The foot of the perpendicular of the point 3,-1,11to the line x2=y−23=z−34 is

Equation of the given line is x2=y−23=z−34 and the given point is A3,-1,11General point on the above line is B2k,3k+2, 4k+3Direction ratios of AB are 2k-3,3k+3,4k-8If B be the foot of the perpendicular of the point A3,-1,11 on the line x2=y−23=z−34then AB is perpendicular to the line Direction ratios of AB are 2k-3,3k+3,4k-8Hence, 22k-3+33k+3+44k-8=04k-6+9k+9+16k-32=029k-29=0k=1Therefore, the foot of the perpendicular of the point A3,-1,11 on the line x2=y−23=z−34 is B21,31+2, 41+3=2,5,7

The foot of the perpendicular of the point 3,-1,11to the line x2=y−23=z−34 is

Equation of the given line is x2=y−23=z−34 and the given point is A3,-1,11General point on the above line is B2k,3k+2, 4k+3Direction ratios of AB are 2k-3,3k+3,4k-8If B be the foot of the perpendicular of the point A3,-1,11 on the line x2=y−23=z−34then AB is perpendicular to the line Direction ratios of AB are 2k-3,3k+3,4k-8Hence, 22k-3+33k+3+44k-8=04k-6+9k+9+16k-32=029k-29=0k=1Therefore, the foot of the perpendicular of the point A3,-1,11 on the line x2=y−23=z−34 is B21,31+2, 41+3=2,5,7

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The foot of the perpendicular of the point $(3, - 1, 11)$ to the line $\frac{x}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ is

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$(0, 2, 3)$

$(2, 3, 4)$

$(3, - 1, 11)$

$(2, 5, 7)$

answer is A.

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Detailed Solution

Equation of the given line is $\frac{x}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and the given point is $A (3, - 1, 11)$

General point on the above line is $B (2 k, 3 k + 2, 4 k + 3)$

Direction ratios of $A B$ are $(2 k - 3, 3 k + 3, 4 k - 8)$

If $B$ be the foot of the perpendicular of the point $A (3, - 1, 11)$ on the line $\frac{x}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ then $A B$ is perpendicular to the line

Direction ratios of $A B$ are $(2 k - 3, 3 k + 3, 4 k - 8)$

Hence,

$\begin{array}{rcl} 2 (2 k - 3) + 3 (3 k + 3) + 4 (4 k - 8) & = & 0 \\ 4 k - 6 + 9 k + 9 + 16 k - 32 & = & 0 \\ 29 k - 29 & = & 0 \\ k & = & 1 \end{array}$

Therefore, the foot of the perpendicular of the point $A (3, - 1, 11)$ on the line $\frac{x}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ is $B (2 (1), 3 (1) + 2, 4 (1) + 3) = (2, 5, 7)$