Vector Algebra – Vector Equations – JEE Main 28 Jan 2026 Shift 1

Solution:

Given $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.
The first condition is:
$$|\vec{a}-\vec{b}|^2 + |\vec{b}-\vec{c}|^2 + |\vec{c}-\vec{a}|^2 = 9$$
Expand using $|\vec{x}-\vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 – 2\vec{x}\cdot\vec{y}$:
$$(2 – 2\vec{a}\cdot\vec{b}) + (2 – 2\vec{b}\cdot\vec{c}) + (2 – 2\vec{c}\cdot\vec{a}) = 9$$
$$6 – 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 9$$
$$-2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 3 \Rightarrow \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} = -\frac{3}{2}$$

We know that $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})$.
$$|\vec{a}+\vec{b}+\vec{c}|^2 = 1 + 1 + 1 + 2\left(-\frac{3}{2}\right) = 3 – 3 = 0$$
Since the magnitude squared is 0, the vector sum must be the zero vector:
$$\vec{a} + \vec{b} + \vec{c} = \vec{0} \Rightarrow \vec{b} + \vec{c} = -\vec{a}$$

Now consider the second condition:
$$|2\vec{a} + k(\vec{b} + \vec{c})| = 3$$
Substitute $\vec{b} + \vec{c} = -\vec{a}$:
$$|2\vec{a} + k(-\vec{a})| = 3$$
$$|(2-k)\vec{a}| = 3$$
$$|2-k| |\vec{a}| = 3$$
Since $|\vec{a}| = 1$:
$$|2-k| = 3$$
$$2-k = 3 \quad \text{or} \quad 2-k = -3$$
$$k = -1 \quad \text{or} \quad k = 5$$

We need the positive value of $k$.
$$k = 5$$

Ans. (4)