Let a-b- and c- be three unit vector such that a-b-c-32b-c- If b- is not parallel to c- then angle between a- and b- is-

The correct option is B 5π6 â×(b̂×ĉ)= (â. ĉ)b̂ (â .b̂)ĉ = √(3)b̂2+√(3)2ĉ ⇒â.ĉ =√(3)2 #160; â.b̂= √(3)2 ⇒|â||ĉ| cosθ_1=√( ...

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Byju's Answer

Standard XII

Physics

Unit Vectors

Let a⃗,b⃗ a...

Question

Let $\to a, \to b$ and $\to c$ be three unit vector such that $\to a \times (\to b \times \to c) = \frac{\sqrt{3}}{2} (\to b + \to c)$ . If $\to b$ is not parallel to $\to c$ . then angle between $\to a$ and $\to b$ is:

\frac{2 π}{3}

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\frac{5 π}{6}

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\frac{3 π}{4}

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\frac{π}{2}

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Solution

The correct option is B $\frac{5 π}{6}$
$^a \times (^b \times^c) = (^a .^c)^b - (^a .^b)^c$

$= \frac{\sqrt{3}^b}{2} + \frac{\sqrt{3}}{2}^c$

$\Rightarrow^a .^c = \frac{\sqrt{3}}{2}^a .^b = \frac{- \sqrt{3}}{2}$

$\Rightarrow |^a | |^c | cos θ_{1} = \frac{\sqrt{3}}{2} a n d |^a | ∣ ∣^b ∣ ∣ cos θ_{2} = - \frac{\sqrt{3}}{2}$

$\Rightarrow cos θ_{1} = \frac{\sqrt{3}}{2} \Rightarrow cos θ_{2} = - \frac{\sqrt{3}}{2}$

$\Rightarrow θ_{1} = 30^{o} = \frac{π}{6} \Rightarrow θ_{2} = 150^{o} = \frac{5 π}{6}$

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Let →a,→b and →c be three unit vector such that →a×(→b×→c)=√32(→b+→c). If →b is not parallel to →c. then angle between →a and →b is:

The correct option is B 5π6^a×(^b×^c)=(^a.^c)^b−(^a.^b)^c=√3^b2+√32^c⇒^a.^c=√32^a.^b=−√32⇒|^a||^c|cosθ1=√32and|^a|∣∣^b∣∣cosθ2=−√32⇒cosθ1=√32⇒cosθ2=−√32⇒θ1=30o=π6⇒θ2=150o=5π6

Let $\to a, \to b$ and $\to c$ be three unit vector such that $\to a \times (\to b \times \to c) = \frac{\sqrt{3}}{2} (\to b + \to c)$ . If $\to b$ is not parallel to $\to c$ . then angle between $\to a$ and $\to b$ is: