Let $\overrightarrow{AB}=2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\overrightarrow{\mathrm{AD}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\lambda \hat{k}, \lambda \in \mathbb{R}$. Let the projection of the vector $\overrightarrow{\mathrm{v}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ on the diagonal $\overrightarrow{\mathrm{AC}}$ of the parallelogram ABCD be of length one unit. If $\alpha, \beta$, where $\alpha>\beta$, be the roots of the equation $\lambda^{2} \mathrm{x}^{2}- 6 \lambda \mathrm{x}+5=0$, then $2 \alpha-\beta$ is equal to

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Vector Algebra

Let

A B = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}

and

AD = \hat{i} + 2 \hat{j} + λ \hat{k}, λ \in R

. Let the projection of the vector

v = \hat{i} + \hat{j} + \hat{k}

on the diagonal

AC

of the parallelogram ABCD be of length one unit. If

α, β

, where

α > β

, be the roots of the equation

λ^{2} x^{2} - 6 λ x + 5 = 0

, then

2 α - β

is equal to

Answer: (C)

In parallelogram

A B C D

the diagonal is

A C = A B + A D = 3 \hat{i} + 6 \hat{j} + (λ - 5) \hat{k}

. The length of the projection of

v = \hat{i} + \hat{j} + \hat{k}

A C

\frac{∣ v \cdot A C ∣}{∣ A C ∣}

, where

v \cdot A C = λ + 4

and

∣ A C ∣ = 45 + (λ - 5)^{2}

. Equating this to

1

and squaring,

(λ + 4)^{2} = 45 + (λ - 5)^{2}

, which reduces to

18 λ = 54

, so

λ = 3

. The equation becomes

9 x^{2} - 18 x + 5 = 0

, with roots

x = \frac{18 \pm 12}{18}

, giving

α = \frac{5}{3}

and

β = \frac{1}{3}

. Hence

2 α - β = \frac{10}{3} - \frac{1}{3} = 3

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option