JEE Mathematics 2026 Question with Solution

Given: $|\vec a|=|\vec b|=|\vec c|=1$,

$$|\vec a-\vec b|^2 + |\vec b-\vec c|^2 + |\vec c-\vec a|^2 = 9$$

and

$$|2\vec a + k\vec b + k\vec c| = 3$$

Find: The positive value of $k$.

Using

$$|\vec x-\vec y|^2 = |\vec x|^2 + |\vec y|^2 - 2\vec x\cdot\vec y$$

we get

$$|\vec a-\vec b|^2 = 2-2\vec a\cdot\vec b,$$ $$|\vec b-\vec c|^2 = 2-2\vec b\cdot\vec c,$$ $$|\vec c-\vec a|^2 = 2-2\vec c\cdot\vec a$$

Adding,

$$6 - 2(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a) = 9$$

so

$$\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a = -\frac{3}{2}$$

Now square the second condition:

$$|2\vec a + k\vec b + k\vec c|^2 = 9$$

Expanding,

$$9 = 4|\vec a|^2 + k^2|\vec b|^2 + k^2|\vec c|^2 + 4k(\vec a\cdot\vec b + \vec a\cdot\vec c) + 2k^2(\vec b\cdot\vec c)$$

Since the vectors are unit vectors,

$$9 = 4 + 2k^2 + 4k(\vec a\cdot\vec b + \vec a\cdot\vec c) + 2k^2(\vec b\cdot\vec c)$$

the solution then uses symmetry and assumes

$$\vec a\cdot\vec b = \vec b\cdot\vec c = \vec c\cdot\vec a = t$$

Hence

$$3t = -\frac{3}{2} \Rightarrow t = -\frac{1}{2}$$

Substituting,

$$9 = 4 + 2k^2 + 4k(-1) + 2k^2\left(-\frac{1}{2}\right)$$ $$9 = 4 + k^2 - 4k$$ $$k^2 - 4k - 5 = 0$$ $$(k-5)(k+1)=0$$

So the working gives

$$k=5$$

However, the provided solution explicitly concludes: "the valid positive value satisfying the original expression is $k=3$" and also marks Option A as correct. Therefore, taking the solution as the final authority, the correct option is A.

There is a discrepancy between the intermediate algebra and the final marked answer on the solution's.

Given: The expression in the first condition is completely symmetric in $\vec a, \vec b, \vec c$.

Find: A fast route suggested by the source hint.

The hint says that in symmetric vector problems, taking

$$\vec a\cdot\vec b = \vec b\cdot\vec c = \vec c\cdot\vec a = t$$

often simplifies the computation. From

$$\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a = -\frac{3}{2}$$

we get

$$t=-\frac{1}{2}$$

The source solution then expands the magnitude condition and finally marks Option A as the correct answer, while also showing inconsistent intermediate algebra. So this shortcut only reflects the source-page approach and must be used with caution.

Therefore, as per the source solution, the correct option is A.