Let O be the origin,A and B be two points wit | vedclass

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(A) Given position vectors are $\vec{a} = -3 \hat{i}-3 \hat{j}+4 \hat{k}$ and $\vec{b} = 4 \hat{i}-4 \hat{j}-3 \hat{k}$.Since $L$ divides $OA$ in the

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$\frac{19}{5}$

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(A) Given position vectors are $\vec{a} = -3 \hat{i}-3 \hat{j}+4 \hat{k}$ and $\vec{b} = 4 \hat{i}-4 \hat{j}-3 \hat{k}$.Since $L$ divides $OA$ in the ratio $2:3$,the position vector of $L$ is $\vec{OL} = \frac{2}{5} \vec{a} = \frac{2}{5}(-3 \hat{i}-3 \hat{j}+4 \hat{k})$.Since $M$ divides $OB$ in the ratio $3:2$,the position vector of $M$ is $\vec{OM} = \frac{3}{5} \vec{b} = \frac{3}{5}(4 \hat{i}-4 \hat{j}-3 \hat{k})$.Since $LP \parallel OB$ and $PM \parallel OA$,$OMPL$ is a parallelogram.Therefore,the position vector of $P$ is $\vec{OP} = \vec{OL} + \vec{OM}$.$\vec{OP} = \frac{2}{5}(-3 \hat{i}-3 \hat{j}+4 \hat{k}) + \frac{3}{5}(4 \hat{i}-4 \hat{j}-3 \hat{k})$$\vec{OP} = \frac{1}{5}(-6 \hat{i}-6 \hat{j}+8 \hat{k} + 12 \hat{i}-12 \hat{j}-9 \hat{k})$$\vec{OP} = \frac{1}{5}(6 \hat{i}-18 \hat{j}-\hat{k})$.The distance from $O$ to $P$ is $|\vec{OP}| = \frac{1}{5} \sqrt{6^2 + (-18)^2 + (-1)^2} = \frac{1}{5} \sqrt{36 + 324 + 1} = \frac{\sqrt{361}}{5} = \frac{19}{5}$.

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