Vector Algebra – Components of Vectors – JEE Main 28 Jan 2025 Shift 2

Solution:

Let $\vec{a}_{\parallel}$ be the component of $\vec{a}$ along $\vec{b}$ and $\vec{a}_{\perp}$ be the component of $\vec{a}$ perpendicular to $\vec{b}$.

Given:

$$\vec{a}_{\parallel} = \frac{16}{11}(3\hat{i}+\hat{j}-\hat{k})$$

$$\vec{a}_{\perp} = \frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k})$$

We know that any vector can be written as the sum of its parallel and perpendicular components:

$$\vec{a} = \vec{a}_{\parallel} + \vec{a}_{\perp}$$

Substituting the given values:

$$\vec{a} = \left[ \frac{16}{11}(3\hat{i}+\hat{j}-\hat{k}) \right] + \left[ \frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k}) \right]$$

$$\vec{a} = \frac{1}{11} \left[ (48\hat{i} + 16\hat{j} – 16\hat{k}) + (-4\hat{i} – 5\hat{j} – 17\hat{k}) \right]$$

$$\vec{a} = \frac{1}{11} \left[ (48-4)\hat{i} + (16-5)\hat{j} + (-16-17)\hat{k} \right]$$

$$\vec{a} = \frac{1}{11} \left[ 44\hat{i} + 11\hat{j} – 33\hat{k} \right]$$

$$\vec{a} = 4\hat{i} + \hat{j} – 3\hat{k}$$

Comparing this with $\vec{a} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$, we get:

$\alpha = 4, \beta = 1, \gamma = -3$

Now, we calculate $\alpha^2 + \beta^2 + \gamma^2$:

$$\alpha^2 + \beta^2 + \gamma^2 = (4)^2 + (1)^2 + (-3)^2$$

$$= 16 + 1 + 9 = 26$$

Ans. (4)