If the direction cosines of two lines satisfy | vedclass

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(A) Given equations are $2l+m-n=0$ $(1)$ and $l^2-2m^2+n^2=0$ $(2)$. From $(1)$, $n = 2l+m$. Substitute $n$ into $(2)$: $l^2 - 2m^2 + (2l+m)^2 = 0$. $

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

(A) Given equations are $2l+m-n=0$ $(1)$ and $l^2-2m^2+n^2=0$ $(2)$. From $(1)$, $n = 2l+m$. Substitute $n$ into $(2)$: $l^2 - 2m^2 + (2l+m)^2 = 0$. $l^2 - 2m^2 + 4l^2 + 4lm + m^2 = 0$. $5l^2 + 4lm - m^2 = 0$. Divide by $m^2$: $5(l/m)^2 + 4(l/m) - 1 = 0$. Let $x = l/m$, then $5x^2 + 4x - 1 = 0$. $(5x-1)(x+1) = 0$, so $x = 1/5$ or $x = -1$. Case $1$: $l/m = 1/5 \implies m = 5l$. Then $n = 2l + 5l = 7l$. Direction ratios are $(l, 5l, 7l)$ or $(1, 5, 7)$. Case $2$: $l/m = -1 \implies m = -l$. Then $n = 2l - l = l$. Direction ratios are $(l, -l, l)$ or $(1, -1, 1)$. Let the two lines have direction ratios $\vec{a} = (1, 5, 7)$ and $\vec{b} = (1, -1, 1)$. $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(1)(1) + (5)(-1) + (7)(1)|}{\sqrt{1^2+5^2+7^2} \sqrt{1^2+(-1)^2+1^2}} = \frac{|1-5+7|}{\sqrt{1+25+49} \sqrt{3}} = \frac{3}{\sqrt{75} \sqrt{3}} = \frac{3}{\sqrt{225}} = \frac{3}{15} = \frac{1}{5}$.

If the direction cosines of two lines satisfy the equations $2l+m-n=0$ and $l^2-2m^2+n^2=0$, and $\theta$ is the angle between the lines, then $\cos \theta=$

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