Let 3 hat{i}+hat{j}-hat{k} be the position ve | vedclass

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(A) The position vector of point $B$ is $\vec{b} = 3 \hat{i} + \hat{j} - \hat{k}$.Since point $A$ lies on the line passing through $B$ and parallel to

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$-9 \hat{i}+7 \hat{j}-13 \hat{k}$

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(A) The position vector of point $B$ is $\vec{b} = 3 \hat{i} + \hat{j} - \hat{k}$.Since point $A$ lies on the line passing through $B$ and parallel to $\vec{v} = 2 \hat{i} - \hat{j} + 2 \hat{k}$, the vector $\overrightarrow{B A}$ can be written as $\overrightarrow{B A} = t \vec{v} = t(2 \hat{i} - \hat{j} + 2 \hat{k})$ for some scalar $t$.Given $|\overrightarrow{B A}| = 18$, we have $|t| \sqrt{2^2 + (-1)^2 + 2^2} = 18$.$|t| \sqrt{4 + 1 + 4} = 18 \Rightarrow |t| \sqrt{9} = 18 \Rightarrow 3|t| = 18 \Rightarrow |t| = 6$.Thus, $t = 6$ or $t = -6$.The position vector of $A$ is $\vec{a} = \vec{b} + \overrightarrow{B A} = (3 \hat{i} + \hat{j} - \hat{k}) + t(2 \hat{i} - \hat{j} + 2 \hat{k})$.For $t = 6$: $\vec{a} = (3 + 12) \hat{i} + (1 - 6) \hat{j} + (-1 + 12) \hat{k} = 15 \hat{i} - 5 \hat{j} + 11 \hat{k}$.For $t = -6$: $\vec{a} = (3 - 12) \hat{i} + (1 + 6) \hat{j} + (-1 - 12) \hat{k} = -9 \hat{i} + 7 \hat{j} - 13 \hat{k}$.Comparing with the given options, the correct position vector is $-9 \hat{i} + 7 \hat{j} - 13 \hat{k}$.

Let $3 \hat{i}+\hat{j}-\hat{k}$ be the position vector of a point $B$. Let $A$ be a point on the line which is passing through $B$ and parallel to the vector $2 \hat{i}-\hat{j}+2 \hat{k}$. If $|\overrightarrow{B A}|=18$, then the position vector of $A$ is

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