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Question -

Find the intercepts cut off by the plane 2x + y – z = 5.



Answer -

Given:

The plane 2x + y – z =5

Let us express theequation of the plane in intercept form

x/a + y/b + z/c = 1

Where a, b, c are theintercepts cut-off by the plane at x, y and z axes respectively.

2x + y – z =5 …. (1)

Now divide both thesides of equation (1) by 5, we get

2x/5 + y/5 – z/5 = 5/5

2x/5 + y/5 – z/5 = 1

x/(5/2) + y/5 + z/(-5)= 1

Here, a = 5/2, b = 5and c = -5

The interceptscut-off by the plane are 5/2, 5 and -5.

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