Question

If \( \vec{a} = (1,1,-1), \vec{b} = (-1,2,1) \) and \( \vec{c} = (-1,2,-1) \), then 

\[ \left| (\vec{a} + \vec{b}) \times (\vec{b} + \vec{c}) \right| \]

is ______.

• A. \( 2 \)
• B. \( 4 \)
• C. \( 6 \)
• D. \( 8 \)
• E. \( 10 \)

Hint:

If both vectors lie in the same plane (here \(z=0\)), their cross product will be perpendicular to that plane.

Answer 1

The Correct Option is C

Concept: The magnitude of the cross product of two vectors gives the area of the parallelogram formed by them: \[ |\vec{u} \times \vec{v}| = \sqrt{(u_2v_3 - u_3v_2)^2 + (u_3v_1 - u_1v_3)^2 + (u_1v_2 - u_2v_1)^2} \]

Step 1: Compute \( \vec{a} + \vec{b} \) and \( \vec{b} + \vec{c} \). \[ \vec{a}+\vec{b} = (1-1,\;1+2,\;-1+1) = (0,3,0) \] \[ \vec{b}+\vec{c} = (-1-1,\;2+2,\;1-1) = (-2,4,0) \]

Step 2: Find the cross product. \[ (\vec{a}+\vec{b}) \times (\vec{b}+\vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & 0 \\ -2 & 4 & 0 \end{vmatrix} \] \[ = \hat{i}(3\cdot0 - 0\cdot4) - \hat{j}(0\cdot0 - 0\cdot(-2)) + \hat{k}(0\cdot4 - 3\cdot(-2)) \] \[ = 0\hat{i} - 0\hat{j} + 6\hat{k} = (0,0,6) \]

Step 3: Find the magnitude. \[ | (0,0,6) | = \sqrt{0^2 + 0^2 + 6^2} = 6 \]