First slide

Dot product or scalar product of vectors

Question

Let $\vec{a} and \vec{b}$ be two vectors such that $| \vec{a} | = 1, | \vec{b} | = 4 and \vec{a} \cdot \vec{b} = 2 . If \vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b}$ then find the angle between

Moderate

A

$\frac{π}{3}$
B

$\frac{π}{6}$
C

$\frac{3 π}{4}$
D

$\frac{5 π}{6}$

Solution

$\begin{array}{l} | \vec{a} | = 1, | \vec{b} | = 4, \vec{a} \cdot \vec{b} = 2 \\ \vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b} \\ or \vec{c} + 3 \vec{b} = 2 \vec{a} \times \vec{b} \\ ∵ \vec{a} \cdot \vec{b} = 2 \\ \Rightarrow | \vec{a} | \cdot | \vec{b} | \cos θ = 2 \end{array}$
$\begin{aligned} or \cos θ = \frac{2}{| \vec{a} | \cdot | \vec{b} |} = \frac{2}{4} \\ = \frac{1}{2} \\ or θ = \frac{π}{3} \end{aligned}$
$\begin{array}{l} or θ = \frac{π}{3} \\ \Rightarrow | \vec{c} + 3 \vec{b} |^{2} = | 2 \vec{a} \times \vec{b} |^{2} \\ or | \vec{c} |^{2} + 9 | \vec{b} |^{2} + 2 \vec{c} \cdot 3 \vec{b} = 4 | \vec{a} |^{2} | \vec{b} |^{2} \sin^{2} θ \\ or | \vec{c} |^{2} + 144 + 6 \vec{b} \cdot \vec{c} = 48 \\ or | \vec{c} |^{2} + 96 + 6 (\vec{b} \cdot \vec{c}) = 0 - - - - - - (i) \end{array}$
$\begin{aligned} or \vec{c} = 2 \vec{a} \times \vec{b} - 3 \vec{b} \\ \Rightarrow \vec{b} \cdot \vec{c} = 0 - 3 \times 16 \\ = - 48 \end{aligned}$
Putting value of b $\vec{b} \vec{c}$ in Eq. (i), we have
$\begin{aligned} | \vec{c} |^{2} + 96 - 6 \times 48 \\ = 0 = 48 \times 4 = 192 \end{aligned}$
Again, putting the value of $|\vec{c}|$ in Eq. (i), we have
$\begin{array}{l} 192 + 96 + 6 | \vec{b} | \cdot | \vec{c} | \cos α = 0 \\ or 6 \times 4 \times 8 \sqrt{3} \cos α = - 288 \\ or \cos α = - \frac{288}{6 \times 4 \times 8 \sqrt{3}} = - \frac{3}{2 \sqrt{3}} \\ = - \frac{\sqrt{3}}{2} \\ or α = \frac{5 π}{6} \end{array}$

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