Find the foot of the perpendicular from the p | vedclass

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Question

(B) Let the given line be $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda$. Any point on this line can be represented as $M(5\lambda-3, 2\lambda+1,

Vedclass · Accepted Answer

$(2,3,-1)$

Vedclass · Answer

(B) Let the given line be $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda$. Any point on this line can be represented as $M(5\lambda-3, 2\lambda+1, 3\lambda-4)$. Let $P$ be the point $(0,2,3)$. The vector $\vec{PM}$ is given by $(5\lambda-3-0, 2\lambda+1-2, 3\lambda-4-3) = (5\lambda-3, 2\lambda-1, 3\lambda-7)$. Since $PM$ is perpendicular to the line,the dot product of $\vec{PM}$ and the direction vector of the line $\vec{v} = (5, 2, 3)$ must be zero. $5(5\lambda-3) + 2(2\lambda-1) + 3(3\lambda-7) = 0$ $25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0$ $38\lambda - 38 = 0$ $\lambda = 1$. Substituting $\lambda = 1$ in the coordinates of $M$,we get: $x = 5(1)-3 = 2$ $y = 2(1)+1 = 3$ $z = 3(1)-4 = -1$ Thus,the foot of the perpendicular is $(2,3,-1)$.

Find the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$.

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