Let u = i ^ - j ^ - 2 k ^ , v = 2 i ^ + j ^ - k ^ , v · w = 2 and v × w = u + v . Then u · w is equal to [2023]

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Solution

Q.
Let $\vec{u} = \hat{i} - \hat{j} - 2 \hat{k},$ $\vec{v} = 2 \hat{i} + \hat{j} - \hat{k},$ $\vec{v} \cdot \vec{w} = 2$ and $\vec{v} \times \vec{w} = \vec{u} + λ \vec{v} .$ Then $\vec{u} \cdot \vec{w}$ is equal to [2023]

1 $1$

2 $2$

3 $\frac{3}{2}$

4 $- \frac{2}{3}$

Ans.
(1)

$\vec{v} \times \vec{w} = (\vec{u} + λ \vec{v}) = \hat{i} - \hat{j} - 2 \hat{k} + λ (2 \hat{i} + \hat{j} - \hat{k})$

$= (2 λ + 1) \hat{i} + (λ - 1) \hat{j} - (2 + λ) \hat{k}$

$Now, \vec{v} \cdot (\vec{v} \times \vec{w}) = 0 \Rightarrow (2 \hat{i} + \hat{j} - \hat{k}) \cdot [(2 λ + 1) \hat{i} + (λ - 1) \hat{j} - (λ + 2) \hat{k}] = 0$

$\Rightarrow 2 (2 λ + 1) + λ - 1 + λ + 2 = 0 \Rightarrow 6 λ + 3 = 0$

$\Rightarrow λ = - \frac{1}{2}$

$\vec{w} \cdot (\vec{v} \times \vec{w}) = \vec{w} \cdot (\vec{u} + λ \vec{v}) = \vec{u} \cdot \vec{w} + λ \vec{v} \cdot \vec{w} = 0$

$\vec{u} \cdot \vec{w} = - λ \vec{v} \cdot \vec{w};$ $\vec{u} \cdot \vec{w} = \frac{1}{2} \times 2 = 1$

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