Then 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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Aicharya · Accepted Answer

Any point on the line $\frac{x+3}{5}=\frac{y-1}{3}=\frac{z+4}{3}=\lambda$ can be taken as $(5 \lambda-3,2 \lambda+1,3 \lambda-4)$ for foot of perpendicular $\lambda=\frac{a\left(\alpha-x_1ight)+b\left(\beta-y_1ight)+c\left(\gamma-z_1ight)}{a^2+b^2+c^2}$ $\begin{aligned} & \Rightarrow \lambda=\frac{5(0+3)+2(2-1)+3(3+4)}{5^2+2^2+3^2} \ & \Rightarrow \lambda=\frac{38}{38}=1 \end{aligned}$ Hence foot of perpendicular is $(5 	imes 1-3,2 	imes 1+1,3 	imes 1-4) \equiv(2,3,-1)$

Then foot of the perpendicular from the point $(0,2,3)$ on the line…

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