Question

Given: $$ \vec{F} = 2\hat{i} + 2\hat{j} + 5\hat{k},\quad A = (1, 2, 5),\quad B = (-1, -2, -3) $$ and: $$ \overrightarrow{BA} \times \vec{F} = 4\hat{i} + 6\hat{j} + 2\lambda \hat{k} $$ Find the value of $ \lambda $.

• A. 0
• B. 1
• C. 2
• D. -2

Hint:

Always compute vector cross product using the determinant formula for 3D vectors.

Answer 1

The Correct Option is D

Step 1: Compute \( \vec{BA} = \vec{A} - \vec{B} \) \[ \vec{BA} = (1 - (-1),\ 2 - (-2),\ 5 - (-3)) = (2, 4, 8) = 2\hat{i} + 4\hat{j} + 8\hat{k} \] Step 2: Cross product \( \vec{BA} \times \vec{F} \) \[ \vec{F} = 2\hat{i} + 2\hat{j} + 5\hat{k} \] Use determinant form: \[ \vec{BA} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\2 & 4 & 8 \\ 2 & 2 & 5 \end{vmatrix} = \hat{i}(4 \cdot 5 - 8 \cdot 2) - \hat{j}(2 \cdot 5 - 8 \cdot 2) + \hat{k}(2 \cdot 2 - 4 \cdot 2) \] \[ = \hat{i}(20 - 16) - \hat{j}(10 - 16) + \hat{k}(4 - 8) = 4\hat{i} + 6\hat{j} -4\hat{k} \] Compare with: \[ 4\hat{i} + 6\hat{j} + 2\lambda \hat{k} \Rightarrow 2\lambda = -4 \Rightarrow \boxed{\lambda = -2} \]