# Cool Beginner Vector Problem from K&K

I thought this was a super cool vector, example problem from An Introduction to Classical Mechanic by K&K. It says:
The problem is to ﬁnd a unit vector lying in the x−y plane that is perpendicular to the vector A = (3, 5, 1).
The solution first begins by recognizing that since the problem is asking to find a unit vector perpendicular to A, then we can conversely say that the dot product of A and B must be equal to zero. And by using the vector component definition of the dot product, $A \cdot B = A_x B_x + A_y B_y = 0$, we can set up our first equation: $3B_x + 5B_y = 0$.

Next, we can say that since it's a unit vector, then the magnitude must be equal to one, and hence $B_x^2 + B_y^2 = 1^2$.

Now we have two equations to solve $B_x$ and $B_y$ with!

I thought this was a great example, neatly combining all the definitions and ideas we'd learned so far.

• vanhees71

jedishrfu
Mentor
Your A.B expression is missing a term namely Az.Bz but since Bz is zero then it drops out. I mention it because it's being solved in x, y and z space and you should show it to complete your solution.

Last edited:
Thanks. The book didn't include it, so I didn't include it either.

jedishrfu
Mentor
The other thing they may not have mentioned is that form of A.B that was used works only when x, y and z are orthogonal.

To me the really cool proof was to show that from the definition of dot product namely

A.B = |A||B|cos(ab-angle)

one can derive the AxBx + AyBy + AzBz expression given the axes are orthogonal.

• mafagafo
I had never written the proof myself. Inspired by what jedishrfu wrote, my attempt follows.

To avoid the problem of showing unnecessary $\LaTeX$, the proof is hidden under spoiler tags.

$$\begin{array}{rcl} A \cdot B &=& |A||B| \cos{\theta} \\ &=& |A||B| \left( \frac{|A|^2+|B|^2-|C|^2}{2|A||B|} \right) \\ &=& \frac{|A|^2+|B|^2-|C|^2}{2} \\ &=& \frac{|A|^2+|B|^2-|B-A|^2}{2} \\ &=& \frac{\left( A_x \right) ^ 2 + \left( A_y \right) ^ 2 + \left( A_z \right) ^ 2 + \left( B_x \right) ^ 2 + \left( B_y \right) ^ 2 + \left( B_z \right) ^ 2 - \left(B_x - A_x \right) ^ 2 - \left(B_y - A_y \right) ^ 2 - \left(B_z - A_z \right) ^ 2}{2} \\ &=& \frac{\left( A_x \right) ^ 2 + \left( A_y \right) ^ 2 + \left( A_z \right) ^ 2 + \left( B_x \right) ^ 2 + \left( B_y \right) ^ 2 + \left( B_z \right) ^ 2 - \left(B_x \right)^2 + 2 \left(A_x B_x \right) - \left(A_x \right) ^ 2 - \left(B_y \right)^2 + 2 \left(A_y B_y \right) - \left(A_y \right) ^ 2 - \left(B_z \right)^2 + 2 \left(A_z B_z \right) - \left(A_z \right) ^ 2}{2} \\ &=& \frac{2 \left(A_x B_x \right) + 2 \left(A_y B_y \right) + 2 \left(A_z B_z \right)}{2} \\ &=& A_x B_x + A_y B_y + A_z B_z \end{array}$$

The other thing they may not have mentioned is that form of A.B that was used works only when x, y and z are orthogonal.

To me the really cool proof was to show that from the definition of dot product namely

A.B = |A||B|cos(ab-angle)

one can derive the AxBx + AyBy + AzBz expression given the axes are orthogonal.
Is there a way to prove it without relying on the law of cosines?

vanhees71