Find the equation of the plane with intercept | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

Find the equation of the plane with intercepts $2, 3$ and $4$ on the $x, y$ and $z$-axis respectively.

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