Let d be the distance of the point of intersection of the lines (x+6)/(3)=(y)/(2)=(z+1)/(1…

Question

Let $d$ be the distance of the point of intersection of the lines $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$ and $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$ from the point $(7,8,9)$. Then ${d}^{2}+6$ is equal to :

Accepted Answer

$$\begin{aligned}
& P_1:(3 k-6,2 k, k-1) \
& P_2(4 \alpha+7,3 \alpha+9,2 \alpha+4) \
& P_1 \equiv P_2 \
& 3 k-6=4 \alpha+7 \Rightarrow 3 k-4 \alpha=13 \
& 2 k=3 \alpha+9 \Rightarrow 2 k-3 \alpha=9 \
& \therefore k=3, \alpha=-1 \
& \therefore P_1:(3,6,2)
\end{aligned}$$

Distance of $(3,6,2)$ and $(7,8,9)$

$$\begin{aligned}
& =\sqrt{16+4+49}=\sqrt{69}=d \
& d^2+6=69+6=75
\end{aligned}$$

Solutions · 1