JEE PYQ: Vector Algebra - Question ID 75b0a537a306 (JEE Main 2024)

In vector algebra, the magnitude of the sum and difference of two vectors $\vec{a}$ and $\vec{b}$ can be expressed using the dot product. The key formulas are:

• Magnitude of sum: $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}$
• Magnitude of difference: $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}$
• Dot product relation: $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

The problem gives the angle between $\vec{a}$ and $\vec{b}$ as $\theta = \cos^{-1}\left(\frac{5}{9}\right)$, so $\cos\theta = \frac{5}{9}$. It also relates the magnitudes of the sum and difference of the vectors, and states that $|\vec{a}| = n|\vec{b}|$, where $n$ is an integer to be found.

Let $|\vec{b}| = b$. Then $|\vec{a}| = nb$.

The problem states: $|\vec{a} + \vec{b}| = \sqrt{2}|\vec{a} - \vec{b}|$ Square both sides to eliminate the square root: $|\vec{a} + \vec{b}|^2 = 2|\vec{a} - \vec{b}|^2$

Using the formulas for $|\vec{a} + \vec{b}|^2$ and $|\vec{a} - \vec{b}|^2$: $|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 2\left(|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}\right)$ Substitute $|\vec{a}| = nb$ and $|\vec{b}| = b$: $(nb)^2 + b^2 + 2\vec{a} \cdot \vec{b} = 2\left((nb)^2 + b^2 - 2\vec{a} \cdot \vec{b}\right)$ Simplify: $n^2b^2 + b^2 + 2\vec{a} \cdot \vec{b} = 2n^2b^2 + 2b^2 - 4\vec{a} \cdot \vec{b}$

Bring all terms to one side: $n^2b^2 + b^2 + 2\vec{a} \cdot \vec{b} - 2n^2b^2 - 2b^2 + 4\vec{a} \cdot \vec{b} = 0$ Simplify: $-n^2b^2 - b^2 + 6\vec{a} \cdot \vec{b} = 0$ Rearrange: $6\vec{a} \cdot \vec{b} = n^2b^2 + b^2$ Factor out $b^2$: $6\vec{a} \cdot \vec{b} = b^2(n^2 + 1)$ Divide both sides by $b^2$ (since $b \neq 0$): $6 \cdot \frac{\vec{a} \cdot \vec{b}}{b^2} = n^2 + 1$

Recall that $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = nb \cdot b \cdot \frac{5}{9} = \frac{5}{9}nb^2$. Substitute this into the equation: $6 \cdot \frac{\frac{5}{9}nb^2}{b^2} = n^2 + 1$ Simplify: $6 \cdot \frac{5}{9}n = n^2 + 1$ $\frac{30}{9}n = n^2 + 1$ Simplify the fraction: $\frac{10}{3}n = n^2 + 1$ Multiply both sides by 3 to eliminate the denominator: $10n = 3n^2 + 3$ Rearrange: $3n^2 - 10n + 3 = 0$

The quadratic equation is: $3n^2 - 10n + 3 = 0$ Use the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 3$, $b = -10$, and $c = 3$: $n = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3}$ $n = \frac{10 \pm \sqrt{100 - 36}}{6}$ $n = \frac{10 \pm \sqrt{64}}{6}$ $n = \frac{10 \pm 8}{6}$ This gives two solutions: $n = \frac{10 + 8}{6} = \frac{18}{6} = 3$ $n = \frac{10 - 8}{6} = \frac{2}{6} = \frac{1}{3}$ Since $n$ must be an integer, the valid solution is $n = 3$.

1. Incorrectly squaring the magnitudes: Students often forget to square the $\sqrt{2}$ when squaring both sides of $|\vec{a} + \vec{b}| = \sqrt{2}|\vec{a} - \vec{b}|$. This leads to incorrect expansion and simplification.
2. Misapplying the dot product formula: Some students confuse the angle in the dot product formula or forget that $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$. Always double-check the angle and the formula.
3. Algebraic errors in simplification: The quadratic equation derived can be tricky to solve. Students may make mistakes in simplifying fractions or solving the quadratic. Always verify the solutions by plugging them back into the original equation.
4. Ignoring the integer constraint: The problem specifies that $n$ must be an integer. Even if you get two solutions, only one will satisfy this condition. Always check the context of the problem for such constraints.