Let L be the line parallel to the vector sqrt | vedclass

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(A) The equation of the line $L$ passing through point $A(\hat{i}+2 \hat{j}-3 \hat{k})$ and parallel to vector $\vec{v} = \sqrt{2} \hat{i}-5 \hat{j}+3

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$(1-3 \sqrt{2}) \hat{i}+17 \hat{j}-12 \hat{k}$

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(A) The equation of the line $L$ passing through point $A(\hat{i}+2 \hat{j}-3 \hat{k})$ and parallel to vector $\vec{v} = \sqrt{2} \hat{i}-5 \hat{j}+3 \hat{k}$ is given by $\vec{r} = (\hat{i}+2 \hat{j}-3 \hat{k}) + \lambda(\sqrt{2} \hat{i}-5 \hat{j}+3 \hat{k})$.Any point $P$ on the line $L$ can be represented as $P = (\sqrt{2} \lambda+1, -5 \lambda+2, 3 \lambda-3)$.The distance $AP$ is given as $18$ units.$AP^2 = (\sqrt{2} \lambda+1-1)^2 + (-5 \lambda+2-2)^2 + (3 \lambda-3+3)^2 = 18^2$.$2 \lambda^2 + 25 \lambda^2 + 9 \lambda^2 = 324$.$36 \lambda^2 = 324$.$\lambda^2 = 9$, which gives $\lambda = \pm 3$.For $\lambda = 3$, $P = (3\sqrt{2}+1)\hat{i} - 13\hat{j} + 6\hat{k}$.For $\lambda = -3$, $P = (-3\sqrt{2}+1)\hat{i} + 15+2\hat{j} - 9-3\hat{k} = (1-3\sqrt{2})\hat{i} + 17\hat{j} - 12\hat{k}$.Comparing with the options, the correct position vector is $(1-3 \sqrt{2}) \hat{i}+17 \hat{j}-12 \hat{k}$.

Let $L$ be the line parallel to the vector $\sqrt{2} \hat{i}-5 \hat{j}+3 \hat{k}$ and passing through the point $A$ given by $\hat{i}+2 \hat{j}-3 \hat{k}$. If the distance between $A$ and a point $P$ on the line $L$ is $18$ units, then the position vector of such a point $P$ is

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