1. ## Gr.12 Vector Question Help Please!?

Two vectors 2a + b and a - 3b are perpendicular. Find the angle between a and b, if |a| = 2|b| .

Step-by-step solution please and thank you!

Been solved but feel free to try!

2. I was about halfway through writing my post when you said you solved it, but I finished it and I'll put it up anyway. The problem itself was easy, but typing out the $\displaystyle \text\LaTeX$ was tedious:

The vectors $\displaystyle 2\textbf a+\textbf b$ and $\displaystyle \textbf a - 3\textbf b$ are orthogonal, so $\displaystyle (2\textbf a+\textbf b)\cdot(\textbf a - 3\textbf b) = 0\text,$

$\displaystyle \Rightarrow2\textbf a\cdot\textbf a-5\textbf a\cdot\textbf b-3\textbf b\cdot\textbf b = 0$

$\displaystyle \Rightarrow2\lVert\textbf a\rVert^2-3\lVert\textbf b\rVert^2-5\textbf a\cdot\textbf b=0$

$\displaystyle \Rightarrow2(2\lVert\textbf b\rVert)^2-3\lVert\textbf b\rVert^2-5\textbf a\cdot\textbf b=0$

$\displaystyle \Rightarrow5\lVert\textbf b\rVert^2=5\textbf a\cdot\textbf b\Rightarrow\lVert\textbf b\rVert^2=\textbf a\cdot\textbf b$

$\displaystyle \Rightarrow1=\frac{\textbf a\cdot\textbf b}{\lVert\textbf b\rVert\lVert\textbf b\rVert}$

$\displaystyle \Rightarrow1=\frac{\textbf a\cdot\textbf b}{\frac12\lVert\textbf a\rVert\lVert\textbf b\rVert}$

$\displaystyle \Rightarrow\frac12=\frac{\textbf a\cdot\textbf b}{\lVert\textbf a\rVert\lVert\textbf b\rVert}$

$\displaystyle \Rightarrow\cos\frac\pi3=\frac{\textbf a\cdot\textbf b}{\lVert\textbf a\rVert\lVert\textbf b\rVert}$

So the angle between $\displaystyle \textbf a$ and $\displaystyle \textbf b$ is $\displaystyle \frac\pi3\text.$