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 $0$.
Was this answer helpful?
Similar Questions
- A vector given by $\vec{P} = f(t)\hat{i} + g(t)\hat{j} + \hat{k}$ moves in such a way that it is always parallel to the vector $\vec{Q} = -f''(t)\hat{i} + f'(t)\hat{j} + \hat{k}$. The magnitude of $\vec{P}$ is
- The position vectors of two adjacent sides $\vec{OA}$ and $\vec{OB}$ of a rectangle $OACB$ are $\vec{a}$ and $\vec{b}$ respectively, where $O$ is the origin. If $16|\vec{a} \times \vec{b}| = 3(|\vec{a}| + |\vec{b}|)^2$ and $\theta$ be the acute angle between the diagonals $OC$ and $AB$, then the value of $\tan\left(\frac{\theta}{2}\right)$ is
- The point of intersection of $\vec{r} \times \vec{a} = \vec{b} \times \vec{a}$ and $\vec{r} \times \vec{b} = \vec{a} \times \vec{b}$, where $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = 2\hat{i} - \hat{k}$ is
- Intercepts of the plane $\vec{r} \cdot \vec{n} = d \, (\neq 0)$ on the coordinate axes respectively are
- Consider a square $ABCD$ of diagonal length $2a$. The square is folded along the diagonal $AC$ so that the plane of $\Delta ABC$ is perpendicular to the plane of $\Delta ADC$. In this case the shortest distance between $AB$ and $CD$ is
- Let $\vec{a} = (x, y, z)$ be the vector with $|\vec{a}| = 2\sqrt{3}$, which makes equal angles with the vector $\vec{b} = (y, -2z, 3x)$ and $\vec{c} = (2z, 3x, -y)$ and is perpendicular to the vector $\vec{d} = (1, -1, 2)$. If the angle between $\vec{a}$ and the unit vector $\hat{j}$ is obtuse, then $\vec{a}$ is
- Let $\vec{r} = \sin x(\vec{a} \times \vec{b}) + \cos y(\vec{b} \times \vec{c}) + 2(\vec{c} \times \vec{a})$, where $\vec{a}, \vec{b}$ and $\vec{c}$ are three non-coplanar vectors. It is given that $\vec{r}$ is perpendicular to $(\vec{a} + \vec{b} + \vec{c})$. Then the possible value(s) of $(x^2 + y^2)$ is/are
- If a vector $\vec{v} = 3\hat{i}$ is rotated in the $x-z$ plane by an angle $\theta$ with respect to $x$-axis in the clockwise direction, then for an observer at $+y$ axis the vector will be
- Three vectors $\vec a,\vec b,\vec c$ are such that $|\vec a|=2,\ |\vec b|=2,\ |\vec c|=4$ and $\vec a+\vec b+\vec c=0$. Then the value of $4\vec a\cdot\vec b+3\vec b\cdot\vec c+3\vec c\cdot\vec a$ will be
Important Questions from Vectors and 3D Geometry
- Let the area of the triangle formed by the lines $x+2=y-1=z$, $\frac{x-3}{5} = \frac{y}{-1} = \frac{z-1}{1}$ and $\frac{x}{-3} = \frac{y-3}{3} = \frac{z-2}{1}$ be $A$. Then $A^2$ is equal to
- Let the line $L_1$ be parallel to the vector $-3\hat{i} + 2\hat{j} + 4\hat{k}$ and pass through the point $(2, 6, 7)$, and the line $L_2$ be parallel to the vector $2\hat{i} + \hat{j} + 3\hat{k}$ and pass through the point $(4, 3, 5)$. If the line $L_3$ is parallel to the vector $-3\hat{i} + 5\hat{j} + 16\hat{k}$ and intersects the lines $L_1$ and $L_2$ at the points $C$ and $D$, respectively, then $| \vec{CD} |^2$ is equal to :
- Let the line $L$ pass through the point $(-3, 5, 2)$ and make equal angles with the positive coordinate axes. If the distance of $L$ from the point $(-2, r, 1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of $r$ is :
- If the image of the point $P(1, 2, a)$ in the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{7-z}{2}$ is $Q(5, b, c)$, then $a^2 + b^2 + c^2$ is equal to
- Let $\vec{AB} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{AD} = \hat{i} + 2\hat{j} + \lambda\hat{k}, \lambda \in \mathbb{R}$. Let the projection of the vector $\vec{v} = \hat{i} + \hat{j} + \hat{k}$ on the diagonal $\vec{AC}$ of the parallelogram ABCD be of length one unit. If $\alpha, \beta$, where $\alpha > \beta$, be the roots of the equation $\lambda^2 x^2 - 6\lambda x + 5 = 0$, then $2\alpha - \beta$ is equal to
Need Expert Advice?
More Questions from WBJEE
