Find the equation of a plane which passes through (2, –3, 1) and is perpendicular to the line through the points (3, 4, –1) and (2,–1, 5).

Solution

The equation of plane through (2, –3, 1) is
A (x – 2) + B ( y + 3) + C (z – 1) = 0 ...(1)
The direction ratios of the line through the points (3, 4, –1) and (2, –1.5) are 2 – 3,–1– 4 ,5 + 1 i.e. -1,-5, 6 i.e. 1, 5,–6.
∴ the line with direction ratios 1, 5, –6 is normal to the plane (1)
∴ 1 (x – 2) + 5(y + 3)–6(z – 1) = 0 $open square brackets because space space straight A over 1 space equals space straight B over 5 space equals space fraction numerator straight C over denominator negative 6 end fraction close square brackets$
or x – 2 + 5y + 15 – 6z + 6 = 0
or x + 5 y – 6 z + 19 = 0
which is required equation of plane.

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Three Dimensional Geometry

Find the equation of a plane which passes through (2, –3, 1) and is perpendicular to the line through the points (3, 4, –1) and (2,–1, 5).

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