If $P$ is a point lying on the line passing through the point $A(\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}})$ and parallel to the vector $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ such that $|\mathbf{A P}|=18$, then a position vector of $P$ is

Question

If $P$ is a point lying on the line passing through the point $A(\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}})$ and parallel to the vector $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ such that $|\mathbf{A P}|=18$, then a position vector of $P$ is, $-13 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}$, $11 \hat{\mathbf{i}}+7 \mathbf{j}-15 \hat{\mathbf{k}}$, $13 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}$, $13 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-9 \hat{\mathbf{k}}$

MARKS App · Accepted Answer

According to the given information, the diagram is shown as below. Given, $\begin{aligned} \mathbf{A P} & =18 \ \mathbf{O A} & =\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned}$ Now, $\mathbf{A P}=18 \times \frac{2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{\sqrt{(2)^2+(l)^2+(-2)^2}}$ $\begin{aligned} & =18 \times \frac{2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{\sqrt{4+1+4}} \ & =18 \times \frac{2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{3} \ & =6(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}})=12 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-12 \hat{\mathbf{k}} \end{aligned}$ By triangle law, $\begin{aligned} \mathbf{O P} & =\mathbf{O A}+\mathbf{A P} \ & =\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}+12 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-12 \hat{\mathbf{k}} \ & =13 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-9 \hat{\mathbf{k}} \end{aligned}$

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