Question

Let $\vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}, \vec{a} \cdot \vec{c}=7,2 \vec{b} \cdot \vec{c}+43=0, \vec{a} \times \vec{c}=\vec{b} \times \vec{c}$. Then $|\vec{a} \cdot \vec{b}|$ is equal to

Answer 1

Correct Answer: 8

Step 1: Use the cross-product property 

From the given condition \( \mathbf{a} \times \mathbf{c} = \mathbf{b} \times \mathbf{c} \), we get:

\[ (\mathbf{a} - \mathbf{b}) \times \mathbf{c} = 0. \]

This implies that \( \mathbf{a} - \mathbf{b} \) is parallel to \( \mathbf{c} \). Let \( \mathbf{c} = k(\mathbf{a} - \mathbf{b}) \), where \( k \) is a scalar.

Step 2: Substitute given dot products

We are given \( \mathbf{a} \cdot \mathbf{c} = 7 \). Substituting \( \mathbf{c} = k(\mathbf{a} - \mathbf{b}) \), we get:

\[ \mathbf{a} \cdot k(\mathbf{a} - \mathbf{b}) = 7. \] Expanding this expression: \[ k(\mathbf{a} \cdot \mathbf{a} - \mathbf{a} \cdot \mathbf{b}) = 7. \] Similarly, we are given \( 2 \mathbf{b} \cdot \mathbf{c} + 43 = 0 \), so: \[ 2 \mathbf{b} \cdot k(\mathbf{a} - \mathbf{b}) = -43. \] 

Step 3: Solve for \( k \) and \( \lambda \)

Substitute \( \mathbf{a} = \hat{i} + 2\hat{j} + \lambda \hat{k} \) and \( \mathbf{b} = 3\hat{i} - 5\hat{j} - \lambda \hat{k} \). After simplifying, we find:

\[ k = \frac{1}{2}, \quad \lambda = \pm 1. \] 

Step 4: Calculate \( | \mathbf{a} \cdot \mathbf{b} | \)

Now, calculate the dot product \( \mathbf{a} \cdot \mathbf{b} \). We have:

\[ \mathbf{a} \cdot \mathbf{b} = (1)(3) + (2)(-5) + (\lambda)(-\lambda) = 3 - 10 - \lambda^2. \] Using \( \lambda^2 = 1 \), we get: \[ \mathbf{a} \cdot \mathbf{b} = -8. \] Therefore, the magnitude is: \[ |\mathbf{a} \cdot \mathbf{b}| = 8. \]