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  • If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP.

16. If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP.

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We have the coordinates of the points O(0,0,0)  and  P(1,2,-3) respectively.

Therefore, the direction ratios of OP are (1-0) = 1, (2-0)=2,\ and\ (-3-0)=-3

And we know that the equation of the plane passing through the point (x_{1},y_{1},z_{1}) is

a(x-x_{1})+b(y-y_{1})+c(z-z_{1})=0 where a,b,c are the direction ratios of normal.

Here, the direction ratios of normal are 1,2, and -3 and the point P is (1,2,-3).

Thus, the equation of the required plane is

1(x-1)+2(y-2)-3(z+3) = 0

\implies x+2y -3z-14 = 0

Posted by

Divya Prakash Singh

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