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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry4 Marks
Question
Find the angle between the lines whose direction cosines are given by the equations:
l + m +n = 0 and l2 + m2 + n2 = 0
✓

Answer

Given that, l + m + n = 0

l2 + m2 + n2 = 0

From equation (1),

l =  -(m + n)

Put the value of l in equation (2),

[-(m + n)]2 + m2 - n2 = 0

(m + n)2 + m2 - n2 = 0

m2 + n2 + 2mn + m2 - n2 = 0

2m2 + 2mn = 0

2m(m + n) = 0

m = 0, m + n = 0

m = -n and m = 0

Put the value of m = -n in equation (1)

l = -(m +n)

= -(0 + n)

l = -n

Thus, the direction ratios are proportional to

0, -n, n and -n, 0, n

⇒ 0, -1 1 and -1, 0, 1

So, vectors parallel to these lines are

$\vec{a}=0\times\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{b}=-\hat{\text{i}}+0\times\hat{\text{j}}+\hat{\text{k}}$ respectively.

Let, $\theta$ be the angle between the $\vec{a}$ and $\vec{b}$

So, $\cos\theta=\frac{\vec{a}\times\vec{b}}{\big|\vec{a}\big|\big|\vec{b}\big|}$

$\vec{a}=0\times\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{b}=-\hat{\text{i}}+0\times\hat{\text{j}}+\hat{\text{k}}$ respectively.

$\cos\theta=\frac{(0\times\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\times(-\hat{\text{i}}+0\times\hat{\text{j}}+\hat{\text{k}})}{\sqrt{0^2+(-1)^2+(1)^2}\sqrt{(-1)^2+(0)^2+(1)^2}}$

$=\frac{(0)(-1)+(-1)(0)+(1)(1)}{\sqrt{1+1}\sqrt{1+1}}$

$=\frac{0+0+1}{\sqrt{2}\times\sqrt{2}}$

$=\frac{1}{2}$

$\theta=\cos^{-1}\Big(\frac{1}{2}\Big)$

$\theta=\frac{\pi}{3}$

So, angle between the lines $=\frac{\pi}{3}$.

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