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

Question

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

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Answer

2x + y - z = 5 ....(1)
Dividing both sides of equation (1) by 5, we obtain
$\frac{2}{5}\text{x}+\frac{\text{y}}{5}-\frac{\text{z}}{5}=1$
$\Rightarrow\frac{\text{x}}{\frac{5}{2}}\text{x}+\frac{\text{y}}{5}+\frac{\text{z}}{-5}=1\ \ ...(2)$
It is known that the equation of a plane in intercept form $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1,$
where a, b, c are the intercepts cut off by the plane at x, y, and z axes respectively.
Therefore, for the given equation,
$\text{a}=\frac{5}{2},\ \text{b}=5,\ \text{and}\ \text{c}=-5$
Thus, the intercepts cut off by the plane are $\frac{5}{2},\ 5,\ \text{and}\ -5.$

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