First slide

Planes in 3D

Question

Let $A (1, 1, 1), B (2, 3, 5) a n d C (- 1, 0, 2)$ be three points, then equation of a plane parallel to the plane ABC which is at a distance 2 units is

Moderate

A

$2 x - 3 y + z + 2 \sqrt{14} = 0$
B

$2 x - 3 y + z - \sqrt{14} = 0$
C

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

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

Solution

$A (1, 1, 1), B (2, 3, 5), C (- 1, 0, 2)$ d.r’s of AB are (1, 2, 4)
D.r’s of AC are $(- 2, - 1, 1)$ .
D.r’s of normal to plane ABC are (2,-3,1) As a result, equation of the plane ABC is $2 x - 3 y + z = 0$ . Let the equation of the required plane is , $2 x - 3 y + z = k$ then $|\frac{k}{\sqrt{4 + 9 + 1}}| = 2 k = \pm 2 \sqrt{14}$

Hence, equation of the required plane is $2 x - 3 y + z + 2 \sqrt{14} = 0$

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