Question

If \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\), \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\), \(\vec{c}=\hat{i}+2\hat{j}-\hat{k}\) then the value of \(\left|\begin{matrix}\vec{a}\cdot\vec{a}&\vec{a}\cdot\vec{b}&\vec{a}\cdot\vec{c}\\ \vec{b}\cdot\vec{a}&\vec{b}\cdot\vec{b}&\vec{b}\cdot\vec{c}\\ \vec{c}\cdot\vec{a}&\vec{c}\cdot\vec{b}&\vec{c}\cdot\vec{c}\end{matrix}\right|\) is equal to:

• A. 64
• B. 0
• C. 14
• D. 16

Hint:

The scalar triple product $\left[\vec{a} \ \vec{b} \ \vec{c}\right]$ represents the geometric volume of the parallelepiped formed by the three vectors. The Gram determinant identity simply states that the determinant of the dot product matrix is equal to the square of this volume value, which saves you from computing nine separate individual dot products!

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The given determinant is the Gramian determinant of three vectors. There is a specific identity relating the determinant of dot products of three vectors to their scalar triple product: \[ \text{Det}(\text{Gramian}) = [\vec{a} \cdot (\vec{b} \times \vec{c})]^2 = [\vec{a} \vec{b} \vec{c}]^2 \] Step 2: Key Formula or Approach:
1. Calculate the scalar triple product \( [\vec{a} \vec{b} \vec{c}] \).
2. Square the result.
Step 3: Detailed Explanation:
\[ [\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & 1 & 1
1 & -1 & 1
1 & 2 & -1 \end{vmatrix} \] Expanding along the first row: \[ = 1((-1)(-1) - (2)(1)) - 1((1)(-1) - (1)(1)) + 1((1)(2) - (1)(-1)) \] \[ = 1(1 - 2) - 1(-1 - 1) + 1(2 + 1) \] \[ = -1 + 2 + 3 = 4 \] The value of the Gramian determinant is: \[ [\vec{a} \vec{b} \vec{c}]^2 = 4^2 = 16 \] Step 4: Final Answer:
The value of the determinant is 16.