A plane P is perpendicular to the vector A=2 hati+3 hatj+6 hatk and contains the terminal point of the vector B= hati+5 hatj+3 hatk. The distance from the origin to the plane P, is

Question

A plane P is perpendicular to the vector A=2 hati+3 hatj+6 hatk and contains the terminal point of the vector B= hati+5 hatj+3 hatk. The distance from the origin to the plane P, is (A) 1 (B) 2 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

Plane P passes through (1,5,3) and normal to A =(2,3,6) Equation is 2(x-1)+3(y-5)+6(z-3)=0 ⇒ 2 x+3 y+6 z=35 ∴ Perpendicular from (0,0,0) to the plane =|(-35/7)|=5

Q. A plane $P$ is perpendicular to the vector $A = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ and contains the terminal point of the vector $B = \hat{i} + 5 \hat{j} + 3 \hat{k}$ . The distance from the origin to the plane $P$ , is

1

2

4

5

Plane P passes through $(1, 5, 3)$ and normal to $A = (2, 3, 6)$
Equation is $2 (x - 1) + 3 (y - 5) + 6 (z - 3) = 0 \Rightarrow 2 x + 3 y + 6 z = 35$
$∴$ Perpendicular from $(0, 0, 0)$ to the plane $= ∣ ∣ \frac{- 35}{7} ∣ ∣ = 5$

Q. A plane P is perpendicular to the vector A=2i^+3j^​+6k^ and contains the terminal point of the vector B=i^+5j^​+3k^. The distance from the origin to the plane P, is

1

2

4

5

Plane P passes through (1,5,3) and normal to A=(2,3,6) Equation is 2(x−1)+3(y−5)+6(z−3)=0⇒2x+3y+6z=35 ∴ Perpendicular from (0,0,0) to the plane =∣∣​7−35​∣∣​=5

Q. A plane $P$ is perpendicular to the vector $A = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ and contains the terminal point of the vector $B = \hat{i} + 5 \hat{j} + 3 \hat{k}$ . The distance from the origin to the plane $P$ , is

Plane P passes through $(1, 5, 3)$ and normal to $A = (2, 3, 6)$
Equation is $2 (x - 1) + 3 (y - 5) + 6 (z - 3) = 0 \Rightarrow 2 x + 3 y + 6 z = 35$
$∴$ Perpendicular from $(0, 0, 0)$ to the plane $= ∣ ∣ \frac{- 35}{7} ∣ ∣ = 5$