If theta is the angle between the lines whose | vedclass

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(A) Given the equations: $6mn - 2nl + 5lm = 0$  $(1)$ $3l + m + 5n = 0 \implies m = -(3l + 5n)$  $(2)$ Substitute $(2)$ into $(1)$: $6n(-(3l + 5n)) -

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$\frac{\sqrt{35}}{6}$

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(A) Given the equations: $6mn - 2nl + 5lm = 0$  $(1)$ $3l + m + 5n = 0 \implies m = -(3l + 5n)$  $(2)$ Substitute $(2)$ into $(1)$: $6n(-(3l + 5n)) - 2nl + 5l(-(3l + 5n)) = 0$ $-18ln - 30n^2 - 2nl - 15l^2 - 25ln = 0$ $-15l^2 - 45ln - 30n^2 = 0$ Divide by $-15$: $l^2 + 3ln + 2n^2 = 0$ $(l + n)(l + 2n) = 0$ Case $1$: $l = -n$. Then $m = -(3(-n) + 5n) = -2n$. Direction ratios are $(-n, -2n, n)$,i.e.,$(1, 2, -1)$. Case $2$: $l = -2n$. Then $m = -(3(-2n) + 5n) = n$. Direction ratios are $(-2n, n, n)$,i.e.,$(-2, 1, 1)$. Let $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = -2\hat{i} + \hat{j} + \hat{k}$. $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|-2 + 2 - 1|}{\sqrt{1+4+1} \sqrt{4+1+1}} = \frac{1}{\sqrt{6} \sqrt{6}} = \frac{1}{6}$. Since $\sin^2 	heta = 1 - \cos^2 	heta = 1 - (\frac{1}{6})^2 = 1 - \frac{1}{36} = \frac{35}{36}$. Therefore,$\sin 	heta = \frac{\sqrt{35}}{6}$.

If $\theta$ is the angle between the lines whose direction cosines $(l, m, n)$ satisfy the equations $6mn - 2nl + 5lm = 0$ and $3l + m + 5n = 0$,then $\sin \theta = $

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