Exercise 11.1 Class 12 Maths Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 Ex 11.1 | LearnCBSEHub

💡 Direction Cosines (DCs)

If a line makes angles $\alpha, \beta, \gamma$ with the $x, y, z$ axes respectively, then the direction cosines are:

$$ l = \cos\alpha, \quad m = \cos\beta, \quad n = \cos\gamma $$

Property: $l^2 + m^2 + n^2 = 1$

Direction Ratios (DRs): Any numbers $a, b, c$ proportional to $l, m, n$.
To find DCs from DRs: $l = \pm \frac{a}{\sqrt{a^2+b^2+c^2}}$, etc.

Question 01

Find direction cosines of a line making angles $90^\circ, 135^\circ, 45^\circ$ with $x, y, z$ axes.

Let direction cosines be $l, m, n$.
$l = \cos 90^\circ = 0$
$m = \cos 135^\circ = \cos(180^\circ – 45^\circ) = -\cos 45^\circ = -\frac{1}{\sqrt{2}}$
$n = \cos 45^\circ = \frac{1}{\sqrt{2}}$

DCs: $0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$

Question 02

Find direction cosines of a line making equal angles with coordinate axes.

Let angles be $\alpha, \beta, \gamma$. Given $\alpha = \beta = \gamma$.
Therefore, $l = \cos\alpha, m = \cos\alpha, n = \cos\alpha$.
Using $l^2 + m^2 + n^2 = 1$:
$\cos^2\alpha + \cos^2\alpha + \cos^2\alpha = 1 \implies 3\cos^2\alpha = 1$.
$\cos^2\alpha = \frac{1}{3} \implies \cos\alpha = \pm \frac{1}{\sqrt{3}}$.

DCs: $\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}$

Question 03

If a line has direction ratios $-18, 12, -4$, find its direction cosines.

Direction ratios: $a=-18, b=12, c=-4$.
Magnitude $r = \sqrt{(-18)^2 + (12)^2 + (-4)^2}$
$= \sqrt{324 + 144 + 16} = \sqrt{484} = 22$.
DCs are $\frac{a}{r}, \frac{b}{r}, \frac{c}{r}$.
$l = \frac{-18}{22} = \frac{-9}{11}, \quad m = \frac{12}{22} = \frac{6}{11}, \quad n = \frac{-4}{22} = \frac{-2}{11}$.

DCs: $\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}$

Question 04

Show points (2, 3, 4), (-1, -2, 1), (5, 8, 7) are collinear.

Let points be A, B, C.
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:
$\frac{a_2}{a_1} = \frac{6}{-3} = -2, \quad \frac{b_2}{b_1} = \frac{10}{-5} = -2, \quad \frac{c_2}{c_1} = \frac{6}{-3} = -2$.
Since DRs are proportional, AB is parallel to BC. Since B is common, points are collinear.

Verified

Question 05

Find direction cosines of sides of triangle with vertices (3, 5, -4), (-1, 1, 2) and (-5, -5, -2).

Side AB

DRs: $(-1-3, 1-5, 2-(-4)) = (-4, -4, 6)$.
Magnitude $= \sqrt{16+16+36} = \sqrt{68} = 2\sqrt{17}$.
DCs: $\frac{-4}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}}, \frac{6}{2\sqrt{17}} = \frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}$.

Side BC

DRs: $(-5-(-1), -5-1, -2-2) = (-4, -6, -4)$.
Magnitude $= \sqrt{16+36+16} = \sqrt{68} = 2\sqrt{17}$.
DCs: $\frac{-4}{2\sqrt{17}}, \frac{-6}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}} = \frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}$.

Side CA

DRs: $(3-(-5), 5-(-5), -4-(-2)) = (8, 10, -2)$.
Magnitude $= \sqrt{64+100+4} = \sqrt{168} = 2\sqrt{42}$.
DCs: $\frac{8}{2\sqrt{42}}, \frac{10}{2\sqrt{42}}, \frac{-2}{2\sqrt{42}} = \frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, \frac{-1}{\sqrt{42}}$.