A line l passing through the origin is perpen | vedclass

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(C) The direction ratios of $l_{1}$ are $\vec{v}_{1} = (1, 2, 2)$ and for $l_{2}$ are $\vec{v}_{2} = (2, 2, 1)$.Since line $l$ is perpendicular to bot

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$44$

Vedclass · Answer

(C) The direction ratios of $l_{1}$ are $\vec{v}_{1} = (1, 2, 2)$ and for $l_{2}$ are $\vec{v}_{2} = (2, 2, 1)$.Since line $l$ is perpendicular to both $l_{1}$ and $l_{2}$,its direction vector is $\vec{v} = \vec{v}_{1} 	imes \vec{v}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 2 \ 2 & 2 & 1 \end{vmatrix} = \hat{i}(2-4) - \hat{j}(1-4) + \hat{k}(2-4) = -2\hat{i} + 3\hat{j} - 2\hat{k}$.Since $l$ passes through the origin,its equation is $\vec{r} = \mu(-2\hat{i} + 3\hat{j} - 2\hat{k})$.For the intersection of $l$ and $l_{1}$:$3+t = -2\mu$,$-1+2t = 3\mu$,$4+2t = -2\mu$.Solving these,we get $t = -1$ and $\mu = -1$. The point of intersection $P$ is $(2, -3, 2)$.Let $Q$ be a point on $l_{2}$,$Q = (3+2s, 3+2s, 2+s)$.Given $PQ = \sqrt{17}$,so $PQ^{2} = 17$.$(3+2s-2)^{2} + (3+2s+3)^{2} + (2+s-2)^{2} = 17$.$(2s+1)^{2} + (2s+6)^{2} + s^{2} = 17$.$4s^{2} + 4s + 1 + 4s^{2} + 24s + 36 + s^{2} = 17$.$9s^{2} + 28s + 20 = 0$.$(9s+10)(s+2) = 0$,so $s = -2$ or $s = -10/9$.For $s = -10/9$,$Q = (3-20/9, 3-20/9, 2-10/9) = (7/9, 7/9, 8/9)$,which is in the first octant.Thus,$a=7/9, b=7/9, c=8/9$.$18(a+b+c) = 18(7/9 + 7/9 + 8/9) = 18(22/9) = 44$.

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