If a- and b- are two vectors with magnitude 1 and 3 respectively and 1-3 a-b-2-2 a-b-3a-b-2-98- Then find the angle between a- and b-A- 32 -B- 2 -D- 5 -6

The correct option is B 2π3Solving the given equation 1^2 6a.b+9(a.b)^2 +4|a|^2+4a.b+|b|^2 +2(2a+b)·(3(a×b)) +9|a×b|^2=98 ⇒ 1 2a.b+81cos^2θ+4+9+0 +81sin^2θ=98 ...

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Byju's Answer

Standard XII

Mathematics

Order of a Differential Equation

If a⃗ and b⃗ ...

Question

If $\to a$ and $\to b$ are two vectors with magnitude $1$ and $3$ respectively and $(1 - 3 \to a . \to b)^{2} + | 2 \to a + \to b + 3 (\to a \times \to b) |^{2} = 98$ .
Then find the angle between $\to a$ and $\to b$

\frac{π}{3}

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\frac{2 π}{3}

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\frac{π}{6}

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\frac{5 π}{6}

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Solution

The correct option is B $\frac{2 π}{3}$
Solving the given equation
$1^{2} - 6 \to a . \to b + 9 (\to a . \to b)^{2} + 4 | \to a |^{2} + 4 \to a . \to b + | \to b |^{2} + 2 (2 \to a + \to b) \cdot (3 (\to a \times \to b)) + 9 | \to a \times \to b |^{2} = 98$
$\Rightarrow 1 - 2 \to a . \to b + 81 {cos}^{2} θ + 4 + 9 + 0 + 81 {sin}^{2} θ = 98$
$(∵ \to a . (\to a \times \to b) = 0 and \to b . (\to a \times \to b) = 0$
$\Rightarrow - 2 \to a . \to b = 3$
$\Rightarrow cos θ = \frac{- 1}{2}$
$\Rightarrow θ = \frac{2 π}{3}$

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If →a and →b are two vectors with magnitude 1 and 3 respectively and (1−3→a.→b)2+|2→a+→b+3(→a×→b)|2=98. Then find the angle between →a and →b

If $\to a$ and $\to b$ are two vectors with magnitude $1$ and $3$ respectively and $(1 - 3 \to a . \to b)^{2} + | 2 \to a + \to b + 3 (\to a \times \to b) |^{2} = 98$ .
Then find the angle between $\to a$ and $\to b$