Question

Find the equation of the plane passing through the point (–1, 2, 1) and perpendicular to the line joining the points (–3, 1, 2) and (2, 3, 4). Also, find the perpendicular distance of the origin from this plane.

Answer 1

Let the equation of plane passing through (-1, 2, 1) be

Now the equation of line joining point (-3, 1, 2) and (2, 3, 4) is

\(\because\) Plane (i) is perpendicular to line (ii)

[\(\because\) Normal vector of (i) is parallel to line (ii)]

Hence, equation of required plane is

If d is the distance from origin to plane then

Comments

can you explain how normal vector of  1 is parallal to line 2?

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Normal vector is perpendicular to the plane and given that line is also perpendicular to the plane. Therefore, normal vector of plane is parallel to the given line.