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

Concept: A determinant formed entirely by the dot products of a set of vectors is known as a Gram determinant. A fundamental identity in vector algebra connects this specific determinant directly to the square of the scalar triple product of the vectors: $$\det(\text{Gram Matrix}) = \left[\vec{a} \ \vec{b} \ \vec{c}\right]^2$$ Step 1: Set up the scalar triple product determinant.
The scalar triple product $\left[\vec{a} \ \vec{b} \ \vec{c}\right]$ can be calculated by evaluating the determinant of the matrix formed by the component coefficients of the three given vectors: $$\left[\vec{a} \ \vec{b} \ \vec{c}\right] = \det \begin{pmatrix} 1 & 1 & 1
1 & -1 & 1
1 & 2 & -1 \end{pmatrix}$$

Step 2: Evaluate the scalar triple product value.
Expand the 3x3 determinant along the first row: $$\left[\vec{a} \ \vec{b} \ \vec{c}\right] = 1 \cdot \big((-1)(-1) - (1)(2)\big) - 1 \cdot \big((1)(-1) - (1)(1)\big) + 1 \cdot \big((1)(2) - (-1)(1)\big)$$ Simplify the inner arithmetic operations step-by-step: $$= 1 \cdot (1 - 2) - 1 \cdot (-1 - 1) + 1 \cdot (2 + 1)$$ $$= 1 \cdot (-1) - 1 \cdot (-2) + 1 \cdot (3)$$ $$= -1 + 2 + 3 = 4$$

Step 3: Square the result to find the Gram determinant value.
Using our core vector identity, square the scalar triple product value to find the final value of the dot product matrix determinant: $$\text{Determinant Value} = \left[\vec{a} \ \vec{b} \ \vec{c}\right]^2 = (4)^2 = 16$$ This matches option (D) perfectly.