Question

Need help! (Vectors)

If the angle between two vectors, m = [3,1] & n = [2,x] is pi/4. Find X

I am using the formula (which I'm sure is the correct one)

cos(theta) = (m crossproduct n)/|m| |n|

I plug in but I am doing something wrong when it comes to simplifying, something, somewhere, goes wrong and it messes it all up. Thanks.

Answer 1

I would use a dot product

Answer 2

\[ m \cdot n = 2\cdot 3 + 1\cdot 1 = x+6 \]
the magnitude of m is sqr(10)
mag of n is \(\sqrt{x^2+4} \)

using
\[ m \cdot n = |m| |n| \cos \frac{\pi}{4}\\

\frac{m \cdot n}{|m| |n|}= \frac{1}{\sqrt{2} }
\]

Answer 3

There's a typo in my post, I meant dot product*
Also, I solved it.. it turned out I was doing everything completely right... but I stupidly forgot to properly expand the (x+6)^2 lol

thanks for your help though!