If - - a - - b - - 3 - - a - b - then the angle between - a and - b is -

The correct option is A π 6 Here, |a⃗.b⃗|=√(3)|a⃗×b⃗| or |a⃗||b⃗|cosθ=√(3)|a⃗||b⃗|sinθ or sinθcosθ=1√(3) or tanθ=tanπ6⇒θ=π6 ...

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Byju's Answer

Standard XII

Physics

Work Done as Dot Product

If | → a . ...

Question

If $∣ ∣ \to a . \to b ∣ ∣ = \sqrt{3} ∣ ∣ \to a \times \to b ∣ ∣$ , then the angle between $\to a$ and $\to b$ is :-

\frac{π}{6}

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\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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Solution

The correct option is A $\frac{π}{6}$
Here, $| \to a . \to b | = \sqrt{3} | \to a \times \to b |$
or $| \to a | | \to b | c o s θ = \sqrt{3} | \to a | | \to b | sin θ$
or $\frac{sin θ}{cos θ} = \frac{1}{\sqrt{3}}$
or $tan θ = tan \frac{π}{6} \Rightarrow θ = \frac{π}{6}$

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Similar questions

Q. If

∣ ∣ \to a ∣ ∣ = 4

∣ ∣ ∣ \to b ∣ ∣ ∣ = 2

and the angle between

\to a

and

\to b

\frac{π}{6}

, then

(\to a \times \to b)

is equal to

Q. If

¯ ¯ c

is perpendicular to

¯ ¯ ¯ a

and

¯ ¯ b

∣ ∣ ¯ ¯ ¯ a ∣ ∣ = 3, ∣ ∣ ¯ ¯ b ∣ ∣ = 4, ∣ ∣ ¯ ¯ c ∣ ∣ = 5

and the angle between

¯ ¯ ¯ a

and

¯ ¯ b

\frac{π}{6}

then

[¯ ¯ ¯ a ¯ ¯ b ¯ ¯ c] =

Let the vectors and be such that and, then is a unit vector, if the angle between and is

(A) (B) (C) (D)

Q. If angle between

\to a

and

\to b

\frac{π}{3}

, then angle between

2 \to a

and

- 3 \to b

Let the vectors a and b be such that $| a | = 3$ and $| b | = \frac{\sqrt{2}}{3}$ , then $a \times b$ is a unit vector, if the angle between a and b is

a) $\frac{π}{6}$

b) $\frac{π}{4}$

c) $\frac{π}{3}$

d) $\frac{π}{2}$

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If ∣∣→a.→b∣∣=√3∣∣→a×→b∣∣, then the angle between →a and →b is :-

The correct option is A π6Here, |→a.→b|=√3|→a×→b|or |→a||→b|cosθ=√3|→a||→b|sinθor sinθcosθ=1√3or tanθ=tanπ6⇒θ=π6

If $∣ ∣ \to a . \to b ∣ ∣ = \sqrt{3} ∣ ∣ \to a \times \to b ∣ ∣$ , then the angle between $\to a$ and $\to b$ is :-

The correct option is A $\frac{π}{6}$
Here, $| \to a . \to b | = \sqrt{3} | \to a \times \to b |$
or $| \to a | | \to b | c o s θ = \sqrt{3} | \to a | | \to b | sin θ$
or $\frac{sin θ}{cos θ} = \frac{1}{\sqrt{3}}$
or $tan θ = tan \frac{π}{6} \Rightarrow θ = \frac{π}{6}$