The foot of the perpendicular from (0,2,3) to | vedclass

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(C) Let $N$ be the foot of the perpendicular from the point $P(0,2,3)$ to the given line. The equation of the line is $\frac{x+3}{5}=\frac{y-1}{2}=\fr

Vedclass · Accepted Answer

$(2,3,-1)$

Vedclass · Answer

(C) Let $N$ be the foot of the perpendicular from the point $P(0,2,3)$ to the given line. The equation of the line is $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=r$. Any general point $N$ on the line is given by $(5r-3, 2r+1, 3r-4)$. The direction ratios of the line $PN$ are $(5r-3-0, 2r+1-2, 3r-4-3)$,which simplifies to $(5r-3, 2r-1, 3r-7)$. Since $PN$ is perpendicular to the given line with direction ratios $(5, 2, 3)$,the dot product of their direction ratios must be zero: $5(5r-3) + 2(2r-1) + 3(3r-7) = 0$. $25r - 15 + 4r - 2 + 9r - 21 = 0$. $38r - 38 = 0$. $r = 1$. Substituting $r=1$ into the coordinates of $N$: $N = (5(1)-3, 2(1)+1, 3(1)-4) = (2, 3, -1)$.

The foot of the perpendicular from $(0,2,3)$ to the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ is

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