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

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Given plane  2x + y-z = 5

We have to find the intercepts that this plane would make so,

Making it look like intercept form first:

By dividing both sides of the equation by 5 (as we have to make the R.H.S =1) , we get then,

\frac{2}{5}x+\frac{y}{5}-\frac{z}{5} =1

\Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5} =1

So, as we know that from the equation of a plane in intercept form,\frac{x}{a}+\frac{y}{b}+\frac{z}{c} = 1 where a,b,c are the intercepts cut off by the plane at x,y, and z-axes respectively.

Therefore after comparison, we get the values of a,b, and c.

a = \frac{5}{2},\ b=5,\ and\ c=-5.

Hence the intercepts are \frac{5}{2},\ 5,\ and\ -5.

Posted by

Divya Prakash Singh

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