Let a and b be two non collinear unit vectors- If u-a- a-b b and v-a-b then - v - is

The correct option is A | u | |v| = |a× #160;b| = |a||b|sinθ = sinθ |u| = |a (a·b)b| = √(a^2 + #160;(a·b)^2b^2 2 #160;(a·b)^2) = ...

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Byju's Answer

Standard XII

Physics

Acceleration in 2D

Let a and ...

Question

Let $\to a$ and $\to b$ be two non collinear unit vectors. If $\to u = \to a - (\to a \cdot \to b) \to b$ and $\to v = \to a \times \to b$ then $∣ ∣ \to v ∣ ∣$ is

∣ ∣ \to u ∣ ∣

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∣ ∣ \to v ∣ ∣ + ∣ ∣ \to v . \to a ∣ ∣

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∣ ∣ \to v ∣ ∣ + ∣ ∣ ∣ \to v . \to b ∣ ∣ ∣

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∣ ∣ \to u ∣ ∣ + \to u . (\to a + \to b)

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Solution

The correct option is A $∣ ∣ \to u ∣ ∣$
$∣ ∣ \to v ∣ ∣ = ∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣ = ∣ ∣ \to a ∣ ∣ ∣ ∣ ∣ \to b ∣ ∣ ∣ sin θ = sin θ$
$∣ ∣ \to u ∣ ∣ = ∣ ∣ ∣ \to a - (\to a \cdot \to b) \to b ∣ ∣ ∣ = \sqrt{a^{2} + (\to a \cdot \to b)^{2} b^{2} - 2 (\to a \cdot \to b)^{2}}$
$= \sqrt{1 + {cos}^{2} θ - 2 {cos}^{2} θ}$
$= \sqrt{1 - {cos}^{2} θ}$
$= \sqrt{{sin}^{2} θ}$
$= sin θ$
Hence option $A$ .

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Similar questions

Q. Let

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\to u = \to a - (\to a . \to b) \to b

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{(1 + ¯ ¯ ¯ u)}^{2} + {(1 + ¯ ¯ ¯ v)}^{2} = {(1 - ¯ ¯ ¯ u . ¯ ¯ ¯ v)}^{2} + {∣ ∣ ¯ ¯ ¯ u + ¯ ¯ ¯ v + (¯ ¯ ¯ u \times ¯ ¯ ¯ v) ∣ ∣}^{2}

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Let →a and →b be two non collinear unit vectors. If →u=→a−(→a⋅→b)→b and →v=→a×→b then ∣∣→v∣∣ is

Let $\to a$ and $\to b$ be two non collinear unit vectors. If $\to u = \to a - (\to a \cdot \to b) \to b$ and $\to v = \to a \times \to b$ then $∣ ∣ \to v ∣ ∣$ is