Short Answer Type
and perpendicular to the vector 
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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).
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 
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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