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(D) Let the position vector of point $A$ be $\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$.The line $L$ passes through $A$ and is paralle

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$-15 \hat{i}-9 \hat{j}+19 \hat{k}$

Vedclass · Answer

(D) Let the position vector of point $A$ be $\vec{a} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$.The line $L$ passes through $A$ and is parallel to $\vec{v} = 2 \hat{i} + \hat{j} - 2 \hat{k}$.The equation of the line $L$ is $\vec{r} = \vec{a} + \lambda \vec{v} = (\alpha + 2\lambda) \hat{i} + (\beta + \lambda) \hat{j} + (\gamma - 2\lambda) \hat{k}$.Point $P$ with position vector $\vec{p} = -7 \hat{i} - 5 \hat{j} + 11 \hat{k}$ lies on $L$, so $\vec{p} = \vec{a} + \lambda \vec{v}$.Thus, $\vec{p} - \vec{a} = \lambda \vec{v}$, which implies $\overline{AP} = \lambda (2 \hat{i} + \hat{j} - 2 \hat{k})$.Given $|\overline{AP}| = 12$, we have $|\lambda| |2 \hat{i} + \hat{j} - 2 \hat{k}| = 12$.Since $|2 \hat{i} + \hat{j} - 2 \hat{k}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4+1+4} = 3$, we get $|\lambda| 	imes 3 = 12$, so $|\lambda| = 4$, which means $\lambda = \pm 4$.Since $\vec{a} = \vec{p} - \lambda \vec{v}$, for $\lambda = 4$:$\vec{a} = (-7 \hat{i} - 5 \hat{j} + 11 \hat{k}) - 4(2 \hat{i} + \hat{j} - 2 \hat{k}) = (-7-8) \hat{i} + (-5-4) \hat{j} + (11+8) \hat{k} = -15 \hat{i} - 9 \hat{j} + 19 \hat{k}$.For $\lambda = -4$:$\vec{a} = (-7 \hat{i} - 5 \hat{j} + 11 \hat{k}) + 4(2 \hat{i} + \hat{j} - 2 \hat{k}) = (-7+8) \hat{i} + (-5+4) \hat{j} + (11-8) \hat{k} = \hat{i} - \hat{j} + 3 \hat{k}$.Comparing with the options, $-15 \hat{i} - 9 \hat{j} + 19 \hat{k}$ is option $D$.

Let $L$ be a line passing through a point $A$ and parallel to the vector $2 \hat{i}+\hat{j}-2 \hat{k}$. Let $-7 \hat{i}-5 \hat{j}+11 \hat{k}$ be the position vector of a point $P$ on $L$ such that $|\overline{AP}|=12$. Then the position vector of $A$ can be

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