Question

Let $\vec{a} = 2\hat{i} + 5\hat{j} - 7\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + 5\hat{k}$. Then $(3\vec{a} - 5\vec{b}) \cdot (4\vec{a} \times 5\vec{b}) =$

• A. $-7$
• B. $0$
• C. $-13$
• D. $1$
• E. $-8$

Hint:

Don't waste time calculating the actual components of $\vec{a}$ and $\vec{b}$. In any expression $(x\vec{a} + y\vec{b}) \cdot (\vec{a} \times \vec{b})$, the answer is always $0$ regardless of the values of $\vec{a}$ and $\vec{b}$ (as long as they are defined).

Answer 1

The Correct Option is B

Concept: This problem can be solved using the properties of the scalar triple product. The cross product $\vec{a} \times \vec{b}$ results in a vector that is perpendicular to both $\vec{a}$ and $\vec{b}$. Consequently, the dot product of any vector in the plane of $\vec{a}$ and $\vec{b}$ with their cross product is always zero.

Step 1: Analyze the structure of the expression.
The expression is a dot product between two vectors:
• Vector 1: $\vec{V_1} = 3\vec{a} - 5\vec{b}$ (a linear combination of $\vec{a}$ and $\vec{b}$).
• Vector 2: $\vec{V_2} = 4\vec{a} \times 5\vec{b} = 20(\vec{a} \times \vec{b})$.

Step 2: Apply properties of the dot and cross product.
By the distributive property of the dot product over subtraction: \[ (3\vec{a} - 5\vec{b}) \cdot 20(\vec{a} \times \vec{b}) = 60[\vec{a} \cdot (\vec{a} \times \vec{b})] - 100[\vec{b} \cdot (\vec{a} \times \vec{b})] \]

Step 3: Evaluate the scalar triple products.
A scalar triple product $[\vec{x} \ \vec{y} \ \vec{z}]$ is zero if any two vectors are identical or if the vectors are coplanar.
• $\vec{a} \cdot (\vec{a} \times \vec{b}) = 0$ because $\vec{a}$ is perpendicular to $(\vec{a} \times \vec{b})$.
• $\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$ because $\vec{b}$ is perpendicular to $(\vec{a} \times \vec{b})$. Therefore: \[ 60(0) - 100(0) = 0 \]