Let , and be three vectors. Let be a unit vector along . If , then is equal to : [2024]

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Solution

Q.
Let $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $\vec{b} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ and $\vec{c} = 3 \hat{i} - \hat{j} + λ \hat{k}$ be three vectors. Let $\vec{r}$ be a unit vector along $\vec{b} + \vec{c}$ . If $\vec{r} \cdot \vec{a} = 3$ , then $3 λ$ is equal to : [2024]

1 21

2 30

3 25

4 27

Ans.
(3)

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

Now, $\vec{b} + \vec{c} = 5 \hat{i} + 2 \hat{j} + (λ - 5) \hat{k}$

$\vec{r}$ is a unit vector along $\vec{b} + \vec{c} \Rightarrow \vec{r} = \frac{\vec{b} + \vec{c}}{| \vec{b} + \vec{c} |}$

$∴ \vec{r} = \frac{5 \hat{i} + 2 \hat{j} + (λ - 5) \hat{k}}{\sqrt{25 + 4 + {(λ - 5)}^{2}}}$

Now, $\vec{r} \cdot \vec{a} = 3$

$\frac{1}{\sqrt{29 + {(λ - 5)}^{2}}} [5 + 4 + 3 (λ - 5)] = 3$

$\Rightarrow \frac{1}{29 + {(λ - 5)}^{2}} {[9 + 3 (λ - 5)]}^{2} = 9$

$\Rightarrow {[3 + (λ - 5)]}^{2} = 29 + {(λ - 5)}^{2}$

$\Rightarrow 9 + {(λ - 5)}^{2} + 6 (λ - 5) = 29 + {(λ - 5)}^{2}$

$\Rightarrow 9 + 6 (λ - 5) = 29 \Rightarrow λ = \frac{25}{3}$

Hence, $3 λ = 3 \times \frac{25}{3} = 25$ .

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