Question

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

• A. 27
• B. 25
• C. 25
• D. 21

Answer 1

Answer: (B)

25

Answer 2

Given vectors:

$\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}, \quad \vec{c} = 3\hat{i} - \hat{j} + \lambda \hat{k}.$

The sum of vectors $\vec{b} + \vec{c}$ is given by:

$\vec{b} + \vec{c} = (2\hat{i} + 3\hat{j} - 5\hat{k}) + (3\hat{i} - \hat{j} + \lambda \hat{k}) = (5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k}).$ The magnitude of $\vec{b} + \vec{c}$ is: $|\vec{b} + \vec{c}| = \sqrt{5^2 + 2^2 + (\lambda - 5)^2} = \sqrt{25 + 4 + (\lambda - 5)^2}.$

Simplifying:

$|\vec{b} + \vec{c}| = \sqrt{29 + (\lambda - 5)^2}.$

The unit vector along $\vec{b} + \vec{c}$ is:

$\vec{r} = \frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|} = \frac{5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k}}{\sqrt{29 + (\lambda - 5)^2}}.$ Given that $\vec{r} \cdot \vec{a} = 3$,

we have: $\frac{1}{\sqrt{29 + (\lambda - 5)^2}} (5 \cdot 1 + 2 \cdot 2 + 3 \cdot (\lambda - 5)) = 3.$

Simplifying:
$\frac{1}{\sqrt{29 + (\lambda - 5)^2}} (5 + 4 + 3\lambda - 15) = 3.$
$\frac{3\lambda - 6}{\sqrt{29 + (\lambda - 5)^2}} = 3.$

Cross-multiplying:
$\lambda - 2 = \sqrt{29 + (\lambda - 5)^2}.$

Squaring both sides:
$(\lambda - 2)^2 = 29 + (\lambda - 5)^2.$

Expanding both sides:
$\lambda^2 - 4\lambda + 4 = 29 + \lambda^2 - 10\lambda + 25.$ Simplifying:
$-4\lambda + 4 = 54 - 10\lambda.$

Rearranging terms:
$6\lambda = 50 \implies \lambda = \frac{50}{6} = \frac{25}{3}.$

Thus:
$3\lambda = 3 \times \frac{25}{3} = 25.$ Therefore:
$25.$