JEE Main 2026 — Jan 24, Shift 2 — Mathematics solution

JEE Main2026Jan 24, Shift 2Mathematics

Let $a = 2 \hat{i} - \hat{j} - \hat{k}$ , $b = \hat{i} + 3 \hat{j} - \hat{k}$ and $c = 2 \hat{i} + \hat{j} + 3 \hat{k}$ . Let $v$ be the vector in the plane of the vectors $a$ and $b$ , such that the length of its projection on the vector $c$ is $\frac{1}{14}$ . Then which of the given options could be a value of $∣ v ∣$ .

A
$\frac{21}{2}$
B
13
C
$\frac{35}{2}$
D
7

Solution

Since $v$ lies in the plane of $a$ and $b$ , write $v = x a + y b$ :

$v = (2 x + y) \hat{i} + (- x + 3 y) \hat{j} + (- x - y) \hat{k}$ .

Length of projection on $c$ : $\frac{∣ v \cdot c ∣}{∣ c ∣} = \frac{1}{14}$ .

$v \cdot c = 2 (2 x + y) + (- x + 3 y) + 3 (- x - y) = 2 y$ .

$∣ c ∣ = 14$ , so $\frac{2 y}{14} = \frac{1}{14} \Rightarrow ∣2 y ∣ = 1 \Rightarrow y = \pm \frac{1}{2}$ .

$∣ v ∣^{2} = (2 x + y)^{2} + (3 y - x)^{2} + (x + y)^{2} = 6 x^{2} + 11 y^{2} = 6 x^{2} + \frac{11}{4}$ .

If $x^{2} = 1$ , then $∣ v ∣^{2} = \frac{35}{4} \Rightarrow ∣ v ∣ = \frac{35}{2}$ , matching option (3).

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