}{2\sqrt{42}} = \frac{4}{\sqrt{42}},\ \frac{5}{\sqrt{42}},\ -\frac{1}{\sqrt{42}}$$

Answer:

• Direction cosines of $AB$: $-\dfrac{2}{\sqrt{17}},\ -\dfrac{2}{\sqrt{17}},\ \dfrac{3}{\sqrt{17}}$
• Direction cosines of $BC$: $-\dfrac{2}{\sqrt{17}},\ -\dfrac{3}{\sqrt{17}},\ -\dfrac{2}{\sqrt{17}}$
• Direction cosines of $CA$: $\dfrac{4}{\sqrt{42}},\ \dfrac{5}{\sqrt{42}},\ -\dfrac{1}{\sqrt{42}}$

Concept: Two lines are perpendicular if $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.

Let the three lines be $L_1$, $L_2$, $L_3$ with the given direction cosines.

Check $L_1 \perp L_2$:
$$\frac{12}{13}\cdot\frac{4}{13} + \frac{-3}{13}\cdot\frac{12}{13} + \frac{-4}{13}\cdot\frac{3}{13} = \frac{48 - 36 - 12}{169} = \frac{0}{169} = 0 \checkmark$$

Check $L_2 \perp L_3$:
$$\frac{4}{13}\cdot\frac{3}{13} + \frac{12}{13}\cdot\frac{-4}{13} + \frac{3}{13}\cdot\frac{12}{13} = \frac{12 - 48 + 36}{169} = \frac{0}{169} = 0 \checkmark$$

Check $L_1 \perp L_3$:
$$\frac{12}{13}\cdot\frac{3}{13} + \frac{-3}{13}\cdot\frac{-4}{13} + \frac{-4}{13}\cdot\frac{12}{13} = \frac{36 + 12 - 48}{169} = \frac{0}{169} = 0 \checkmark$$

Since each pair of lines satisfies the perpendicularity condition, the three lines are mutually perpendicular. $\hfill\blacksquare$

Direction ratios of line $L_1$ through $(1,-1,2)$ and $(3,4,-2)$:
$$(3-1,\ 4-(-1),\ -2-2) = (2,\ 5,\ -4)$$

Direction ratios of line $L_2$ through $(0,3,2)$ and $(3,5,6)$:
$$(3-0,\ 5-3,\ 6-2) = (3,\ 2,\ 4)$$

Check perpendicularity: $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$
$$2(3) + 5(2) + (-4)(4) = 6 + 10 - 16 = 0 \checkmark$$

Hence, the two lines are perpendicular to each other. $\hfill\blacksquare$

Direction ratios of line $L_1$ through $(4,7,8)$ and $(2,3,4)$:
$$(2-4,\ 3-7,\ 4-8) = (-2,\ -4,\ -4)$$

Direction ratios of line $L_2$ through $(-1,-2,1)$ and $(1,2,5)$:
$$(1-(-1),\ 2-(-2),\ 5-1) = (2,\ 4,\ 4)$$

Check proportionality:
$$\frac{-2}{2} = \frac{-4}{4} = \frac{-4}{4} = -1$$

The direction ratios are proportional, hence the two lines are parallel. $\hfill\blacksquare$

Given: Point $A(1,2,3)$, so position vector $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$; direction vector $\vec{b} = 3\hat{i} + 2\hat{j} - 2\hat{k}$.

Vector equation of the line:
$$\vec{r} = \vec{a} + \lambda\vec{b}$$
$$\boxed{\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 2\hat{k})}$$

Cartesian form:
$$\frac{x-1}{3} = \frac{y-2}{2} = \frac{z-3}{-2}$$

Given: $\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}$, $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$.

Vector equation:
$$\vec{r} = \vec{a} + \lambda\vec{b}$$
$$\boxed{\vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})}$$

Cartesian equation: The point is $(2, -1, 4)$ and direction ratios are $(1, 2, -1)$.
$$\boxed{\frac{x-2}{1} = \frac{y+1}{2} = \frac{z-4}{-1}}$$

Given: Point $(-2, 4, -5)$; the given line has direction ratios $(3, 5, 6)$.

Since the required line is parallel to the given line, it has the same direction ratios $(3, 5, 6)$.

Cartesian equation:
$$\boxed{\frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6}}$$

Given: Cartesian equation $\dfrac{x-5}{3} = \dfrac{y+4}{7} = \dfrac{z-6}{2}$.

The line passes through the point $(5, -4, 6)$ and has direction ratios $(3, 7, 2)$.

So $\vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k}$ and $\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}$.

Vector equation:
$$\boxed{\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k})}$$

Formula: $\cos\theta = \left|\dfrac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1||\vec{b}_2|}\right|$

(i) $\vec{b}_1 = 3\hat{i}+2\hat{j}+6\hat{k}$, $\vec{b}_2 = \hat{i}+2\hat{j}+2\hat{k}$

$$\vec{b}_1 \cdot \vec{b}_2 = 3(1)+2(2)+6(2) = 3+4+12 = 19$$
$$|\vec{b}_1| = \sqrt{9+4+36} = \sqrt{49} = 7,\quad |\vec{b}_2| = \sqrt{1+4+4} = \sqrt{9} = 3$$
$$\cos\theta = \left|\frac{19}{7 \times 3}\right| = \frac{19}{21}$$
$$\boxed{\theta = \cos^{-1}\left(\frac{19}{21}\right)}$$

(ii) $\vec{b}_1 = \hat{i}-\hat{j}-2\hat{k}$, $\vec{b}_2 = 3\hat{i}-5\hat{j}-4\hat{k}$

$$\vec{b}_1 \cdot \vec{b}_2 = 1(3)+(-1)(-5)+(-2)(-4) = 3+5+8 = 16$$
$$|\vec{b}_1| = \sqrt{1+1+4} = \sqrt{6},\quad |\vec{b}_2| = \sqrt{9+25+16} = \sqrt{50} = 5\sqrt{2}$$
$$\cos\theta = \left|\frac{16}{\sqrt{6}\cdot 5\sqrt{2}}\right| = \frac{16}{5\sqrt{12}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}$$
$$\boxed{\theta = \cos^{-1}\left(\frac{8}{5\sqrt{3}}\right)}$$