Let the plane P contain the line 2x+y-z-3=0=5x-3y+4z+9 and be parallel to the line x+22=3-…

Question

Let the plane $P$ contain the line $2x+y-z-3=0=5x-3y+4z+9$ and be parallel to the line $\dfrac{x+2}{2}=\dfrac{3-y}{-4}=\dfrac{z-7}{5}$. Then the distance of the point $A(8,-1,-19)$ from the plane $P$, measured parallel to the line $\dfrac{x}{-3}=\dfrac{y-5}{4}=\dfrac{2-z}{-12}$, is equal to:

Answer: 26

Accepted Answer

Family: $(5\lambda+2)x+(1-3\lambda)y+(4\lambda-1)z+9\lambda-3=0$. Parallel to the given line's direction $(2,4,5)$:
$$2(5\lambda+2)+4(1-3\lambda)+5(4\lambda-1)=0 \implies \lambda=-\frac16.$$
$$P:\ 7x+9y-10z-27=0.$$
Line through $A(8,-1,-19)$ with direction $(-3,4,12)$: $B=(8-3\lambda,-1+4\lambda,-19+12\lambda)$ on $P$:
$$7(8-3\lambda)+9(4\lambda-1)-10(12\lambda-19)=27 \implies \lambda=2.$$
$B=(2,7,5).$
$$AB=\sqrt{6^2+8^2+24^2}=\sqrt{36+64+576}=\sqrt{676}=26.$$

Solutions · 1