Question on the right hand rule

Question

Say I'm taking the cross product of vectors $a$ and $b$. Say that $b$ is totally in the $z$ direction and has length $7$, so $b = 7k$. Say that $a$ is in the $xy$-plane with positive coefficients, $a = 3x + 4y$.

I want to understand the sign of the components of $a \times b$ using the right hand rule. Now, surele since $a \times b$ is orthogonal to both $a$ and $b$, it's $z$ component will be zero. But will the $x$ and $y$ components be positive or negative, and how can i see this with the right hand rule? Thank you for your time

Answer 1

Take a piece of paper, and place a pencil on top of it. That pencil would represent the $z$-axis.

Your thumb should follow the direction of vector $a$, your index finger is pointing towards the $b$ direction and you will notice that your midddle finger should be pointing to the fourth quadrant, hence the $x$ coordinate is positive and the $y$ coordinate is negative.

Answer 2

Yep, the cross product will have no z component. The cross product of vector in the xy plane with a vector perpendicular to the plane has the effect of rotating the vector plus or minus 90 degrees then scaling it. If the perpendicular vector has a positive component, the plane's vector rotates clockwise, otherwise couter-clockwise, by the right hand rule.

This applies just as well to the unit vectors. So the positive $\hat{y}$ unit vector becomes the positive $\hat{x}$ unit vector.

Rotating a vector doesn't necessarily change the sine of its component. $<\frac{-\sqrt{2}}{2},\frac{\sqrt{2}}{2}>$ doesn't change the sign of its y component upon taking the cross product.

Answer 3

Remember that $a$ points in direction $(3,4)$ which is little above $y=x$ ($45^{\circ}$) line.

Open your palm so your fingers point in direction of a. Now curl your four fingers towards b. Your stretched out thumb will point in direction of $a \times b$.

Easy to see $x$-component is positive and $y$-component negative. More precisely, the components will be proportional to $(4,-3)$ so that $$(3,4)\cdot(4,-3)=0$$