First slide

Planes in 3D

Question

Let A(1,1,1,), B(2,3,5) and C(-1,0,2) be three points, then equation of a plane parallel to the plane ABC which is at distance 2 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) direction ratios of AB are < 1,2, 4 >.
Direction ratios of AC are < -2, -1, l >.
Therefore, direction ratios of normal to plane ABC are < 2, -3, 1 >
As a result, equation of the plane ABC is 2x - 3y + z=0.
Let the equation of the required plane be 2x - 3y +z= k.
Then $|\frac{k}{\sqrt{4 + 9 + 1}}| = 2 or k = \pm 2 \sqrt{14}$
Hence, equation of the required plane is $2 x - 3 y + z + 2 \sqrt{14} = 0$

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