If the plane 56x + 4y + 9z = 2016 meets the c | vedclass

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Question

(C) The given equation of the plane is $56x + 4y + 9z = 2016$. Dividing by $2016$,we get the intercept form: $\frac{56x}{2016} + \frac{4y}{2016} + \fr

Vedclass · Accepted Answer

$\left(12, 168, \frac{224}{3}\right)$

Vedclass · Answer

(C) The given equation of the plane is $56x + 4y + 9z = 2016$. Dividing by $2016$,we get the intercept form: $\frac{56x}{2016} + \frac{4y}{2016} + \frac{9z}{2016} = 1$ $\frac{x}{36} + \frac{y}{504} + \frac{z}{224} = 1$. The coordinates of the points where the plane meets the axes are $A(36, 0, 0)$,$B(0, 504, 0)$,and $C(0, 0, 224)$. The centroid of $	riangle ABC$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}ight)$. Centroid $= \left(\frac{36+0+0}{3}, \frac{0+504+0}{3}, \frac{0+0+224}{3}ight) = \left(12, 168, \frac{224}{3}ight)$.

If the plane $56x + 4y + 9z = 2016$ meets the coordinate axes at points $A$,$B$,and $C$,then the centroid of the $\triangle ABC$ is

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