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 $\begin{vmatrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \\ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c} \end{vmatrix}$ is equal to
This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
$0$
Vector Dot Products
The given vectors are:
- $\vec{a} = \hat{i} + \hat{j} + \hat{k}$
- $\vec{b} = \hat{i} - \hat{j} + \hat{k}$
- $\vec{c} = \hat{i} + 2\hat{j} - \hat{k}$
Calculate the required dot products:
- $\vec{a} \cdot \vec{a} = (1)(1) + (1)(1) + (1)(1) = 3$
- $\vec{b} \cdot \vec{b} = (1)(1) + (-1)(-1) + (1)(1) = 3$
- $\vec{c} \cdot \vec{c} = (1)(1) + (2)(2) + (-1)(-1) = 1 + 4 + 1 = 6$
- $\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (1)(1) = 1 - 1 + 1 = 1$
- $\vec{a} \cdot \vec{c} = (1)(1) + (1)(2) + (1)(-1) = 1 + 2 - 1 = 2$
- $\vec{b} \cdot \vec{c} = (1)(1) + (-1)(2) + (1)(-1) = 1 - 2 - 1 = -2$
Gram Determinant Matrix
The determinant required is the Gram determinant, formed by the dot products:
$ D = \begin{vmatrix} \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{vmatrix} $Substituting the calculated values, we get:
$ D = \begin{vmatrix} 3 & 1 & 2 \\ 1 & 3 & -2 \\ 2 & -2 & 6 \end{vmatrix} $Determinant Value
The Gram determinant of a set of vectors is zero if and only if those vectors are linearly dependent.
The value of the determinant is $0$.
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