Example 1 from Class

Find $\vec{i}\times\vec{j}$.

Solution:

You could do this both algebraically and geometrically. Geometrically, notice that the right hand rule gives the direction of $\vec{i}\times\vec{j}$ as pointing along the positive $z$-axis. Also, since $\Vert\vec{i}\times\vec{j}\Vert$ is the area of the parallelogram formed by $\vec{i}\times\vec{j}$, we can conclude immediately that $\Vert\vec{i}\times\vec{j}\Vert=1$. Thus, $\vec{i}\times\vec{j}=\vec{k}$.

Algebraically, this is also possible, but much harder. We have

Example 2 from Class

Find $\vec{i}\times(\vec{i}+\vec{j})$

Solution:

We know which direction this vector will be in. For the magnitude, we find

\[\Vert\vec{i}\times(\vec{i}+\vec{j})\Vert=\Vert\vec{i}\Vert\Vert\vec{j}\Vert\sin\theta=\sqrt{2}\cdot\frac{\sqrt{2}}{2}=1\]

This implies that $\vec{i}\times\vec{j}=\vec{i}\times(\vec{i}+\vec{j})$. This makes sense, because we know that

\[\vec{i}\times(\vec{i}+\vec{j})=\vec{i}\times\vec{i}+\vec{i}\times\vec{j}\]

and we know that $\vec{i}\times\vec{i}=\vec{0}$.

Example 3 from Class

Are the four points

\[a=(0,0,0),\ b=(2,1,-3),\ c=(-1,-1,1),\text{ and }d=(0,2,2)\]

coplanar? That is, do they all lie on the same line?

Solution:

There are many ways to do this problem, as mentioned in class. A relatively straightforward approach is to consider the pair of vectors $\vec{b}$ and $\vec{c}$. Forming their cross product, we obtain $\vec{b}\times\vec{c}=\langle -2,1,-1\rangle$. Now, we find the cross product from $\vec{c}$ to $\vec{d}$, and we have $\vec{c}\times\vec{d}=\langle -4,2,-2\rangle$. Notice that the cross products are parallel (they are scalars of each other), and so the planes corresponding to those normal vectors are parallel. But, each plane contains $a=(0,0,0)$, and so the planes are identical. In other words, yes, all four points are coplanar.

The Volume of a Parallelopiped

We know that the cross product of two vectors is the area of the parallelogram those vectors form. A parallelopiped is a 3-dimensional generalization of a parallelogram, having 6 sides that are all parallelograms. If we wish to calculate the volume of the parallelopiped, we simply calculate the area of any side and multiply by the height (when viewing the side in question as the base).

For instance, in the following diagram the angle $\alpha$ and the length of the side $a$ are used to calculate the height of the parallelopiped.

In this situation, the area of the base is $\Vert\vec{b}\times\vec{c}\Vert$ and the height of the parallelopiped is $h=\Vert a\Vert\cos\alpha$. Putting these together, we have

\[V=\Vert\vec{b}\times\vec{c}\Vert\Vert a\Vert\cos\alpha .\]

Since $\vec{u}\cdot\vec{v}=\Vert\vec{u}\Vert\Vert\vec{v}\Vert\cos\theta$, we can rewrite the above volume equation by realizing that the cross product $\vec{b}\times\vec{c}$ is parallel to the vector from which the angle $\alpha$ emanates from. So, we really have

\[V=\Vert\vec{b}\times\vec{c}\Vert\Vert a\Vert\cos\alpha=\left|(\vec{b}\times\vec{c})\cdot\vec{a}\right|.\]

 

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