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Question

(C) The intercept form of the equation of a plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b$,and $c$ are the intercepts on

Vedclass · Accepted Answer

$-3x + 6y + 4z = 12$

Vedclass · Answer

(C) The intercept form of the equation of a plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b$,and $c$ are the intercepts on the $x, y$,and $z$ axes respectively.Given intercepts are $a = -4, b = 2$,and $c = 3$.Substituting these values into the formula:$\frac{x}{-4} + \frac{y}{2} + \frac{z}{3} = 1$To simplify,find the least common multiple of the denominators $4, 2$,and $3$,which is $12$.Multiplying the entire equation by $12$:$12 	imes (\frac{x}{-4}) + 12 	imes (\frac{y}{2}) + 12 	imes (\frac{z}{3}) = 12 	imes 1$$-3x + 6y + 4z = 12$Thus,the equation of the plane is $-3x + 6y + 4z = 12$.

Find the equation of the plane whose intercepts on the coordinate axes are $-4, 2$,and $3$.

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