JEE Main 2025 Apr 2 Shift 1, Mathematics Q11: Scalar (or Dot) Product of Two Vectors

Let $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ as a unit vector, so $a_1^2+a_2^2+a_3^2=1$.

Define the three reference vectors:

$\vec{b}=2\hat{i}-\hat{j}+2\hat{k}$ with $|\vec{b}|=3$

$\vec{c}=\hat{i}+2\hat{j}-2\hat{k}$ with $|\vec{c}|=3$

$\vec{d}=\hat{k}$ with $|\vec{d}|=1$

Equality of projections of $\vec{a}$ on these vectors:

$$\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}=\frac{\vec{a}\cdot \vec{c}}{|\vec{c}|}=\frac{\vec{a}\cdot \vec{d}}{|\vec{d}|}$$

$$\frac{2a_1-a_2+2a_3}{3}=\frac{a_1+2a_2-2a_3}{3}=a_3$$

From the first two equal: $2a_1-a_2+2a_3=a_1+2a_2-2a_3$, i.e. $a_1-3a_2+4a_3=0$ ... (i)

From the first equal to the third: $2a_1-a_2+2a_3=3a_3$, i.e. $2a_1-a_2-a_3=0$ ... (ii)

Solve (i)–(ii) parametrically: from (ii), $a_3=2a_1-a_2$. Substitute into (i):

$$a_1-3a_2+4(2a_1-a_2)=0\Rightarrow 9a_1-7a_2=0\Rightarrow a_2=\frac{9}{7}{a}_1$$

Then $a_3=2a_1-\frac{9}{7}{a}_1=\frac{5}{7}{a}_1$. So $(a_1,a_2,a_3)\propto (7,9,5)$.

Magnitude $=\sqrt{49+81+25}=\sqrt{155}$, giving:

$$\hat{a}=\frac{1}{\sqrt{155}}(7\hat{i}+9\hat{j}+5\hat{k})$$

$\therefore$ Option (3) is correct.