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Three Dimensional Geometry

Question
CBSEENMA12033384

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
5y + 8 = 0

Solution

The equation of plane is
5 y + 8 = 0 or 5 y = – 8 or – 5 y = 8
Dividing both sides by square root of left parenthesis 0 right parenthesis squared plus left parenthesis negative 5 right parenthesis squared plus left parenthesis 0 right parenthesis squared end root space equals space 5 comma space we space get comma space space straight y space equals space 8 over 5     ...(1)
∴    direction ratios of the normal OP to the plane are 0, – 1, 0 where O is origin and P(x1, y1, z1) is foot of perpendicular.
Direction ratios of OP are x1– 0, y1 – 0, z1 – 0 i.e. x1, y1, z1.
Since direction cosines and direction ratios of a line are proportional.
therefore space space space space space space straight x subscript 1 over 0 space equals space fraction numerator straight y subscript 1 over denominator negative 1 end fraction space equals space straight z subscript 1 over 0 space equals space straight k comma space space space say.
therefore space space space space space space space straight x subscript 1 space equals space 0 comma space space space space straight y subscript 1 space equals space minus straight k comma space space space straight z subscript 1 space equals space 0
therefore space space space space straight P space is space left parenthesis 0 comma space space minus straight k comma space space 0 right parenthesis
Since P lies on plane (1)
 therefore space space space space space space straight k space equals space 8 over 5
therefore space space space space space straight P space is space open parentheses 0 comma space space minus 8 over 5 comma space 0 close parentheses.

 

Some More Questions From Three Dimensional Geometry Chapter

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