Key Concepts and Formulas

• Vector Addition: Summing vectors by adding their corresponding components.
• Magnitude of a Vector: Calculating the length of a vector using the Pythagorean theorem in 3D.
• Unit Vector: Finding a vector of unit length in a specific direction.
• Dot Product: Computing the scalar product of two vectors.

Step-by-Step Solution

Step 1: Calculate the sum of vectors $\vec{b}$ and $\vec{c}$. The problem states that $\vec{r}$ is a unit vector along $\vec{b}+\vec{c}$. We first need to find this resultant vector by adding their components. Given: $\vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}$ $\vec{c} = 3\hat{i} - \hat{j} + \lambda\hat{k}$ Adding them: $\vec{b}+\vec{c} = (2+3)\hat{i} + (3-1)\hat{j} + (-5+\lambda)\hat{k}$ $\vec{b}+\vec{c} = 5\hat{i} + 2\hat{j} + (\lambda-5)\hat{k}$

Step 2: Determine the unit vector $\vec{r}$. A unit vector in the direction of a vector $\vec{V}$ is given by $\vec{r} = \frac{\vec{V}}{|\vec{V}|}$. Here, $\vec{V} = \vec{b}+\vec{c}$. First, we find the magnitude of $\vec{b}+\vec{c}$: $|\vec{b}+\vec{c}| = \sqrt{(5)^2 + (2)^2 + (\lambda-5)^2}$ $|\vec{b}+\vec{c}| = \sqrt{25 + 4 + (\lambda^2 - 10\lambda + 25)}$ $|\vec{b}+\vec{c}| = \sqrt{\lambda^2 - 10\lambda + 54}$ Now, the unit vector $\vec{r}$ is: $\vec{r} = \frac{5\hat{i} + 2\hat{j} + (\lambda-5)\hat{k}}{\sqrt{\lambda^2 - 10\lambda + 54}}$

Step 3: Apply the dot product condition $\vec{r} \cdot \vec{a} = 3$. We are given $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{r} \cdot \vec{a} = 3$. $\left( \frac{5\hat{i} + 2\hat{j} + (\lambda-5)\hat{k}}{\sqrt{\lambda^2 - 10\lambda + 54}} \right) \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) = 3$ Multiplying the corresponding components: $\frac{(5)(1) + (2)(2) + (\lambda-5)(3)}{\sqrt{\lambda^2 - 10\lambda + 54}} = 3$ $\frac{5 + 4 + 3\lambda - 15}{\sqrt{\lambda^2 - 10\lambda + 54}} = 3$ $\frac{3\lambda - 6}{\sqrt{\lambda^2 - 10\lambda + 54}} = 3$ Divide both sides by 3: $\frac{\lambda - 2}{\sqrt{\lambda^2 - 10\lambda + 54}} = 1$

Step 4: Solve the equation for $\lambda$. To eliminate the square root, we square both sides of the equation: $(\lambda - 2)^2 = \left( \sqrt{\lambda^2 - 10\lambda + 54} \right)^2$ $\lambda^2 - 4\lambda + 4 = \lambda^2 - 10\lambda + 54$ The $\lambda^2$ terms cancel out: $-4\lambda + 4 = -10\lambda + 54$ Rearrange the terms to solve for $\lambda$: $10\lambda - 4\lambda = 54 - 4$ $6\lambda = 50$ $\lambda = \frac{50}{6} = \frac{25}{3}$

Step 5: Calculate the value of $3\lambda$. The question asks for the value of $3\lambda$. $3\lambda = 3 \times \frac{25}{3}$ $3\lambda = 25$

Common Mistakes & Tips

• Squaring Equations: Be cautious when squaring both sides of an equation, as it can introduce extraneous solutions. Always check your final answer in the original equation if time permits.
• Algebraic Errors: Carefully expand squared terms (e.g., $(\lambda-5)^2$) and simplify expressions to avoid arithmetic mistakes.
• Magnitude Calculation: Ensure the correct formula for the magnitude of a 3D vector is used: $\sqrt{x^2+y^2+z^2}$.

Summary

The problem required us to first find the sum of vectors $\vec{b}$ and $\vec{c}$, then determine the unit vector $\vec{r}$ in that direction. By applying the given dot product condition $\vec{r} \cdot \vec{a} = 3$, we derived an equation involving $\lambda$. Solving this equation, after squaring both sides to remove the radical, yielded $\lambda = \frac{25}{3}$. Finally, we calculated $3\lambda$ to find the answer.

The final answer is $\boxed{25}$, which corresponds to option (B).