JEE PYQ: Vector Algebra - Question ID a1344b9740b7 (JEE Main 2023)

When two vectors are perpendicular (orthogonal), their dot product is zero. The dot product of two vectors $\vec{A} = A_x \widehat{i} + A_y \widehat{j} + A_z \widehat{k}$ and $\vec{B} = B_x \widehat{i} + B_y \widehat{j} + B_z \widehat{k}$ is given by: $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$ If $\vec{A} \cdot \vec{B} = 0$, the vectors are perpendicular.

In this problem, we are given two vectors: $\vec{A} = a \widehat{i} + b \widehat{j} + \widehat{k} \quad \text{and} \quad \vec{B} = 2 \widehat{i} - 3 \widehat{j} + 4 \widehat{k}$ They are perpendicular, so their dot product must be zero. Additionally, we are given a linear constraint on $a$ and $b$: $3a + 2b = 7$ We are also told that the ratio of $a$ to $b$ is $\frac{x}{2}$, and we must find the value of $x$.

Step 1: Compute the Dot Product

Since $\vec{A}$ and $\vec{B}$ are perpendicular: $\vec{A} \cdot \vec{B} = 0$ Substitute the components: $(a)(2) + (b)(-3) + (1)(4) = 0$ Simplify: $2a - 3b + 4 = 0 \implies 2a - 3b = -4 \quad \text{(Equation 1)}$

Step 2: Use the Given Constraint

We are given: $3a + 2b = 7 \quad \text{(Equation 2)}$ Now, we have a system of two linear equations: $$\begin{cases} 2a - 3b = -4 \quad \text{(1)} \\ 3a + 2b = 7 \quad \text{(2)} \end{cases}$$

Step 3: Solve the System of Equations

We solve for $a$ and $b$ using substitution or elimination. Here, we use elimination. Multiply Equation (1) by 3 and Equation (2) by 2 to align coefficients of $a$: $$\begin{cases} 6a - 9b = -12 \quad \text{(1a)} \\ 6a + 4b = 14 \quad \text{(2a)} \end{cases}$$ Subtract (1a) from (2a): $(6a + 4b) - (6a - 9b) = 14 - (-12) \implies 13b = 26 \implies b = 2$ Substitute $b = 2$ into Equation (2): $3a + 2(2) = 7 \implies 3a + 4 = 7 \implies 3a = 3 \implies a = 1$

Step 4: Find the Ratio $a : b$ and Determine $x$

We have $a = 1$ and $b = 2$. The ratio of $a$ to $b$ is: $\frac{a}{b} = \frac{1}{2}$ But the problem states that the ratio is $\frac{x}{2}$. Therefore: $\frac{1}{2} = \frac{x}{2} \implies x = 1$

1. Misapplying the Dot Product Condition: Some students forget that the dot product must be zero for perpendicular vectors and instead set the cross product to zero, which is incorrect for this context.

2. Incorrectly Solving the System of Equations: Students often make arithmetic errors when solving the system of equations, especially when using elimination. Always double-check calculations.

3. Misinterpreting the Ratio: The problem states that the ratio of $a$ to $b$ is $\frac{x}{2}$, not $\frac{a}{b} = x$. Students may confuse the order and set $\frac{a}{b} = x$, leading to $x = \frac{1}{2}$ instead of $x = 1$.

Exam Tip: Always verify your solution by plugging the values back into the original equations. Here, $a = 1$ and $b = 2$ satisfy both $2a - 3b = -4$ and $3a + 2b = 7$, confirming correctness.