Mathematics · Vector Algebra

JEE Main 2026 — 22 January, Morning Shift — Question 1

Let $\overrightarrow{\mathrm{AB}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-5 \mathrm{k}$ and $\overrightarrow{\mathrm{AD}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\lambda \mathrm{k}, \lambda \in \mathbb{R}$ . Let the projection of the vector $\vec{v}=\hat{i}+\hat{j}+\hat{k}$ 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

Option A:
$1$
Option B:
$4$
Option C:
$3$
Correct
Option D:
$6$

Answer: C

Step-by-step solution

$\overrightarrow{\mathrm{AC}}=3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+(\lambda-5) \mathrm{k}$

\overrightarrow{\mathrm{v}} \cdot \mathrm{AC}=1

\Rightarrow 3+6+\lambda-5=\sqrt{9+36+(\lambda-5)^{2}}

$\Rightarrow \lambda^{2}+8 \lambda+16=\lambda^{2}-10 \lambda+70$

$\Rightarrow \lambda=\frac{54}{18}=3$

∴ Quadratic : $9 \mathrm{x}^{2}-18 \mathrm{x}+5=0$

$\Rightarrow \mathrm{x}=\frac{1}{3}, \frac{5}{3}$

$\therefore 2 \alpha-\beta=\frac{10-1}{3}=3$

Answer key and solution verified before publishing.

Practise Vector Algebra

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 22 January, Morning Shift
Subject: Mathematics
Chapter: Vector Algebra
Topic: Projection & component of a vector along another vector.