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

$\begin{array}{1 1}(A) \large\frac{-5}{2},5,5 \\ (B) \large\frac{-5}{2},-5,5 \\(C) \large\frac{5}{2},5,-5 \\ (D) \large\frac{2}{5},5,-5\end{array}$

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## 1 Answer

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Toolbox:
• Equation of a plane in intercept form is $\large\frac{x}{a}$$+\large\frac{y}{b}$$+\large\frac{z}{c}$$=1 • Where a,b,c are the intercepts cut off by the plane at x,y and z respectively. Step 1: The given equation is 2x+y-z=5 Divide throughout by 5,we get \large\frac{x}{\Large\frac{2}{5}}$$+\large\frac{y}{5}$$+\large\frac{z}{-5}$$=1$
Step 2:
Equation of a plane in intercept form is $\large\frac{x}{a}$$+\large\frac{y}{b}$$+\large\frac{z}{c}$$=1 Where a,b,c are the intercepts cut off by the plane at x,y and z axes respectively. Here for the given equation a=\large\frac{2}{5}$$,b=5$ and $c=-5$
Thus the intercepts cutoff by the plane are $\large\frac{5}{2}$$,5$ and $-5$
answered May 30, 2013

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