Volume of a Pyramid Linear Algebra Find the volume of a pyramid with triangular base bounded by vectors (1,-1, 2) and (1, 1, 1)
and vertex located at (3, 2, 5).
I am not sure how I would solve this. I know the volume of a pyramid is: 1/3 (base * height), but with the given vectors, I am unsure how to solve this.
Hope someone can help me, would really appreciate it.
Thanks.
 A: I'm guessing since you're only given two vectors $\vec{a}, \vec{b}$ on the base that the third is found by connecting the heads.  In that case, the area of the base is
$$A_{base} = \frac{1}{2}|\vec{a} \times \vec{b}| = \frac{1}{2}|-3\hat{x} + \hat{y} + 2 \hat{z}| = \sqrt{14}/2.$$
The cross product also defines a normal to the plane containing $\vec{a}, \vec{b}$.  The equation of this plane is $-3x + y + 2z = 0$ since the origin is contained in the plane.
The distance to the plane, which is the height of the pyramid, is then
$$h = \frac{-3(3) + 1(2) + 2(5)}{\sqrt{14}} = \frac{3}{\sqrt{14}}.$$
Then your volume is
$$V = \frac{1}{3}A_{base}h = \frac{1}{3} \frac{\sqrt{14}}{2} \frac{3}{\sqrt{14}} = \frac{1}{2}.$$
A: Hint 1: For two arbitrary vectors $v_1$ and $v_2$, $\|v_1 \times v_2\| = \|v_1\| \|v_2\|\sin(\theta)$.  You should be able to find the area of the base using this fact.
Hint 2: $v_1 \times v_2 $ will give you a vector perpendicular to the base.
Hint 3: Let $u = \frac {v_1 \times v_2}{\|v_1 \times v_2\| }$, then $u \cdot v_3$ will give you the height of the pyramid, where $v_3$ is the vector to the vertex from the corner at the origin.
Hint 4 (bonus): The vector triple product $|v_3 \cdot (v_1 \times v_2)|$ gives you the volume of the parallelepiped defined by the three vectors.  How many pyramids will fit in the parallelepiped?
