Let O be the origin,and M and N be the points | vedclass

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(B) Let the lines be $L_1: \frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}=\lambda$ and $L_2: \frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}=\mu$.Any point $M$

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) Let the lines be $L_1: \frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}=\lambda$ and $L_2: \frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}=\mu$.Any point $M$ on $L_1$ is $(4\lambda+5, \lambda+4, 3\lambda+5)$ and any point $N$ on $L_2$ is $(12\mu-8, 5\mu-2, 9\mu-11)$.The vector $\overrightarrow{MN} = (12\mu-4\lambda-13, 5\mu-\lambda-6, 9\mu-3\lambda-16)$.The direction vectors are $\vec{b}_1 = (4, 1, 3)$ and $\vec{b}_2 = (12, 5, 9)$.The shortest distance vector $\overrightarrow{MN}$ must be perpendicular to both $\vec{b}_1$ and $\vec{b}_2$.$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 1 & 3 \ 12 & 5 & 9 \end{vmatrix} = -6\hat{i} + 0\hat{j} + 8\hat{k}$.Since $\overrightarrow{MN}$ is parallel to $\vec{b}_1 	imes \vec{b}_2$,we have $\frac{12\mu-4\lambda-13}{-6} = \frac{5\mu-\lambda-6}{0} = \frac{9\mu-3\lambda-16}{8}$.From the middle term,$5\mu-\lambda-6=0 \implies \lambda = 5\mu-6$.Substituting into the ratio $\frac{12\mu-4\lambda-13}{-6} = \frac{9\mu-3\lambda-16}{8}$:$8(12\mu-4(5\mu-6)-13) = -6(9\mu-3(5\mu-6)-16) \implies 8(-8\mu+11) = -6(-6\mu+2) \implies -64\mu+88 = 36\mu-12 \implies 100\mu = 100 \implies \mu=1$.Then $\lambda = 5(1)-6 = -1$.Thus,$M = (4(-1)+5, -1+4, 3(-1)+5) = (1, 3, 2)$ and $N = (12(1)-8, 5(1)-2, 9(1)-11) = (4, 3, -2)$.Finally,$\overrightarrow{OM} \cdot \overrightarrow{ON} = (1)(4) + (3)(3) + (2)(-2) = 4 + 9 - 4 = 9$.

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