First slide

Dot product or scalar product of vectors

Question

If unit vectors $\vec{a} and \vec{b}$ are inclined at an angle 2 $θ$ such $| \vec{a} - \vec{b} | < 1 and 0 \leq θ \leq π, then θ$ lies in the interval

Moderate

A

$[0, π / 6)$
B

$(5 π / 6, π]$
C

$[π / 6, π / 2)$
D

$(π / 2, 5 π / 6]$

Solution

we have $| \vec{a} - \vec{b} |^{2} = | \vec{a} |^{2} + | \vec{b} |^{2} - 2 (\vec{a} \cdot \vec{b})$
$\begin{array}{l} or | \vec{a} - \vec{b} |^{2} = | \vec{a} |^{2} + | \vec{b} |^{2} - 2 | \vec{a} | | \vec{b} | \cos 2 θ \\ or | \vec{a} - \vec{b} |^{2} = 2 - 2 \cos 2 θ (∵ | \vec{a} | = | \vec{b} | = 1) \\ = 4 \sin^{2} θ \\ \Rightarrow | \vec{a} - \vec{b} | = 2 | \sin θ | \\ now | \vec{a} - \vec{b} | < 1 \\ \Rightarrow 2 | \sin θ | < 1 \\ or | \sin θ | < \frac{1}{2} \\ \Rightarrow θ \in [0, π / 6) or θ \in (5 π / 6, π] \end{array}$

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