Question

The distance of the point `(1,3,-7)` from the plane passing through the point `(1,-1,-1),` having normal perpendicular to both the lines `(x-1)/1=(y+2)/(-2)=(z-4)/3a n d(x-2)/2=(y+1)/(-1)=(z+7)/(-1)i s:` `5/(sqrt(83))` (2) `(10)/(sqrt(74))` (3) `(20)/(sqrt(74))` (4) `(10)/(sqrt(83))`

Answer 1

Let `vecn` is the vector normal to the plane.
Then, as `vecn` is perpendicular to given two lines.
`:. vec n = |[hati,hatj,hatk],[1,-2,3],[2,-1,-1]|`
`=> vec n = 5hati + 7hat j+3hat k`
Now, as `vec n` is normal to the plane passing through point `(1,-1,1)`,
`:. (xhati+yhatj+zhatk - hati+hatj+hatk).(5hati + 7hat j+3hat k) = 0`
`=>5x-5+7y+7+3z+3 = 0`
`=>5x+7y+3z+5 = 0`
This is the equation of the plane and we have to find the distance of point `(1,3,-7)` from this plane.
So, required distance `= |5*1+7*3+3*(-7)+5|/(sqrt(5^2+7^2+3^2)) = 10/sqrt83.`