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An angle between the lines whose direction cosines are given by the equations, $l+3m+5n=0$ and $5lm-2mn+6nl=0$ , is :

Option 1)
$\cos ^{-1}\left ( \frac{1}{3} \right )$

Option 2)
$\cos ^{-1}\left ( \frac{1}{4} \right )$

Option 3)
$\cos ^{-1}\left ( \frac{1}{6} \right )$

Option 4)
$\cos ^{-1}\left ( \frac{1}{8} \right )$

Answers (1)

best_answer

As we learned,

Direction Cosines -

If $\alpha, \beta, \gamma$ are the angles which a vector makes with positive X-axis,Y-axis and Z-axis respectively then $\cos \alpha, \cos \beta, \cos \gamma$ are known as diresction cosines, generally denoted by (l,m,n).

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

$l^{2}+m^{2}+n^{2}= 1$

$\cos^{2} \alpha+ \cos^{2} \beta+\cos^{2} \gamma= 1$

- wherein

$l+3m+5n=0\: \Rightarrow \: l=-\left ( 3m+5n \right )\cdots \cdots \left ( i \right )$

$5lm-2mn+6nl=0\cdots \cdots \left ( ii \right )$

Substitute (i) in (ii)

$-\left ( 5m+6n \right )\left ( 3m+5n \right )-2mn=0$

$-15m^{2}-30n^{2}-43mn-2mn=0$

$m^{2}+2n^{2}+3mn=0$

$m^{2}+2mn+mn+2n^{2}=0$

$m\left ( m+2n \right )+n\left ( m+2n \right )=0$

$m=-n$ and $m=-2n$

m=-n m=-2n

l+5n-3n=0 l=n

l=-2n $l^{2}+m^{2}+n^{2}=1$

$l^{2}+m^{2}+n^{2}=1$ $6n^{2}=1\: \Rightarrow \: n=\pm \frac{1}{\sqrt{6}}$

$6n^{2}=1\: \Rightarrow \: n=\pm \frac{1}{\sqrt{6}}$

DCs are $+\frac{2}{\sqrt{6}}\hat{i}+\frac{1}{\sqrt{6}}\hat{j}-\frac{1}{\sqrt{6}}\hat{k}\: ;\: \frac{1}{\sqrt{6}}\hat{i}-\frac{2}{\sqrt{6}}\hat{j}+\frac{1}{\sqrt{6}}\hat{k}$

Angle $\Rightarrow \: \cos \theta =\frac{2}{6}-\frac{2}{6}-\frac{1}{6}=\frac{-1}{6}\theta =\cos ^{-1}\left ( \frac{1}{6} \right )$

Option 1)

$\cos ^{-1}\left ( \frac{1}{3} \right )$

Option 2)

$\cos ^{-1}\left ( \frac{1}{4} \right )$

Option 3)

$\cos ^{-1}\left ( \frac{1}{6} \right )$

Option 4)

$\cos ^{-1}\left ( \frac{1}{8} \right )$

Posted by

Himanshu

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