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Three Dimensional Geometry

Question
CBSEENMA12033434

Find the equations of the planes that passes through three points:
(1, 1,–1), (6, 4, –5),(–4, –2, 3)

Solution
The equation of line through (1, 1, –1), (6, 4, – 5) is
            fraction numerator straight x minus 1 over denominator 6 minus 1 end fraction space equals space fraction numerator straight y minus 1 over denominator 4 minus 1 end fraction space equals space fraction numerator straight z plus 1 over denominator negative 5 plus 1 end fraction
or            fraction numerator straight x minus 1 over denominator 5 end fraction space equals space fraction numerator straight y minus 1 over denominator 3 end fraction space equals space fraction numerator straight z plus 1 over denominator negative 4 end fraction
The point (– 4, –2, 3) will lie on this line
if  fraction numerator negative 4 minus 1 over denominator 5 end fraction space equals space fraction numerator negative 2 minus 1 over denominator 3 end fraction space equals space fraction numerator 3 plus 1 over denominator negative 4 end fraction
i.e. if –1 = –1 = –1, which is true
∴ given points (1, 1, –1), (6, 4, – 5), (– 4, – 2, 3) are collinear.
∴ infinite number of planes pass through the given points.

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.