Question
Find the distance between the point and the plane (see figure). The distance $D$ between a point $\left(x_{0}, y_{0}, z_{0}\right)$ and the plane $a x+b y+c z+d=0$ is$$D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$$$$(2,-1,0), 3 x+3 y+2 z=6$$
Step 1
The point is $(x_0, y_0, z_0) = (2, -1, 0)$ and the coefficients of the plane equation are $a = 3$, $b = 3$, $c = 2$, and $d = -6$. Show more…
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