The angle between the straight lines,whose di | vedclass

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(A) Given equations are $n = 2(l + m)$ and $mn + nl + lm = 0$. Substituting $n = 2(l + m)$ into the second equation: $m(2l + 2m) + 2l(l + m) + lm = 0$

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$\frac{\pi}{2}$

Vedclass · Answer

(A) Given equations are $n = 2(l + m)$ and $mn + nl + lm = 0$. Substituting $n = 2(l + m)$ into the second equation: $m(2l + 2m) + 2l(l + m) + lm = 0$ $2lm + 2m^2 + 2l^2 + 2lm + lm = 0$ $2l^2 + 5lm + 2m^2 = 0$ Dividing by $m^2$,we get $2t^2 + 5t + 2 = 0$ where $t = \frac{l}{m}$. Solving the quadratic equation: $(2t + 1)(t + 2) = 0$,so $t = -\frac{1}{2}$ or $t = -2$. Case $1$: If $\frac{l}{m} = -2$,then $l = -2m$. Substituting into $n = 2(l + m)$,we get $n = 2(-2m + m) = -2m$. The direction ratios are $(-2m, m, -2m)$,which simplifies to $(-2, 1, -2)$. Case $2$: If $\frac{l}{m} = -\frac{1}{2}$,then $m = -2l$. Substituting into $n = 2(l + m)$,we get $n = 2(l - 2l) = -2l$. The direction ratios are $(l, -2l, -2l)$,which simplifies to $(1, -2, -2)$. Let the direction ratios be $\vec{a} = (-2, 1, -2)$ and $\vec{b} = (1, -2, -2)$. The cosine of the angle $	heta$ is given by $\cos 	heta = \frac{|a_1b_1 + a_2b_2 + a_3b_3|}{\sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}}$. $\cos 	heta = \frac{|(-2)(1) + (1)(-2) + (-2)(-2)|}{\sqrt{4+1+4} \sqrt{1+4+4}} = \frac{|-2 - 2 + 4|}{3 	imes 3} = 0$. Therefore,$	heta = \frac{\pi}{2}$.

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

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