1. Tardigrade
  2. Question
  3. Mathematics
  4. Consider the lines L 1: ( x -1/2)=( y -3/2)=( z -2/-1) and L 2: ( x -2/1)=( y -6/-1)=( z +2/3). The distance of the point P (√10, √10, √10) from the plane passing through origin and whose normal is perpendicular to both the lines L 1 and L 2, is

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Q. Consider the lines $L _1: \frac{ x -1}{2}=\frac{ y -3}{2}=\frac{ z -2}{-1}$ and $L _2: \frac{ x -2}{1}=\frac{ y -6}{-1}=\frac{ z +2}{3}$. The distance of the point $P (\sqrt{10}, \sqrt{10}, \sqrt{10})$ from the plane passing through origin and whose normal is perpendicular to both the lines $L _1$ and $L _2$, is

Vector Algebra

Solution:

The normal vector of plane is parallel to vector $=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 2 & -1 \\ 1 & -1 & 3\end{vmatrix}=5 \hat{ i }-7 \hat{ j }-4 \hat{ k }$
$\therefore$ Equation of plane is $5 x-7 y-4 z=0$ .....(1)
So, distance of plane in equation (1) from $P (\sqrt{10}, \sqrt{10}, \sqrt{10})$
$=\frac{|5(\sqrt{10})-7(\sqrt{10})-4(\sqrt{10})|}{\sqrt{(5)^2+(-7)^2+(-4)^2}}=\frac{6 \sqrt{10}}{3 \sqrt{10}}=2 .$