Two parallel planes ${{\mathrm{p}}_{1}}$ and ${{\mathrm{p}}_{3}}$ have equations \(\mathbf{r}\cd

Question

Two parallel planes ${{\mathrm{p}}_{1}}$ and ${{\mathrm{p}}_{3}}$ have equations $\mathbf{r}\cdot \left( \begin{matrix}3 \-2 \-3 \\end{matrix} \right)=4$ and $\mathbf{r}\cdot \left( \begin{matrix} -3 \2 \ 3 \\end{matrix} \right)=7$ respectively. Find the perpendicular distance between ${{\mathrm{p}}_{1}}$ and ${{\mathrm{p}}_{3}}$, the exact cartesian equations of parallel planes which are $\sqrt{66}$ units away from ${{p}_{1}}$.

Accepted Answer

(i) $\frac{11}{\sqrt{22}}$units (ii) $3x-2y-3z=4-22\sqrt{3}$ and $3x-2y-3z=4+22\sqrt{3}$

Vectors - Distance from a Point to a Plane

Vectors - Distance from a Point to a Plane

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