If the direction cosines of two lines satisfy | vedclass

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(B) Given equations are $l-2m+n=0$ $(1)$ and $lm+10mn-2nl=0$ $(2)$. From $(1)$,$l = 2m-n$. Substitute $l$ in $(2)$: $(2m-n)m + 10mn - 2n(2m-n) = 0$. $

Vedclass · Accepted Answer

$8/\sqrt{70}$

Vedclass · Answer

(B) Given equations are $l-2m+n=0$ $(1)$ and $lm+10mn-2nl=0$ $(2)$. From $(1)$,$l = 2m-n$. Substitute $l$ in $(2)$: $(2m-n)m + 10mn - 2n(2m-n) = 0$. $2m^2 - mn + 10mn - 4mn + 2n^2 = 0$. $2m^2 + 5mn + 2n^2 = 0$. Divide by $n^2$: $2(m/n)^2 + 5(m/n) + 2 = 0$. $(2m/n + 1)(m/n + 2) = 0$. Case $1$: $m/n = -1/2 \implies m = -k, n = 2k$. $l = 2(-k) - 2k = -4k$. Direction ratios $(l_1, m_1, n_1) = (-4, -1, 2)$. Case $2$: $m/n = -2 \implies m = -2k, n = k$. $l = 2(-2k) - k = -5k$. Direction ratios $(l_2, m_2, n_2) = (-5, -2, 1)$. $\cos 	heta = \frac{|l_1 l_2 + m_1 m_2 + n_1 n_2|}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}}$. $\cos 	heta = \frac{|(-4)(-5) + (-1)(-2) + (2)(1)|}{\sqrt{16+1+4} \sqrt{25+4+1}} = \frac{|20+2+2|}{\sqrt{21} \sqrt{30}} = \frac{24}{\sqrt{630}} = \frac{24}{3\sqrt{70}} = \frac{8}{\sqrt{70}}$.

If the direction cosines of two lines satisfy the equations $l-2m+n=0$ and $lm+10mn-2nl=0$,and $\theta$ is the angle between the lines,then $\cos \theta=$

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