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Direction Cosines and Direction Ratios — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDirection Cosines and Direction Ratios4 Marks
Question
Find the angle between the lines whose direction cosines are given by the equations:
2l - m + 2n = 0 and mn + nl + lm = 0
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Answer

Given that,
2l - m + 2n = 0 .....(1)
mn + nl + lm = 0 .....(2)
From equation (1),
2l - m + 2n = 0
m = 2l + 2n  
Put the value of m in equation (2),
mn + nl + lm = 0
(2l + 2n) n + nl + l(2l + 2n) = 0
2ln + 2n2 +nl +2l2 + 2ln = 0
2l2 + 5ln + 2n2 = 0
2l2 + 4ln + ln + 2ln2 = 0
2l (l + 2n) + n(l + 2n) = 0
(1 + 2n) (2l = n) = 0
l + 2n = 0 or 2l + n = 0
l = -2n or  $\text{l}=-\frac{\text{n}}{2}$
Put the value of l = -2n in equation (1)
2l - m + 2n = 0
2 (-2n) - m + 2n = 0
-4n - m + 2n = 0
-2n - m = 0
-2n = m
m = -2n
Again, put the value of $\text{l}=-\frac{1}{2}$ in equation (1)
2l - m + 2n = 0
$2\Big(-\frac{1}{2}\text{n}\Big)-\text{m}+2\text{n}=0$
-n - m + 2n = 0
-m + n = 0
-m = -n
m = n
So, direction cosines of the lines are given by,
-2n, -2n, n or $-\frac{1}{2},\text{n},\text{n},\text{n}$
-2, -2, 1 or $-\frac{1}{2},1,1$
So, vectors parallel to these lines
$\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=-\frac{1}{2}\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ respectively.
Let, $\theta$ be the angle between the $\vec{\text{a}}$ and $\vec{\text{b}},$
 $\cos\theta=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{\big(-2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)\times\Big(-\frac{1}{2}\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)}{\sqrt{(-2)^2+(-2)^2+(1)^2}\sqrt{\Big(-\frac{1}{2}^2\Big)+(1)^2+(1)^2}}$
$=\frac{(-2)\big(-\frac{1}{2}\big)+(-2)(1)+(1)(1)}{\sqrt{4+4+1}\sqrt{\frac{1}{4}1+1}}$
$=\frac{1-2+1}{\sqrt{9}\sqrt{\frac{9}{4}}}$
$=\frac{0}{3\times\frac{3}{2}}$
$\cos\theta=0$
$\theta=\cos^{-1}(0)$
 $\theta=\frac{\pi}{2}$
So, angle between the lines $=\frac{\pi}{2}$.

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