If v- and w- are two mutually perpendicular unit vectors and u-av-bw- where a and b are non zero real numbers- then the angle between u- and w- is-

The correct option is A cos^ 1(b√(a^2+b^2)) u⃗.w⃗ = a v⃗.w⃗ + b w⃗ . w⃗ u⃗.w⃗ = b #160; |u⃗| |w⃗| cosθ = b √(a^2 + b^2)cosθ = b ...

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Byju's Answer

Standard XII

Physics

Unit Vectors

If v⃗ and ...

Question

If $\to v$ and $\to w$ are two mutually perpendicular unit vectors and $\to u = a \to v + b \to w$ , where a and b are non zero real numbers, then the angle between $\to u$ and $\to w$ is?

{cos}^{- 1} (\frac{b}{\sqrt{a^{2} + b^{2}}})

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{cos}^{- 1} (\frac{a}{\sqrt{a^{2} + b^{2}}})

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{cos}^{- 1} (b)

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{cos}^{- 1} (a)

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Solution

The correct option is A ${cos}^{- 1} (\frac{b}{\sqrt{a^{2} + b^{2}}})$
$\to u . \to w = a \to v . \to w + b \to w . \to w$
$\to u . \to w = b$
$| \to u | | \to w | cos θ = b$
$\sqrt{a^{2} + b^{2}} cos θ = b$
$cos θ = \frac{b}{\sqrt{a^{2} + b^{2}}}$
$θ = {cos}^{- 1} \frac{b}{\sqrt{a^{2} + b^{2}}}$

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Similar questions

\to u, \to v, \to w

be such that

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\to u, \to v, \to w

be such that

| \to u | = 1, | \to v | = 2, | \to w | = 3

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If →v and →w are two mutually perpendicular unit vectors and →u=a→v+b→w, where a and b are non zero real numbers, then the angle between →u and →w is?

The correct option is A cos−1(b√a2+b2)→u.→w=a→v.→w+b→w.→w→u.→w=b |→u||→w|cosθ=b√a2+b2cosθ=bcosθ=b√a2+b2θ=cos−1b√a2+b2

If $\to v$ and $\to w$ are two mutually perpendicular unit vectors and $\to u = a \to v + b \to w$ , where a and b are non zero real numbers, then the angle between $\to u$ and $\to w$ is?