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Three Dimensional Geometry

Question
CBSEENMA12033424

Find the equation of plane passing through the line of intersection of the planes 2x – y = 0 and 3z – y = 0 and perpendicular to the plane 4x + 5y – 3Z = 8.

Solution

The equation of any plane through the line of intersection of he planes 2 x – y = 0 and 3 z – y = 0 or y – 3 z = 0 is
(2 x – y) + k (y – 3 z) = 0    ...(1)
or 2x – y + ky – 3 k z = 0
or 2x + (k – 1) y – 3 k z = 0
∴ this plane is perpendicular to the plane 4 x + 5 y – 3 z = 8.
∴ (2) (4) + (k – 1) (5) + (–3 k) (– 3) = 0    [∴ aa2 + bb2 + c1 c2 = 0]
∴ 8 + 5 k – 5 + 9 k = 0 or 14 k = –3
therefore space space space space straight k space equals space minus 3 over 14
Putting this value of k in (1),  we get left parenthesis 2 straight x minus straight y right parenthesis space minus space 3 over 14 left parenthesis straight y minus 3 straight z right parenthesis space equals 0
or 14 (2 x – y) – 3 (y – 3 z) = 0
or  28 x – 14 y – 3 y + 9 z =0
or    28 x – 17 y + 9 z = 0
which is the required equation of the plane.


 

Some More Questions From Three Dimensional Geometry Chapter

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