Let a = i ^ + 2 j ^ + k ^ , b = 3 i ^ - 5 j ^ - k ^ , a · c = 7 , 2 b · c + 43 = 0 , a × c = b × c . The | a · b | is equal to _______ . [2023]

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Solution

Q.
Let $\vec{a} = \hat{i} + 2 \hat{j} + λ \hat{k},$ $\vec{b} = 3 \hat{i} - 5 \hat{j} - λ \hat{k},$ $\vec{a} \cdot \vec{c} = 7,$ $2 \vec{b} \cdot \vec{c} + 43 = 0,$ $\vec{a} \times \vec{c} = \vec{b} \times \vec{c} .$ The $| \vec{a} \cdot \vec{b} |$ is equal to _______ . [2023]

Ans.
(8)

$\vec{a} = \hat{i} + 2 \hat{j} + λ \hat{k},$ $\vec{b} = 3 \hat{i} - 5 \hat{j} - λ \hat{k},$ $\vec{a} \cdot \vec{c} = 7$

$Now, \vec{a} \times \vec{c} - \vec{b} \times \vec{c} = 0 \Rightarrow (\vec{a} - \vec{b}) \times \vec{c} = 0$

$\Rightarrow (\vec{a} - \vec{b}) is parallel to \vec{c}$

$∴$ $\vec{a} - \vec{b} = μ \vec{c} (where μ is a scalar)$

$\Rightarrow (\hat{i} + 2 \hat{j} + λ \hat{k}) - (3 \hat{i} - 5 \hat{j} - λ \hat{k}) = μ \cdot \vec{c}$

$\Rightarrow (1 - 3) \hat{i} + (2 + 5) \hat{j} + (λ + λ) \hat{k} = μ \cdot \vec{c}$

$\Rightarrow - 2 \hat{i} + 7 \hat{j} + 2 λ \hat{k} = μ \cdot \vec{c}$

$Now, \vec{a} \cdot \vec{c} = 7 \Rightarrow 2 λ^{2} + 12 = 7 μ$

$and \vec{b} \cdot \vec{c} = \frac{- 43}{2} gives 4 λ^{2} + 82 = 43 μ$

$∴$ $μ = 2$ and $λ^{2} = 1$

$So,$ $| \vec{a} \cdot \vec{b} |$ $= 8$

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