Find the intercepts made by the plane bar{r} | vedclass

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(B) The given equation of the plane is $\bar{r} \cdot (2, -3, 4) = 12$. Substituting $\bar{r} = (x, y, z)$,we get $2x - 3y + 4z = 12$. To find the int

Vedclass · Accepted Answer

$6, -4, 3$

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(B) The given equation of the plane is $\bar{r} \cdot (2, -3, 4) = 12$. Substituting $\bar{r} = (x, y, z)$,we get $2x - 3y + 4z = 12$. To find the intercepts,we write the equation in the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Dividing the equation $2x - 3y + 4z = 12$ by $12$,we get $\frac{2x}{12} - \frac{3y}{12} + \frac{4z}{12} = 1$. This simplifies to $\frac{x}{6} + \frac{y}{-4} + \frac{z}{3} = 1$. Comparing this with the standard intercept form,the intercepts are $a = 6$,$b = -4$,and $c = 3$.

Find the intercepts made by the plane $\bar{r} \cdot (2, -3, 4) = 12$ on the coordinate axes.

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