Vectors – Linear combinations – JEE Main 24 January 2025 Shift 1

Solution:

Since $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$, we can assume:
$$ \vec{c} = x\vec{a} + y\vec{b} $$
Given that $\vec{c} \perp \vec{b}$, we have $\vec{c} \cdot \vec{b} = 0$:
$$ (x\vec{a} + y\vec{b}) \cdot \vec{b} = 0 $$
$$ x(\vec{a} \cdot \vec{b}) + y(\vec{b} \cdot \vec{b}) = 0 $$

Calculate the dot products:
$$ \vec{a} \cdot \vec{b} = (1)(3) + (2)(1) + (3)(-1) = 3 + 2 – 3 = 2 $$
$$ \vec{b} \cdot \vec{b} = (3)^2 + (1)^2 + (-1)^2 = 9 + 1 + 1 = 11 $$
Substituting these values:
$$ 2x + 11y = 0 \Rightarrow x = -\frac{11}{2}y \quad \dots(1) $$

Also given that $\vec{a} \cdot \vec{c} = 5$:
$$ \vec{a} \cdot (x\vec{a} + y\vec{b}) = 5 $$
$$ x(\vec{a} \cdot \vec{a}) + y(\vec{a} \cdot \vec{b}) = 5 $$
Calculate $\vec{a} \cdot \vec{a}$:
$$ \vec{a} \cdot \vec{a} = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 $$
Substituting values:
$$ 14x + 2y = 5 \quad \dots(2) $$

Substitute (1) into (2):
$$ 14(-\frac{11}{2}y) + 2y = 5 $$
$$ -77y + 2y = 5 $$
$$ -75y = 5 \Rightarrow y = -\frac{5}{75} = -\frac{1}{15} $$

Now find $x$:
$$ x = -\frac{11}{2} \left(-\frac{1}{15}\right) = \frac{11}{30} $$

Now, find vector $\vec{c}$:
$$ \vec{c} = \frac{11}{30}\vec{a} – \frac{1}{15}\vec{b} = \frac{1}{30}(11\vec{a} – 2\vec{b}) $$
$$ 11\vec{a} = 11(\hat{i}+2\hat{j}+3\hat{k}) = 11\hat{i} + 22\hat{j} + 33\hat{k} $$
$$ 2\vec{b} = 2(3\hat{i}+\hat{j}-\hat{k}) = 6\hat{i} + 2\hat{j} – 2\hat{k} $$
$$ 11\vec{a} – 2\vec{b} = (11-6)\hat{i} + (22-2)\hat{j} + (33-(-2))\hat{k} $$
$$ = 5\hat{i} + 20\hat{j} + 35\hat{k} = 5(\hat{i} + 4\hat{j} + 7\hat{k}) $$

Substituting back into $\vec{c}$:
$$ \vec{c} = \frac{1}{30} [5(\hat{i} + 4\hat{j} + 7\hat{k})] = \frac{1}{6}(\hat{i} + 4\hat{j} + 7\hat{k}) $$

Finally, calculate magnitude $|\vec{c}|$:
$$ |\vec{c}| = \frac{1}{6} \sqrt{1^2 + 4^2 + 7^2} = \frac{1}{6} \sqrt{1 + 16 + 49} $$
$$ |\vec{c}| = \frac{\sqrt{66}}{6} = \sqrt{\frac{66}{36}} = \sqrt{\frac{11}{6}} $$

Ans. (4)