The distance, of the point (7,-2,11) from the line (x-6)/(1)=(y-4)/(0)=(z-8)/(3) along the…

Question

The distance, of the point $(7,-2,11)$ from the line $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is :

Accepted Answer

$$\mathrm{B}=(2 \lambda+7,-3 \lambda-2,6 \lambda+11)$$

Point B lies on $$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$$

$$\frac{2 \lambda+7-6}{1}=\frac{-3 \lambda-2-4}{0}=\frac{6 \lambda+11-8}{3}$$

$$-3 \lambda-6=0$$

$$\lambda=-2$$

$$\mathrm{B} \Rightarrow(3,4,-1)$$

$$\mathrm{AB}=\sqrt{(7-3)^2+(4+2)^2+(11+1)^2}$$

$$=\sqrt{16+36+144}$$

$$=\sqrt{196}=14$$

Solutions · 1