Find $|\boldsymbol{a \times b}|$ (graph included) if $|\boldsymbol{a}| = 2$, $|\boldsymbol{b}| = 5$

Question

I don't see how I have enough information to figure this one out.

Here's what I'm thinking: \begin{align*} |\boldsymbol{a \times b}| & = |\boldsymbol{a}|~|\boldsymbol{b}|\sin(\theta)\\ & = 2 \cdot 5 \cdot \sin(\theta)\\ & =10\sin(\theta) \end{align*} except how do I find out what theta is??? Am I barking up the wrong tree?

(Also about the second part of the question. I got it right, but I guessed on the $x$/$y$ component. Couldn't it also be $x$-component is negative, and $y$-component is positive? So basically, the other way around, except $z$ is still $0$?)

Answer 1

In 3 dimensions the absolute value of the cross product is equal to the area of the parallelogram co-planar to the 2 vectors.

The area is $A=B\times H$ so $||a|| \times ||b||$ which you know. As for finding if the components are positive or not, try actually using the right hand rule as the question recommends.

The image is By Oleg Alexandrov (Own work) [Public domain], via Wikimedia Commons.

Answer 2

a) That's simple. $a$ is in xy-plane and $b$ is in the direction of $k$ which implies $a$ and $b$ are perpendicular and hence $\theta=\frac\pi2=90^{\circ}$. So $|a \times b|=|a||b|sin(\theta)=|2||5|sin(90)=10$

b) In this case x coordinate is positive and y negative because of the order of the Cross Product, to get x negative and y positive you would do $b \times a$. To see it clearer it is the same as the unit vector Cross $i=j \times k$ and $-i=k \times j$

Answer 3

Since $\boldsymbol{a}$ lies in the $xy$-plane and $\boldsymbol{b}$ lies on the $z$-axis, the vectors are perpendicular. Hence, $$||\boldsymbol{a \times b}|| = ||\boldsymbol{a}||~||\boldsymbol{b}||\sin\theta = 2 \cdot 5 \cdot \sin\left(\frac{\pi}{2}\right) = 2 \cdot 5 \cdot 1 = 10$$ To apply the right-hand rule, point the fingers of your right hand in the direction of $\boldsymbol{a}$, then curl them toward $\boldsymbol{b}$. Your thumb then points in the direction of $\boldsymbol{a \times b}$. In this case, your thumb will point to the left and out of the page, which is positive in the $x$ direction and negative in the $y$ direction.

We can also see this algebraically. By definition, if $\boldsymbol{a} = a_x\boldsymbol{i} + a_y\boldsymbol{j} + a_z\boldsymbol{k}$ and $\boldsymbol{b} = b_x\boldsymbol{i} + b_y\boldsymbol{j} + b_z\boldsymbol{k}$, then $$\boldsymbol{a \times b} = (a_yb_z - a_zb_y)\boldsymbol{i} + (a_zb_x - a_xb_z)\boldsymbol{j} + (a_xb_y - a_yb_x)\boldsymbol{k}$$ In our example, $a_x > 0$, $a_y > 0$, $a_z = 0$, $b_x = 0$, $b_y = 0$, and $b_z > 0$. Thus, \begin{align*} \vec{a} \times \vec{b} & = (a_yb_z - a_zb_y)\boldsymbol{i} + (a_zb_x - a_xb_z)\boldsymbol{j} + (a_xb_y - a_yb_x)\boldsymbol{k}\\ & = (a_yb_z - 0 \cdot 0)\boldsymbol{i} + (0 \cdot 0 - a_xb_z)\boldsymbol{j} + (a_x \cdot 0 - a_y \cdot 0)\boldsymbol{k}\\ & = a_yb_z\boldsymbol{i} - a_xb_z\boldsymbol{j} + 0\boldsymbol{k}\\ & = a_yb_z\boldsymbol{i} - a_xb_z\boldsymbol{j} \end{align*} Since $a_y > 0$ and $b_z > 0$, $a_yb_z > 0$. Since $a_x > 0$ and $b_z > 0$, $-a_xb_z < 0$, so we conclude that the $x$-coordinate of the cross product is positive, its $y$-coordinate is negative, and its $z$-coordinate is zero.