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Three Dimensional Geometry

Question

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Find the distance from P (2, 1, – 1 ) to the plane x – 2 y + 4 z = 9.

Solution

The equation of plane is x – 2y + 4z - 9 = 0.
Let p be the length of perpendicular from P (2, 1, – 1) to the plane.

therefore space space space straight p space equals space fraction numerator open vertical bar 2 minus 2 space left parenthesis 1 right parenthesis space plus space 4 space left parenthesis negative 1 right parenthesis space minus space 9 close vertical bar over denominator square root of left parenthesis 1 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared plus left parenthesis 4 right parenthesis squared end root end fraction space equals space fraction numerator open vertical bar 2 minus 2 minus 4 minus 9 close vertical bar over denominator square root of 1 plus 4 plus 16 end root end fraction space equals fraction numerator 13 over denominator square root of 21 end fraction

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

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