Question

Let: $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=5 \hat{i}-3 \hat{j}+3 \hat{k}$ be there vectors If $\vec{r}$ is a vector such that, $\vec{r} \times \vec{b}=\vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a}=0$, then $25|\vec{r}|^2$ is equal to

• A. 560
• B. 339
• C. 449
• D. 336

Hint:

When dealing with vector cross products and dot products, ensure that you correctly substitute values and solve for the unknowns. Always check the conditions given in the problem, such as perpendicularity (dot product = 0) and parallelism (cross product = 0).

Answer 1

The Correct Option is B

We are given the following vectors: \[ \mathbf{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \mathbf{b} = \hat{i} - \hat{j} + 2\hat{k}, \quad \mathbf{c} = 5\hat{i} - 3\hat{j} + 3\hat{k}. \] Step 1: We know that \( \mathbf{r} \times \mathbf{b} = \mathbf{c} \times \mathbf{b} \) and \( \mathbf{r} \cdot \mathbf{a} = 0 \). From the condition \( \mathbf{r} \times \mathbf{b} = \mathbf{c} \times \mathbf{b} \), we have: \[ \mathbf{r} - \mathbf{c} = \lambda \mathbf{b} \quad \text{(where \( \lambda \) is a constant)}. \] Step 2: Next, substitute the expression for \( \mathbf{r} \): \[ \mathbf{r} = \mathbf{c} + \lambda \mathbf{b}. \] Substituting the values of \( \mathbf{c} \) and \( \mathbf{b} \), we get: \[ \mathbf{r} = 5\hat{i} - 3\hat{j} + 3\hat{k} + \lambda (\hat{i} - \hat{j} + 2\hat{k}). \] Thus, the vector \( \mathbf{r} \) is: \[ \mathbf{r} = (5 + \lambda)\hat{i} + (-3 - \lambda)\hat{j} + (3 + 2\lambda)\hat{k}. \] Step 3: Using the condition \( \mathbf{r} \cdot \mathbf{a} = 0 \), we calculate the dot product: \[ \mathbf{r} \cdot \mathbf{a} = (5 + \lambda)(1) + (-3 - \lambda)(2) + (3 + 2\lambda)(3). \] Simplifying: \[ \mathbf{r} \cdot \mathbf{a} = 5 + \lambda - 6 - 2\lambda + 9 + 6\lambda = 8 + 5\lambda = 0. \] Thus, solving for \( \lambda \), we get: \[ 5\lambda = -8 \quad \Rightarrow \quad \lambda = -\frac{8}{5}. \] Step 4: Substitute \( \lambda = -\frac{8}{5} \) into the expression for \( \mathbf{r} \): \[ \mathbf{r} = \left( 5 - \frac{8}{5} \right) \hat{i} + \left( -3 + \frac{8}{5} \right) \hat{j} + \left( 3 - \frac{16}{5} \right) \hat{k}. \] Simplifying: \[ \mathbf{r} = \frac{17}{5} \hat{i} - \frac{7}{5} \hat{j} + \frac{1}{5} \hat{k}. \] Step 5: Now, calculate \( |\mathbf{r}|^2 \): \[ |\mathbf{r}|^2 = \left( \frac{17}{5} \right)^2 + \left( \frac{-7}{5} \right)^2 + \left( \frac{1}{5} \right)^2 = \frac{289}{25} + \frac{49}{25} + \frac{1}{25} = \frac{339}{25}. \] Thus: \[ 25|\mathbf{r}|^2 = 25 \times \frac{339}{25} = 339. \]