First slide

Dot product or scalar product of vectors

Question

If the vectors $3 \vec{a} - 5 \vec{b} and 2 \vec{a} + \vec{b}$ are mutually perpendicular and if $\vec{a} + 4 \vec{b} and \vec{b} - \vec{a}$ are also mutually perpendicular, then the cosine of the angle between $\vec{a} and \vec{b}$ is

Moderate

A

$\frac{19}{5 \sqrt{43}}$
B

$\frac{19}{3 \sqrt{43}}$
C

$\frac{19}{2 \sqrt{43}}$
D

$\frac{19}{6 \sqrt{43}}$

Solution

$\begin{array}{l} (3 \vec{a} - 5 \vec{b}) \cdot (2 \vec{a} + \vec{b}) = 0 \\ or 6 | \vec{a} |^{2} - 5 | \vec{b} |^{2} = 7 \vec{a} \cdot \vec{b} \end{array}$
$Also, (\vec{a} + 4 \vec{b}) \cdot (\vec{b} - \vec{a}) = 0$
$\begin{array}{l} or - | \vec{a} |^{2} + 4 | \vec{b} |^{2} = 3 \vec{a} \cdot \vec{b} \\ or \frac{6}{7} | \vec{a} |^{2} - \frac{5}{7} | \vec{b} |^{2} = - \frac{1}{3} | \vec{a} |^{2} + \frac{4}{3} | \vec{b} |^{2} \\ or 25 | \vec{a} |^{2} = 43 | \vec{b} |^{2} \\ \Rightarrow 3 \vec{a} \cdot \vec{b} = - | \vec{a} |^{2} + 4 | \vec{b} |^{2} = \frac{57}{25} | \vec{b} |^{2} \end{array}$
$\begin{array}{l} or 3 | \vec{a} | | \vec{b} | \cos θ = \frac{57}{25} | \vec{b} |^{2} \\ or 3 \sqrt{\frac{43}{25}} | \vec{b} |^{2} \cos θ = \frac{57}{25} | \vec{b} |^{2} \\ or \cos θ = \frac{19}{5 \sqrt{43}} \end{array}$

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