Vector Algebra – Projection of Vectors – JEE Main 2 Apr 2025 Shift 1

Solution:

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$.

Let the three vectors be:
$\vec{b} = 2\hat{i}-\hat{j}+2\hat{k} \Rightarrow |\vec{b}| = \sqrt{4+1+4} = 3$

$\vec{c} = \hat{i}+2\hat{j}-2\hat{k} \Rightarrow |\vec{c}| = \sqrt{1+4+4} = 3$

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

Given that projections are equal:
$$ \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{\vec{a} \cdot \vec{c}}{|\vec{c}|} = \frac{\vec{a} \cdot \vec{d}}{|\vec{d}|} $$

$$ \frac{2a_1 – a_2 + 2a_3}{3} = \frac{a_1 + 2a_2 – 2a_3}{3} = \frac{a_3}{1} $$

From the second and third parts:
$$ \frac{a_1 + 2a_2 – 2a_3}{3} = a_3 \Rightarrow a_1 + 2a_2 – 2a_3 = 3a_3 \Rightarrow a_1 + 2a_2 = 5a_3 \quad \dots(1) $$

From the first and third parts:
$$ \frac{2a_1 – a_2 + 2a_3}{3} = a_3 \Rightarrow 2a_1 – a_2 + 2a_3 = 3a_3 \Rightarrow 2a_1 – a_2 = a_3 \quad \dots(2) $$

From (2), $a_2 = 2a_1 – a_3$. Substitute this into (1):
$$ a_1 + 2(2a_1 – a_3) = 5a_3 $$
$$ a_1 + 4a_1 – 2a_3 = 5a_3 $$
$$ 5a_1 = 7a_3 \Rightarrow a_1 = \frac{7}{5}a_3 $$

Now find $a_2$:
$$ a_2 = 2\left(\frac{7}{5}a_3\right) – a_3 = \frac{14}{5}a_3 – \frac{5}{5}a_3 = \frac{9}{5}a_3 $$

So, $\vec{a} = \frac{7}{5}a_3\hat{i} + \frac{9}{5}a_3\hat{j} + a_3\hat{k}$.
Multiplying by $\frac{5}{a_3}$, the direction ratios are proportional to $(7, 9, 5)$.

The unit vector is:
$$ \hat{a} = \frac{7\hat{i} + 9\hat{j} + 5\hat{k}}{\sqrt{7^2 + 9^2 + 5^2}} = \frac{7\hat{i} + 9\hat{j} + 5\hat{k}}{\sqrt{49 + 81 + 25}} = \frac{7\hat{i} + 9\hat{j} + 5\hat{k}}{\sqrt{155}} $$

Ans. (3)