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 the distance of the mid-point of the line segment \[\mathrm{AB}\] from the plane 2x−2y+z=14 is :
Solutions · 1
Official solution
We have, 1x=26−y=5z+8 intersect the line
4x−5=3y−7=1z+2 and 6x+3=33−y=1z−6
\[
\[\begin{gathered} \text { Now, } \frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}=\lambda ...........(i) \\ \Rightarrow x=\lambda, y=6-2 \lambda, z=5 \lambda-8 \end{gathered}\]
\]
\[
\[\begin{gathered} \text { Also, } \frac{x-5}{4} =\frac{y-7}{3}=\frac{z+2}{1}=k ...........(ii) \\ \Rightarrow x =4 k+5, y=3 k+7, z=k-2 \end{gathered}\]
\]
\[
\[\begin{gathered} \frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}=\mu ..........(iii) \\ \Rightarrow x = 6 \mu-3, y=3-3 \mu, z=\mu+6 \end{gathered}\]
\]
On solving Eqs. (i) and (ii), we get λ=1,k=−1∴ Co-ordinate of A is (1,4,−3)
On solving Eqs. (i) and (iii), we get λ=3,μ=1∴ Co-ordinate of β is (3,0,7)
Co-ordinate of mid-point of AB is (21+3,24+0,2−3+7) or (2,2,2)
Perpendicular distance of mid-point of AB from the plane 2x−2y+z=14 is
\[
\frac{|2(2)-2(2)+2-14|}{\sqrt{2^2+(-2)^2+1^2}}=4
\]