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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 $$2 x-2 y+z=14$$ is :

JEE · Math · previous-year question

  1. A.3
  2. B.$$\frac{10}{3}$$
  3. C.4correct
  4. D.$$\frac{11}{3}$$

Answer

C. 4

Explanation

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 $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$ $$ \begin{array}{ll} & \text { Now, } \frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}=\lambda ...........(i)\\\\ & \Rightarrow x=\lambda, y=6-2 \lambda, z=5 \lambda-8 \end{array} $$ $$ \begin{array}{ll} &\text { Also, } \frac{x-5}{4} =\frac{y-7}{3}=\frac{z+2}{1}=k ...........(ii)\\\\ &\Rightarrow x =4 k+5, y=3 k+7, z=k-2 \end{array} $$ $$ \begin{array}{rlrl} & \frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}=\mu ..........(iii)\\\\ &\Rightarrow x = 6 \mu-3, y=3-3 \mu, z=\mu+6 \end{array} $$ On solving Eqs. (i) and (ii), we get $\lambda=1, k=-1$ $\therefore$ Co-ordinate of $A$ is $(1,4,-3)$ On solving Eqs. (i) and (iii), we get $\lambda=3, \mu=1$ $\therefore$ Co-ordinate of $\beta$ is $(3,0,7)$ Co-ordinate of mid-point of $A B$ is $\left(\frac{1+3}{2}, \frac{4+0}{2}, \frac{-3+7}{2}\right)$ or $(2,2,2)$ Perpendicular distance of mid-point of $A B$ from the plane $2 x-2 y+z=14$ is $$ \frac{|2(2)-2(2)+2-14|}{\sqrt{2^2+(-2)^2+1^2}}=4 $$

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