Volume of a Pyramid Linear Algebra

Question

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: $\displaystyle\frac{1}{3}(\text{base} *\text{height})$, but with the given vectors, I am unsure how to solve this.

Hope someone can help me, would really appreciate it.

Thanks.

Answer 1

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

$$V = \frac{1}{3}A_{base}h = \frac{1}{3} \frac{\sqrt{14}}{2} \frac{3}{\sqrt{14}} = \frac{1}{2}.$$

Answer 2

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?