Question

Find the coordinates of the foot q of the perpendicular drawn from the point p (1,3,4) to the plane 2x-y z 3

Answer 1

Therefore the coordinate of the intersection point is $=(\frac{1}{3} ,\frac{10}{3}, \frac{11}{3})$

Step-by-step explanation:

Given equation of plane is 2x -y+z=3

The equation of line which passes through the point (1,3,4) and perpendicular to the given plane is

$\frac{x-1}{2} =\frac{y-3}{-1} =\frac{z-4}{1}$              [ The direction ratio of straight is same with the                                                              

                                         direction ratio given plane]

Any point on the above line be Q (2r+1,-r+3,r+4)

Let Q be the intersection point of the line and the plane

Then Q will be satisfy the equation of plane

∴ 2(2r+1) -(-r+3)+r+4=3

$\Leftrightarrow 6r+5=3$

$\Leftrightarrow r=-\frac{1}{3}$

Therefore the coordinate of the intersection point is $(-\frac{2}{3} +1,\frac{1}{3} +3,-\frac{1}{3} +4)$$=(\frac{1}{3} ,\frac{10}{3}, \frac{11}{3})$

                                                                                           

Answer 2

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