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Applications of Dot Product

If a and ...

Question

If $\to a$ and $\to b$ are unit vectors such that $| \to a + \to b | = 1$ , then find the value of $| \to a - \to b |$

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Solution

Consider the problem

$\begin{matrix} | \to a | = ∣ ∣ \to b ∣ ∣ = 1 ∣ ∣ \to a + \to b ∣ ∣ = 1 \Rightarrow {∣ ∣ \to a + \to b ∣ ∣}^{2} = 1 \Rightarrow (\to a + \to b) . (\to a + \to b) = 1 (\to a . \to a = {| \to a |}^{2}) \Rightarrow \to a . \to a + \to a . \to b + \to b . \to a + \to b . \to b = 1 \Rightarrow {| \to a |}^{2} + 2 \to a . \to b + {∣ ∣ \to b ∣ ∣}^{2} = 1 (∵ \to a . \to b = \to b . \to a) \Rightarrow 1 + 2 \to a . \to b + 1 = 1 \Rightarrow 2 \to a . \to b = - 1 \Rightarrow \to a . \to b = - \frac{1}{2} . . . (1) \end{matrix}$

Now

$\begin{matrix} {∣ ∣ \to a - \to b ∣ ∣}^{2} = (\to a - \to b) . (\to a - \to b) = \to a . \to a - \to a . \to b - \to b . \to a + \to b . \to b = {| \to a |}^{2} - 2 \to a . \to b + {∣ ∣ \to b ∣ ∣}^{2} = 1 - 2 \times \frac{- 1}{2} + 1 (f r o m (1)) = 1 + 1 + 1 = 3 \end{matrix}$

$\Rightarrow ∣ ∣ \to a - \to b ∣ ∣ = \sqrt{3}$

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