Find the intercepts cut off by the plane 2x + | vedclass

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Question

(A) The given equation of the plane is $2x + y - z = 5$. Dividing both sides of the equation by $5$,we get: $\frac{2x}{5} + \frac{y}{5} - \frac{z}{5}

Vedclass · Accepted Answer

$\frac{5}{2}, 5, -5$

Vedclass · Answer

(A) The given equation of the plane is $2x + y - z = 5$. Dividing both sides of the equation by $5$,we get: $\frac{2x}{5} + \frac{y}{5} - \frac{z}{5} = 1$ This can be rewritten in the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ as: $\frac{x}{\frac{5}{2}} + \frac{y}{5} + \frac{z}{-5} = 1$ Comparing this with the standard intercept form,the intercepts $a, b, c$ are: $a = \frac{5}{2}$,$b = 5$,and $c = -5$. Thus,the intercepts cut off by the plane are $\frac{5}{2}, 5, -5$.

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

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