Question

Let $\vec{a}=\hat{i}+2\hat{j}+4\hat{k}$, $\vec{b}=2\hat{i}+4\hat{j}+8\hat{k}$ and $\vec{c}=2\hat{i}+4\hat{j}+3\hat{k}$. Then $(\vec{a}\times\vec{b})\cdot\vec{c}=$ ________.

• A. 4
• B. 6
• C. 8
• D. 2
• E. 0

Hint:

The scalar triple product is 0 if vectors are linearly dependent (parallel or coplanar).

Answer 1

Answer: (E)

To solve the given problem, we need to find the value of \((\vec{a}\times\vec{b})\cdot\vec{c}\) where:

• \(\vec{a} = \hat{i} + 2\hat{j} + 4\hat{k}\)
• \(\vec{b} = 2\hat{i} + 4\hat{j} + 8\hat{k}\)
• \(\vec{c} = 2\hat{i} + 4\hat{j} + 3\hat{k}\)

Step 1: Compute the cross product \(\vec{a} \times \vec{b}\)

The cross product \(\vec{a} \times \vec{b}\) is calculated using the determinant method as follows:

\(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 4 & 8 \end{vmatrix}\)

Calculating the determinant, we get:

• Coefficient of \(\hat{i}\): \((2 \times 8 - 4 \times 4) = (16 - 16) = 0\)
• Coefficient of \(\hat{j}\): \(-(1 \times 8 - 4 \times 2) = -(8 - 8) = 0\)
• Coefficient of \(\hat{k}\): \((1 \times 4 - 2 \times 2) = (4 - 4) = 0\)

Thus, \(\vec{a} \times \vec{b} = 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0}\).

Step 2: Compute the dot product \((\vec{a} \times \vec{b}) \cdot \vec{c}\)

Since \(\vec{a} \times \vec{b} = \vec{0}\), the dot product \((\vec{a} \times \vec{b}) \cdot \vec{c}\) is:

• \((\vec{0}) \cdot \vec{c} = 0 \cdot 2 + 0 \cdot 4 + 0 \cdot 3 = 0\)

Therefore, the value of \((\vec{a} \times \vec{b}) \cdot \vec{c}\) is 0.

Hence, the correct answer is 0.