First slide

Dot product or scalar product of vectors

Question

$If \vec{a}, \vec{b}, \vec{c} are three vectors such that \{\begin{array}{l} | \vec{a} | = | \vec{c} | = 1 \\ | \vec{b} | = 4 \\ | \vec{b} \times \vec{c} | = 2 \\ 2 \vec{b} = \vec{c} + λ \vec{a} \end{array} where λ is a scalar. If the value of λ$
$is equal to \sqrt{α - β \sqrt{3}} where α and β are natural numbers, then find the value of α + β$

Moderate

Solution

$\begin{array}{l} Given | \vec{b} \times \vec{c} | = 2 \\ | \vec{b} | | \vec{c} | \sin θ = 2 \Rightarrow \sin θ = \frac{1}{2} \\ θ = 30^{\circ} \end{array}$
$Hence, angle between \vec{b} and \bar{c} is 30^{\circ} .$
$Now, 2 \vec{b} - \vec{c} = λ \vec{a} \Rightarrow | 2 \vec{b} - \vec{c} | = | λ \vec{a} | \Rightarrow | 2 \vec{b} - \vec{c} |^{2} = λ^{2} | \vec{a} |^{2}$
$Hene, λ^{2} = 65 - 8 \sqrt{3}$
$\begin{array}{l} λ = \sqrt{65 - 8 \sqrt{3}} \equiv \sqrt{α - β \sqrt{3}} \\ Hence, α = 65 and β = 8 \\ α + β = 73 \end{array}$

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