If O is the origin and the coordinates of P is $(1, 2, - 3)$ , then find the equation of the plane passing through P and perpendicular to OP.

x - 2 y - 3 z = - 15

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x + 2 y - 3 z = 14

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x - 2 y + 3 z = 15

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x - 2 y - 3 z = 15

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $x + 2 y - 3 z = 14$
Given:

Origin $O (0, 0, 0)$ and point $P (1, 2, - 3)$

To find: Equation of the plane passing through $P (1, 2, - 3) = (x_{1}, y_{1}, z_{1})$ .

The direction ratios of normal OP to the plane are $1 - 0, 2 - 0, - 3 - 0$ .

$(1, 2, - 3) = (a, b, c)$

The equation of the required plane is $a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$ .

$1 (x - 1) + 2 (y - 2) - 3 (z + 3)$

$x - 1 + 2 y - 4 - 3 z - 9 = 0$

$x + 2 y - 3 z - 14 = 0$

Thus, the equation of the plane is $x + 2 y - 3 z = 14$ .

Suggest Corrections

Similar questions

The equation of the plane passing through the origin and perpendicular to the line x=2 y=3 z is

(- 1, 2, 0)

L_{1} : \frac{x}{3} = \frac{y + 1}{0} = \frac{z - 2}{- 1}

L_{2} : \frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}

x + 2 y + 3 z = 14

2 x - y + 5 z = 15

3 x - 2 y - 4 z = - 13

x + 2 y + 3 z = 5, x + 2 y + 3 z - 7 = 0

(2, 5, - 3)

x + 2 y + 2 z = 1

x - 2 y + 3 z = 4

Join BYJU'S Learning Program