A- and b- are two vectors such that - a- -1- b- -4 and a- -b- -2 If c- - 2a-b- -3b- then find the angle between b- and c-

The correct option is D 5π 6 Given | a⃗ |=1, | b⃗ | = 4 a⃗.b⃗ = 2 and c⃗ = (2a⃗×b⃗) 3b⃗ a⃗.b⃗ = 2 ⇒ | a | | b |. cos (a,b) = 2 ...

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Byju's Answer

Standard XII

Physics

Unit Vectors

a⃗ and b⃗ a...

Question

$\to a$ and $\to b$ are two vectors such that $| \to a | = 1, ∣ ∣ \to b ∣ ∣ = 4$ and $\to a . \to b = 2$
If $\to c = (2 \to a \times \to b) - 3 \to b$ , then find the angle between $\to b$ and $\to c$ .

\frac{π}{3}

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\frac{π}{6}

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\frac{3 π}{4}

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\frac{5 π}{6}

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Solution

The correct option is D $\frac{5 π}{6}$
Given $| \to a | = 1, ∣ ∣ \to b ∣ ∣ = 4 \to a . \to b = 2$ and $\to c = (2 \to a \times \to b) - 3 \to b$
$\to a . \to b = 2 \Rightarrow | a | | b | . c o s (a, b) = 2$
$1 \times 4 \times c o s θ = 2$
$c o s θ = \frac{1}{2} \Rightarrow θ = 60^{\circ}$ (Angle b/w $\to a, \to b)$
$\to a \times \to b = | a | | b | . s i n θ = 1 \times 4 \times s i n 60 = 4 \times \frac{\sqrt{2}}{2} = 2 \sqrt{3}$
Given $\to c = 2 \to a \times \to b - 3 \to b$
squaring on both sides
${| \to c |}^{2} = {∣ ∣ 2 ¯ a \times ¯ b ∣ ∣}^{2} + 9 {∣ ∣ \to b ∣ ∣}^{2} - 2 \times (3 \to a (2 \to a \times \to b))$
${| \to c |}^{2} = 4 {∣ ∣ \to a \times \to b ∣ ∣}^{2} + 9 {∣ ∣ \to b ∣ ∣}^{2} - 0$
$(∵ \to b, 2 \to a \times \to b$ are perpendicular to each other)
${| \to c |}^{2} = 4 \times (2 \sqrt{3})^{2} + 9 (4)^{2}$
${| c |}^{2} = 4 \times 12 + 9 \times 16 = 48 + 144 = 192$
${| c |}^{2} = 192 \Rightarrow | c | = 8 \sqrt{3}$
$\to c = (2 \to a \times \to b) - 3 \to b$
dot product on both sides with $\to b$
$\to c . \to b = (2 \to a \times \to b) . \to b - 3 \to b . \to b$
$\to c . \to b = 0 - 3 {| b |}^{2}$ $(∵ \to b 12 \to a \times \to b$ and $c o s 90 = 0)$
$∴ | c | . | b | . c o s (α) = - 3 | b |$
$c o s (α) = \frac{- 3 \times 4}{8 \sqrt{3}} = \frac{- \sqrt{3}}{2}$
$- α = \frac{5 π}{6}$

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Similar questions

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→a and →b are two vectors such that |→a|=1,∣∣→b∣∣=4 and →a.→b=2 If →c=(2→a×→b)−3→b, then find the angle between →b and →c.

$\to a$ and $\to b$ are two vectors such that $| \to a | = 1, ∣ ∣ \to b ∣ ∣ = 4$ and $\to a . \to b = 2$
If $\to c = (2 \to a \times \to b) - 3 \to b$ , then find the angle between $\to b$ and $\to c$ .