The distance of the point ( 1 , 3 , − 7 ) from the plane passing through the point ( 1 , − 1 , − 1 ) having normal perpendicular to both the lines 1 x − 1 = − 2 y + 2 = 3 z − 4 and 2 x − 2 = −

Question

The distance of the point ( 1 , 3 , − 7 ) from the plane passing through the point ( 1 , − 1 , − 1 ) having normal perpendicular to both the lines 1 x − 1 ​ = − 2 y + 2 ​ = 3 z − 4 ​ and 2 x − 2 ​ = − 1 y + 1 ​ = − 1 z + 7 ​ is

TestHub · Accepted Answer

The direction vectors of the given lines are $b_{1} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $b_{2} = 2 \hat{i} - \hat{j} - \hat{k}$ .

The normal vector $n$ to the plane is perpendicular to both lines, so it is given by their cross product:

$n = b_{1} \times b_{2} = \hat{i} 12 \hat{j} - 2 - 1 \hat{k} 3 - 1 = 5 \hat{i} + 7 \hat{j} + 3 \hat{k}$

The equation of the plane passing through the point $(1, - 1, - 1)$ with normal vector $n$ is:

$5 (x - 1) + 7 (y + 1) + 3 (z + 1) = 0$

$5 x + 7 y + 3 z + 5 = 0$

The perpendicular distance $d$ of the point $(1, 3, - 7)$ from this plane is:

$d = \frac{∣5 ( 1 ) + 7 ( 3 ) + 3 ( - 7 ) + 5∣}{5 ^{2} + 7 ^{2} + 3 ^{2}}$

$d = \frac{∣5 + 21 - 21 + 5∣}{25 + 49 + 9}$

$d = \frac{10}{83}$ units.

Answer: $\frac{10}{83}$ units.

Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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