The square of the distance of the image of th | vedclass

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(C) Let the given line be $L: \frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4} = \lambda$. Any point $M$ on the line is given by $M(3\lambda+1, 2\lambda, 4

Vedclass · Accepted Answer

$62$

Vedclass · Answer

(C) Let the given line be $L: \frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4} = \lambda$. Any point $M$ on the line is given by $M(3\lambda+1, 2\lambda, 4\lambda+2)$.Since $AM$ is perpendicular to the line $L$,the direction vector $\vec{AM} = (3\lambda+1-6, 2\lambda-1, 4\lambda+2-5) = (3\lambda-5, 2\lambda-1, 4\lambda-3)$ must be perpendicular to the direction vector of the line $\vec{b} = (3, 2, 4)$.Thus,$\vec{AM} \cdot \vec{b} = 0 \implies 3(3\lambda-5) + 2(2\lambda-1) + 4(4\lambda-3) = 0$.$9\lambda - 15 + 4\lambda - 2 + 16\lambda - 12 = 0 \implies 29\lambda - 29 = 0 \implies \lambda = 1$.Substituting $\lambda = 1$,we get $M(4, 2, 6)$.Let $I(x, y, z)$ be the image of $A(6, 1, 5)$ in the line. Since $M$ is the midpoint of $AI$,we have $\frac{x+6}{2} = 4, \frac{y+1}{2} = 2, \frac{z+5}{2} = 6$.$x+6 = 8 \implies x = 2$; $y+1 = 4 \implies y = 3$; $z+5 = 12 \implies z = 7$.So,the image point is $I(2, 3, 7)$.The square of the distance of $I(2, 3, 7)$ from the origin $(0, 0, 0)$ is $2^2 + 3^2 + 7^2 = 4 + 9 + 49 = 62$.

The square of the distance of the image of the point $A(6, 1, 5)$ in the line $\frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4}$ from the origin is:

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