Let P be the plane, passing through the point (1,-1,-5) and perpendicular to the line joining the points (4,1,-3) and (2,4,3). Then the distance of P from the point (3,-2,2) is

Question

Let P be the plane, passing through the point (1,-1,-5) and perpendicular to the line joining the points (4,1,-3) and (2,4,3). Then the distance of P from the point (3,-2,2) is (A) 6 (B) 4 (C) 5 (D) 7

Tardigrade Admin · Accepted Answer

Equation of Plane: 2( x -1)-3( y +1)-6( z +5)=0 Or 2 x-3 y-6 z=35 ⇒ Required distance = (|2(3)-3(-2)-6(2)-35|/√4+9+36) =5

Q. Let $P$ be the plane, passing through the point $(1, - 1, - 5)$ and perpendicular to the line joining the points $(4, 1, - 3)$ and $(2, 4, 3)$ . Then the distance of $P$ from the point $(3, - 2, 2)$ is

6

4

5

7

Equation of Plane :
$2 (x - 1) - 3 (y + 1) - 6 (z + 5) = 0$
Or $2 x - 3 y - 6 z = 35$
$\Rightarrow$ Required distance $=$
$\frac{∣2 ( 3 ) - 3 ( - 2 ) - 6 ( 2 ) - 35∣}{4 + 9 + 36}$
$= 5$

Q. Let P be the plane, passing through the point (1,−1,−5) and perpendicular to the line joining the points (4,1,−3) and (2,4,3). Then the distance of P from the point (3,−2,2) is

6

4

5

7

Equation of Plane : 2(x−1)−3(y+1)−6(z+5)=0 Or 2x−3y−6z=35 ⇒ Required distance = 4+9+36​∣2(3)−3(−2)−6(2)−35∣​ =5

Q. Let $P$ be the plane, passing through the point $(1, - 1, - 5)$ and perpendicular to the line joining the points $(4, 1, - 3)$ and $(2, 4, 3)$ . Then the distance of $P$ from the point $(3, - 2, 2)$ is

Equation of Plane :
$2 (x - 1) - 3 (y + 1) - 6 (z + 5) = 0$
Or $2 x - 3 y - 6 z = 35$
$\Rightarrow$ Required distance $=$
$\frac{∣2 ( 3 ) - 3 ( - 2 ) - 6 ( 2 ) - 35∣}{4 + 9 + 36}$
$= 5$