If a- and b- are two vectors perpendicular to each other and -a-b-1- Determine the angle between a-b- and 2 a-b-

The correct option is C θ = cos^ 1(3√(10))Given nbsp;|a| = |b|=1 cos θ = (a+b) . (2a + b)|a + b||2a+b| = 2a^2+b^2√(a^2 + b^2 + 2ab cos90^o)√((2a)^2 + b^2 + ...

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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

If a⃗ and b⃗ ...

Question

If $\to a$ and $\to b$ are two vectors perpendicular to each other and $| \to a | = | \to b | = 1$ . Determine the angle between $\to a + \to b$ and $- \to 2 a + \to b$

θ = {cos}^{- 1} (\frac{2}{\sqrt{5}})

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θ = {cos}^{- 1} (\frac{3}{2 \sqrt{5}})

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θ = {cos}^{- 1} (\frac{3}{\sqrt{10}})

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θ = {cos}^{- 1} (\frac{2}{3 \sqrt{5}})

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Solution

The correct option is C $θ = {cos}^{- 1} (\frac{3}{\sqrt{10}})$
Given $| \to a | = | \to b | = 1$
$cos θ = \frac{(\to a + \to b) . (2 \to a + \to b)}{| \to a + \to b | | - \to 2 a + \to b |}$
$= \frac{2^{2} +^{2}}{\sqrt{a^{2} + b^{2} + 2 a b cos 90^{o}} \sqrt{(2 a)^{2} + b^{2} + 2 (2 a) b cos 90^{o}}}$
$\frac{2^{2} +^{2}}{\sqrt{2} \sqrt{5}} = \frac{3}{\sqrt{10}}$
$θ = {cos}^{- 1} (\frac{3}{\sqrt{10}})$

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If →a and →b are two vectors perpendicular to each other and |→a|=|→b|=1 . Determine the angle between →a+→b and −→2a+→b

The correct option is C θ=cos−1(3√10)Given |→a|=|→b|=1 cos θ=(→a+→b).(2→a+→b)|→a+→b||−→2a+→b| =2→a2+→b2√a2+b2+2ab cos90o√(2a)2+b2+2(2a)b cos90o 2→a2+→b2√2√5=3√10 θ=cos−1(3√10)

If $\to a$ and $\to b$ are two vectors perpendicular to each other and $| \to a | = | \to b | = 1$ . Determine the angle between $\to a + \to b$ and $- \to 2 a + \to b$