Question

If $\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}$, $\vec{b} = -3\hat{i} + 5\hat{j} - 4\hat{k}$, $\vec{c} = 6\hat{i} - 4\hat{j} + 5\hat{k}$, then $(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) = $

• A. $-216$
• B. $243$
• C. $81$
• D. $-27$

Hint:

Vector Quadruple Product. Use Lagrange identity for scalar triple products to simplify vector expressions efficiently.

Answer 1

The Correct Option is A

Use Lagrange identity: \[ (\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c}) - (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{b}) \] \[ \vec{a} \cdot \vec{b} = -23,\quad \vec{b} \cdot \vec{c} = -58,\quad \vec{a} \cdot \vec{c} = 31,\quad \vec{b} \cdot \vec{b} = 50 \] \[ (-23)(-58) - (31)(50) = 1334 - 1550 = -216 \]

Answer 2

Step 1: Understand the problem
We are given three vectors:
\[ \vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}, \quad \vec{b} = -3\hat{i} + 5\hat{j} - 4\hat{k}, \quad \vec{c} = 6\hat{i} - 4\hat{j} + 5\hat{k} \]
We need to find \((\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c})\).

Step 2: Use the vector identity
Recall the identity:
\[ (\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c}) - (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{b}) \]

Step 3: Calculate dot products
\[ \vec{a} \cdot \vec{b} = (2)(-3) + (-1)(5) + (3)(-4) = -6 -5 -12 = -23 \]
\[ \vec{b} \cdot \vec{c} = (-3)(6) + (5)(-4) + (-4)(5) = -18 -20 -20 = -58 \]
\[ \vec{a} \cdot \vec{c} = (2)(6) + (-1)(-4) + (3)(5) = 12 + 4 + 15 = 31 \]
\[ \vec{b} \cdot \vec{b} = (-3)^2 + 5^2 + (-4)^2 = 9 + 25 + 16 = 50 \]

Step 4: Substitute values in identity
\[ (\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) = (-23)(-58) - (31)(50) = 1334 - 1550 = -216 \]

Final answer: \(-216\)