Vector Algebra – Vectors – JEE Main 29 Jan 2025 Shift 1

Solution:

Given $\vec{a}\times\vec{c} = \vec{c}\times\vec{b}$

$\vec{a}\times\vec{c} – \vec{c}\times\vec{b} = 0$

$\vec{a}\times\vec{c} + \vec{b}\times\vec{c} = 0$

$(\vec{a}+\vec{b})\times\vec{c} = 0$

This implies $\vec{c}$ is parallel to $(\vec{a}+\vec{b})$.

$\vec{c} = \lambda(\vec{a}+\vec{b})$

Calculate $\vec{a}+\vec{b} = (2+3)\hat{i} + (-1-5)\hat{j} + (3+1)\hat{k} = 5\hat{i} – 6\hat{j} + 4\hat{k}$.

Let $\vec{v} = 5\hat{i} – 6\hat{j} + 4\hat{k}$. Then $\vec{c} = \lambda\vec{v}$.

$|\vec{v}|^{2} = 5^{2} + (-6)^{2} + 4^{2} = 25 + 36 + 16 = 77$.

So, $|\vec{c}|^{2} = \lambda^{2}(77)$.

Given relation: $(\vec{a}+\vec{c})\cdot(\vec{b}+\vec{c}) = 168$

$\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} + \vec{c}\cdot\vec{b} + |\vec{c}|^{2} = 168$

$\vec{a}\cdot\vec{b} + \vec{c}\cdot(\vec{a}+\vec{b}) + |\vec{c}|^{2} = 168$

Calculate $\vec{a}\cdot\vec{b} = (2)(3) + (-1)(-5) + (3)(1) = 6 + 5 + 3 = 14$.

Substitute values:

$14 + (\lambda\vec{v})\cdot\vec{v} + 77\lambda^{2} = 168$

$14 + \lambda|\vec{v}|^{2} + 77\lambda^{2} = 168$

$14 + 77\lambda + 77\lambda^{2} = 168$

$77\lambda^{2} + 77\lambda – 154 = 0$

Divide by 77:

$\lambda^{2} + \lambda – 2 = 0$

$(\lambda+2)(\lambda-1) = 0$

$\lambda = -2$ or $\lambda = 1$.

We need the maximum value of $|\vec{c}|^{2} = 77\lambda^{2}$.

For $\lambda = 1$, $|\vec{c}|^{2} = 77(1)^{2} = 77$.

For $\lambda = -2$, $|\vec{c}|^{2} = 77(-2)^{2} = 77(4) = 308$.

The maximum value is 308.

Ans. (3)