The equation of the plane through the point ( 2 , − 1 , − 3 ) and parallel to the lines 3 x − 1 = 2 y + 2 = − 4 z and 2 x = − 3 y − 1 = 2 z − 2 is

Question

The equation of the plane through the point ( 2 , − 1 , − 3 ) and parallel to the lines 3 x − 1 ​ = 2 y + 2 ​ = − 4 z ​ and 2 x ​ = − 3 y − 1 ​ = 2 z − 2 ​ is

TestHub · Accepted Answer

The plane is parallel to the given lines, so its normal vector is perpendicular to the direction vectors of both lines. The direction ratios of the lines are $3, 2, - 4$ and $2, - 3, 2$ .

The normal vector $n$ to the plane is given by the cross product of these direction vectors:

$n = \hat{i} 32 \hat{j} 2 - 3 \hat{k} - 4 2$

$n = \hat{i} (4 - 12) - \hat{j} (6 - (- 8)) + \hat{k} (- 9 - 4)$

$n = - 8 \hat{i} - 14 \hat{j} - 13 \hat{k}$

The direction ratios of the normal are proportional to $8, 14, 13$ .

The equation of the plane passing through the point $(2, - 1, - 3)$ with normal direction ratios $8, 14, 13$ is:

$8 (x - 2) + 14 (y + 1) + 13 (z + 3) = 0$

$8 x - 16 + 14 y + 14 + 13 z + 39 = 0$

$8 x + 14 y + 13 z + 37 = 0$

Answer: $8 x + 14 y + 13 z + 37 = 0$

Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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