The sum of intercepts of the plane 4x + 3y + | vedclass

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(A) The given equation of the plane is $4x + 3y + 2z = 2$. To find the intercepts,we convert this equation into the intercept form $\frac{x}{a} + \fra

Vedclass · Accepted Answer

$\frac{13}{6}$

Vedclass · Answer

(A) The given equation of the plane is $4x + 3y + 2z = 2$. To find the intercepts,we convert this equation into the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Dividing the entire equation by $2$,we get: $\frac{4x}{2} + \frac{3y}{2} + \frac{2z}{2} = \frac{2}{2}$ $\frac{x}{1/2} + \frac{y}{2/3} + \frac{z}{1} = 1$. Comparing this with the intercept form,the intercepts are $a = \frac{1}{2}$,$b = \frac{2}{3}$,and $c = 1$. The sum of the intercepts is $a + b + c = \frac{1}{2} + \frac{2}{3} + 1$. Taking the least common multiple $(LCM)$ of $2$ and $3$,which is $6$: $a + b + c = \frac{3 + 4 + 6}{6} = \frac{13}{6}$.

The sum of intercepts of the plane $4x + 3y + 2z = 2$ on the coordinate axes is

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