The equation of the plane, passing through the point ( − 1 , 2 , − 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, passing through the point ( − 1 , 2 , − 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 direction ratios of the given lines are $b_{1} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k}$ and $b_{2} = 2 \hat{i} - 3 \hat{j} + 2 \hat{k}$ .

Since the plane is parallel to both lines, its normal vector $n$ is perpendicular to both $b_{1}$ and $b_{2}$ .

$n = b_{1} \times b_{2} = \hat{i} 32 \hat{j} 2 - 3 \hat{k} - 4 2$

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

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

The direction ratios of the normal to the plane can be taken as $(8, 14, 13)$ .

The equation of the plane passing through $(- 1, 2, - 3)$ is

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

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

$8 x + 8 + 14 y - 28 + 13 z + 39 = 0$

$8 x + 14 y + 13 z + 19 = 0$

Answer: $8 x + 14 y + 13 z + 19 = 0$

Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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