Two vectors 2a+b and a-3b are perpendicular. Find the angle between a and b if |a|=2|b|
I will type the question again directly from my textbook without checking the question above.
Two vectors 2a+b and a-3b are perpendicular. Find the angle between a and b, if |a|=2|b|.
Their answer: 60 degrees.
They want the angle between vector a and b, not vectors 2a+b and a-3b.
Sorry, I misread the question.
We get from the given that $\displaystyle a\cdot a = 4b\cdot b$ and $\displaystyle 2a\cdot a-5a\cdot b-3b\cdot b=0$
From which it follows that $\displaystyle b\cdot b=a\cdot b$
But that means $\displaystyle \left(\frac{a\cdot b}{||a||||b||}=\frac{1}{2}\right)$
Hell0: this is the solution
$\displaystyle \left( {2a + b} \right) \bot \left( {a - 3b} \right) \Leftrightarrow \left( {2a + b} \right).\left( {a - 3b} \right) = 0
$
Calculate :
$\displaystyle \left( {2a + b} \right).\left( {a - 3b} \right) = 0 \Leftrightarrow 2\left\| a \right\|^2 - 6a.b + b.a - 3\left\| b \right\|^2 = 0$
But :
$\displaystyle \left\{ \begin{array}{l}
\left\| a \right\| = 2\left\| b \right\| \\
a.b = b.a \\
a.b = \left\| a \right\| \times \left\| b \right\|\cos \theta \\
\theta = \left( {a,b} \right) \\
\end{array} \right.
$
Subst:
$\displaystyle \left( {2a + b} \right).\left( {a - 3b} \right) = 0 \Leftrightarrow 2\left\| a \right\|^2 - 6a.b + b.a - 3\left\| b \right\|^2 = 0$
Calculate:
$\displaystyle \left( {2a + b} \right).\left( {a - 3b} \right) = 0 \Leftrightarrow 2\left( {2\left\| b \right\|} \right)^2 - 5\left( {2\left\| b \right\|} \right) \times \left\| b \right\|\cos \theta - 3\left\| b \right\|^2 = 0$
I'have :
$\displaystyle \begin{array}{l}
8\left\| b \right\|^2 - 10\left\| b \right\|^2 \cos \theta - 3\left\| b \right\|^2 = 0 \\
\left\{ \begin{array}{l}
5\left\| b \right\|^2 \left( {1 - 2\cos \theta } \right) = 0 \\
b \ne \overrightarrow 0 \\
\end{array} \right. \Leftrightarrow 1 - 2\cos \theta = 0 \\
\end{array}$
Conclusion:
$\displaystyle \cos \theta = \frac{1}{2} \Leftrightarrow \theta = \frac{\pi }{3}$