Question

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

• A. $-30$
• B. $84$
• C. $70$
• D. $984$

Hint:

Using vector identity dot products ($\vec{a}\cdot\vec{b}$) is drastically faster and significantly less prone to sign errors than setting up and computing two separate $3 \times 3$ cross product determinants manually!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question requires us to compute the dot product of two cross products, $(\vec{a} \times \vec{b})$ and $(\vec{a} \times \vec{c})$, using three given vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$.

Step 2: Key Formula or Approach:
We can solve this either by calculating the individual vector cross products using determinants and then finding their dot product, or by using the standard vector identity for the dot product of cross products:
$$(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})$$ Replacing $\vec{c}$ with $\vec{a}$ and $\vec{d}$ with $\vec{c}$, the identity becomes:
$$(\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{c}) = (\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{c}) - (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{a})$$

Step 3: Detailed Explanation:
Let's calculate the scalar dot products required for the vector identity:
1. Calculate $\vec{a} \cdot \vec{a}$:
$$\vec{a} \cdot \vec{a} = (2)^2 + (3)^2 + (-1)^2 = 4 + 9 + 1 = 14$$ 2. Calculate $\vec{b} \cdot \vec{c}$:
$$\vec{b} \cdot \vec{c} = (-1)(1) + (2)(1) + (-4)(-2) = -1 + 2 + 8 = 9$$ 3. Calculate $\vec{a} \cdot \vec{c}$:
$$\vec{a} \cdot \vec{c} = (2)(1) + (3)(1) + (-1)(-2) = 2 + 3 + 2 = 7$$ 4. Calculate $\vec{b} \cdot \vec{a}$:
$$\vec{b} \cdot \vec{a} = (-1)(2) + (2)(3) + (-4)(-1) = -2 + 6 + 4 = 8$$ Now, substitute these scalar values into our vector identity expression:
$$(\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{c}) = (14 \times 9) - (7 \times 8)$$ $$(\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{c}) = 126 - 56 = 70$$ This perfectly matches option (C).

Step 4: Final Answer:
The scalar value of the expression is $70$, which corresponds to option (C).