Let the line [(x)/(1)=(6-y)/(2)=(z+8)/(5)] intersect the lines [(x-5)/(4)=(y-7)/(3)=(z+2)/…

Question

Let the line $$\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$$ intersect the lines $$\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$$ and $$\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$$ at the points $$\mathrm{A}$$ and $$\mathrm{B}$$ respectively. Then ...

Accepted Answer

We have, $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the line

$\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$

$\begin{gathered} \frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}=\lambda ...........(i) \ \Rightarrow x=\lambda ,y=6-2\lambda ,z=5\lambda -8 \end{gathered}$

$\begin{gathered} \frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}=k...........(ii) \ \Rightarrow x=4k+5,y=3k+7,z=k-2 \end{gathered}$

$\begin{gathered} \frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}=\mu ..........(iii) \ \Rightarrow x=6\mu -3,y=3-3\mu ,z=\mu +6 \end{gathered}$

On solving Eqs. (i) and (ii), we get $\lambda =1,k=-1$

$	herefore $ Co-ordinate of $A$ is $(1,4,-3)$

On solving Eqs. (i) and (iii), we get $\lambda =3,\mu =1$

$	herefore $ Co-ordinate of $\beta $ is $(3,0,7)$

Co-ordinate of mid-point of $AB$ is $(\frac{1+3}{2},\frac{4+0}{2},\frac{-3+7}{2})$ or $(2,2,2)$

Perpendicular distance of mid-point of $AB$ from the plane $2x-2y+z=14$ is

$\frac{|2(2)-2(2)+2-14|}{\sqrt{{2}^{2}+(-2{)}^{2}+{1}^{2}}}=4$

Solutions · 1