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- Thread starter jaychay
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- Mar 5, 2012

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The volume of the pyramid is $V=\frac 13 Bh$, where $B$ is the area of the base, and $h$ is the height perpendicular to the base.

The normal vector $\vec n$ of the plane can be deduced from its equation $-x+z=0$, meaning it is $\vec n=(-1,0,1)$.

The height $h$ of the pyramid is the projection of the vector $\overrightarrow{ad}$ onto the normal vector $\vec n$.

The formula for that projection is $h=\frac{\overrightarrow{\mathstrut ad} \cdot \overrightarrow{\mathstrut n}}{\|\vec n\|}$.

So we have:

$$\begin{cases}B=\frac 12\sqrt 2 \\

V=\frac 13 Bh = 4 \\

\vec n = (-1,0,1) \\

h=\frac{\overrightarrow{\mathstrut ad} \cdot \overrightarrow{\mathstrut n}}{\|\vec n\|} = \frac{(-1,1,1)t \cdot (-1,0,1)}{\|(-1,0,1)\|} = \frac{2}{\sqrt 2}t=t\sqrt 2

\end{cases}

\implies V = \frac 13 \cdot \frac 12\sqrt 2 \cdot t\sqrt 2 = 4

\implies t = 12

$$

So point $d$ is $(0,0,0)+(-1,1,1)12=(-12,12,12)$.

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- Mar 1, 2012

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$d = \dfrac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$

$V = \dfrac{Bh}{3} = \dfrac{1}{\sqrt{2}} \cdot \dfrac{h}{3} = 4 \implies h = 12 \sqrt{2}$

$12\sqrt{2} = \dfrac{|-1(-t) + 0(t) + 1(t) + 0|}{\sqrt{(-1)^2 + 0^2 + 1^2}}$

$12\sqrt{2} = \dfrac{2t}{\sqrt{2}} \implies t = 12$