Question

If $\vec{a} = 2\vec{i} + 3\vec{j} - \vec{k}$, $\vec{b} = -\vec{i} + 2\vec{j} - 4\vec{k}$ and $\vec{c} = \vec{i} + \vec{j} + \vec{k}$, then $(\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{c}) =$

• A. $-74$
• B. $74$
• C. $-42$
• D. $42$

Hint:

Using Lagrange's identity for cross products saves significant time compared to computing the cross products $\vec{a}\times\vec{b}$ and $\vec{a}\times\vec{c}$ individually.

Answer 1

The Correct Option is A

Step 1: A handy box-product identity.
Instead of finding two cross products, use \[ (\vec a \times \vec b)\cdot(\vec a \times \vec c) = (\vec a\cdot\vec a)(\vec b\cdot\vec c) - (\vec a\cdot\vec c)(\vec b\cdot\vec a) \] This turns everything into easy dot products.

Step 2: Compute $\vec a\cdot\vec a$.
\[ \vec a\cdot\vec a = 4 + 9 + 1 = 14 \]

Step 3: Compute $\vec b\cdot\vec c$.
\[ \vec b\cdot\vec c = -1 + 2 - 4 = -3 \]

Step 4: Compute $\vec a\cdot\vec c$ and $\vec a\cdot\vec b$.
\[ \vec a\cdot\vec c = 2 + 3 - 1 = 4, \qquad \vec a\cdot\vec b = -2 + 6 + 4 = 8 \]

Step 5: Substitute.
\[ (14)(-3) - (4)(8) = -42 - 32 \]

Step 6: Final value.
\[ = -74 \] \[ \boxed{ -74 } \]