The square of the distance of the point left( | vedclass

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(C) Let the given point be $P\left(\frac{15}{7}, \frac{32}{7}, 7\right)$ and the line $L$ be $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} = k$.Any point

Vedclass · Accepted Answer

$66$

Vedclass · Answer

(C) Let the given point be $P\left(\frac{15}{7}, \frac{32}{7}, 7\right)$ and the line $L$ be $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} = k$.Any point $Q$ on the line $L$ is given by $Q(3k-1, 5k-3, 7k-5)$.The vector $\vec{PQ}$ is given by $\vec{PQ} = \left(3k-1-\frac{15}{7}\right)\hat{i} + \left(5k-3-\frac{32}{7}\right)\hat{j} + (7k-5-7)\hat{k} = \left(3k-\frac{22}{7}\right)\hat{i} + \left(5k-\frac{53}{7}\right)\hat{j} + (7k-12)\hat{k}$.Since the line $PQ$ is parallel to the vector $\vec{v} = \hat{i} + 4\hat{j} + 7\hat{k}$,the components of $\vec{PQ}$ must be proportional to the components of $\vec{v}$:$\frac{3k-\frac{22}{7}}{1} = \frac{5k-\frac{53}{7}}{4} = \frac{7k-12}{7} = \lambda$.From $\frac{7k-12}{7} = \lambda$,we have $7k-12 = 7\lambda \Rightarrow k = \lambda + \frac{12}{7}$.Substituting $k$ into the first equality: $3(\lambda + \frac{12}{7}) - \frac{22}{7} = \lambda \Rightarrow 3\lambda + \frac{36-22}{7} = \lambda \Rightarrow 2\lambda = -2 \Rightarrow \lambda = -1$.Thus,$k = -1 + \frac{12}{7} = \frac{5}{7}$.The point $Q$ is $Q(3(\frac{5}{7})-1, 5(\frac{5}{7})-3, 7(\frac{5}{7})-5) = Q(\frac{8}{7}, \frac{4}{7}, 0)$.The distance $PQ = \sqrt{(\frac{15}{7}-\frac{8}{7})^2 + (\frac{32}{7}-\frac{4}{7})^2 + (7-0)^2} = \sqrt{1^2 + 4^2 + 7^2} = \sqrt{1+16+49} = \sqrt{66}$.Therefore,the square of the distance is $(PQ)^2 = 66$.

The square of the distance of the point $\left(\frac{15}{7}, \frac{32}{7}, 7\right)$ from the line $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ in the direction of the vector $\hat{i}+4 \hat{j}+7 \hat{k}$ is :

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