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# Find the direction cosines of the normal to the plane 2x + 3y - z = 4.

Answer :

Given :

Equation of plane : 2x + 3y – z = 4

To Find : Direction cosines of the normal i.e.

Formula :

1) Direction cosines :

If a, b & c are direction ratios of the vector then its direction cosines are given by

Answer :

For the given equation of plane

2x + 3y – z = 4

Direction ratios of normal vector are (2, 3, -1)

Therefore, direction cosines are

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Find the equation of the plane which contains the line of intersection of the planes

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