Q1 of 181 Page 1196

Find the distance of the point from the plane


Formula :


where () is point from which distance is to be calculated


Therefore ,


Plane can be written in cartesian form as


3x - 4y + 12z = 9


3x - 4y + 12z - 9 = 0


Point = (2i - j - 4k)


Which can be also written as


Point = ( 2 , - 1 , - 4 )


Distance





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29

A vector of magnitude 8 units is inclined to the x-axis at 45o, y-axis at 60o and an acute angle with the z-axis, if a plane passes through a point (√2, -1, 1) and is normal to find its equation in vector form.

 30

Find the vector equation of a line passing through the point and perpendicular to the plane

Also, find the point of intersection of this line and the plane.


 2

Find the distance of the point from the plane

3

Find the distance of the point (3, 4, 5) from the plane