Let and be two vectors such that and . If and the angle between and is , then 192 is equal to __________. [2024]

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Solution

Q.
Let $\vec{a}$ and $\vec{b}$ be two vectors such that $| \vec{a} | = 1, | \vec{b} | = 4$ and $\vec{a} \cdot \vec{b} = 2$ . If $\vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b}$ and the angle between $\vec{b}$ and $\vec{c}$ is $α$ , then 192 $\sin^{2} α$ is equal to __________. [2024]

Ans.
(48)

Given $| \vec{a} |$ $= 1,$ $| \vec{b} |$ $= 4 and \vec{a} \cdot \vec{b} = 2$

$\Rightarrow | \vec{a} \times \vec{b} |^{2} = | \vec{a} |^{2} | \vec{b} |^{2} - {(\vec{a} \cdot \vec{b})}^{2}$

$| \vec{a} \times \vec{b} |^{2} = 16 - 4 = 12$

Now $\vec{c} = 2 \vec{a} \times \vec{b} - 3 \vec{b}$

$\Rightarrow | \vec{c} |^{2} = 4 | \vec{a} \times \vec{b} |^{2} + 9 | \vec{b} |^{2}$

$| \vec{c} |^{2} = 4 \times 12 + 9 \times 16 = 192$

$\vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b}$ [Taking dot product with $\vec{b}$ ]

$\vec{b} \cdot \vec{c} = 0 - 3 | \vec{b} |^{2} = - 48$

Let $α$ be the angle between $\vec{b}$ and $\vec{c}$ then

$\cos α = \frac{\vec{b} \cdot \vec{c}}{| \vec{b} | | \vec{c} |} = \frac{- 48}{4 \times 8 \sqrt{3}} = \frac{- \sqrt{3}}{2} \Rightarrow α = \frac{5 π}{6}$

then $192 \sin^{2} α = 192 \sin^{2} (150 °) = 192 \times \sin^{2} (180 ° - 30 °)$

$= 192 \times \sin^{2} 30 ° = 192 \times \frac{1}{4} = 48$ .

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