In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
5y + 8 = 0

Solution

The equation of plane is
5y + 8 = 0 or 5y = – 8
or 0x – 5 y + 0z = 8
Dividing both sides by $square root of left parenthesis 0 right parenthesis squared plus left parenthesis 5 right parenthesis squared plus left parenthesis 0 right parenthesis squared end root space equals space 5 comma space space space we space get comma$
0x - y + 0 z = $8 over 5$
It is of the form lx + my + nz = p, where l = 0, m = -1, n = 0, $straight p space equals space 8 over 5$
∴ direction cosines of the normal to the plane are 0, – 1, 0 and distance from origin = $8 over 5.$

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Three Dimensional Geometry

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
5y + 8 = 0

Some More Questions From Three Dimensional Geometry Chapter

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

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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For Daily Free Study Material Join wiredfaculty

Three Dimensional Geometry

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.5y + 8 = 0

Some More Questions From Three Dimensional Geometry Chapter

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

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
5y + 8 = 0