JEE PYQ: Vector Algebra - Question ID 847b23cdf307 (JEE Main 2018)

In vector algebra, the dot product (scalar product) of two vectors $\vec{A}$ and $\vec{B}$ is defined as: $\vec{A} \cdot \vec{B} = |\vec{A}|\,|\vec{B}|\,\cos\theta$ where $\theta$ is the angle between them. When two vectors are given in Cartesian form, say $\vec{A} = a_x\,\hat{i} + a_y\,\hat{j}$ and $\vec{B} = b_x\,\hat{i} + b_y\,\hat{j}$, their dot product is computed component-wise: $\vec{A} \cdot \vec{B} = a_x b_x + a_y b_y.$

In this problem we are given three coplanar vectors $\vec{A}$, $\vec{B}$, and $\vec{C}$ and the conditions $\vec{A} \cdot \vec{C} = \vec{B} \cdot \vec{C} = \vec{A} \cdot \vec{B}.$ These conditions translate into two linear equations for the components of $\vec{C}$. Solving those equations yields $\vec{C}$ up to a scalar multiple; its magnitude then follows from the Pythagorean theorem.

1. Write the given vectors in component form.
Given $\vec{A} = \hat{i} + \hat{j} \quad\Longrightarrow\quad A_x = 1,\quad A_y = 1;$ $\vec{B} = 2\,\hat{i} - \hat{j} \quad\Longrightarrow\quad B_x = 2,\quad B_y = -1.$

2. Compute $\vec{A}\cdot\vec{B}$.
$\vec{A}\cdot\vec{B} = A_x B_x + A_y B_y = (1)(2) + (1)(-1) = 2 - 1 = 1.$ Hence the common value of the dot products is $1$.

3. Set up the two linear equations for $\vec{C}=(C_x,C_y)$.
Condition 1: $\vec{A}\cdot\vec{C} = 1 \quad\Longrightarrow\quad 1\cdot C_x + 1\cdot C_y = 1 \quad\Longrightarrow\quad C_x + C_y = 1.$ Condition 2: $\vec{B}\cdot\vec{C} = 1 \quad\Longrightarrow\quad 2\,C_x - 1\,C_y = 1 \quad\Longrightarrow\quad 2\,C_x - C_y = 1.$

4. Solve the system of equations.
We have $$\begin{cases} C_x + C_y = 1,\\ 2\,C_x - C_y = 1. \end{cases}$$ Add the two equations to eliminate $C_y$: $3\,C_x = 2 \quad\Longrightarrow\quad C_x = \tfrac{2}{3}.$ Substitute back into the first equation: $\tfrac{2}{3} + C_y = 1 \quad\Longrightarrow\quad C_y = 1 - \tfrac{2}{3} = \tfrac{1}{3}.$ Thus $\vec{C} = \tfrac{2}{3}\,\hat{i} + \tfrac{1}{3}\,\hat{j}.$

5. Compute the magnitude of $\vec{C}$.
$|\vec{C}| = \sqrt{\bigl(\tfrac{2}{3}\bigr)^2 + \bigl(\tfrac{1}{3}\bigr)^2} = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}.$

6. Match with the given options.
The magnitude is $\sqrt{\tfrac{5}{9}}$, which corresponds to option B.

1. Misinterpreting the dot-product conditions. Students sometimes treat $\vec{A}\cdot\vec{C}=\vec{B}\cdot\vec{C}$ as an orthogonality condition rather than an equality of two scalars. 2. Sign errors in components. In $\vec{B}=2\hat{i}-\hat{j}$, the negative sign on the $j$–component is easy to overlook when setting up the second equation. 3. Forgetting to take the square root. After computing $C_x^2+C_y^2$, one must remember to take the square root to obtain the magnitude. 4. Assuming $\vec{C}$ is unique. In fact the system has exactly one solution, but if the two conditions were not independent one would have to consider a family of solutions.

Exam Tip: Always write down the component form of the dot product explicitly. That turns a vector condition into a simple linear equation, which is much easier to solve.