Consider the vectors vec x =hat i +2 hat j +3 hat k , vec y =2 hat i +3 hat j +hat k , and vec z =3 hat i +hat j +2 hat k . For two distinct positive real numbers α and β , define vec X =α vec x +β vec y -vec z , vec Y =α vec y +β vec z -vec x ,and vec Z =α vec z +β vec x -vec y . If the vectors vec X , vec Y , and vec Z lie in a plane, the value of α+β-3 is

Mathematics · Vector Algebra

JEE Advanced 2025 — Paper 2 — Question 28

Consider the vectors $\vec{x}=\hat{i}+2 \hat{j}+3 \hat{k}$ , $\vec{y}=2 \hat{i}+3 \hat{j}+\hat{k}$ , and $\vec{z}=3 \hat{i}+\hat{j}+2 \hat{k} .$ For two distinct positive real numbers $\alpha$ and $\beta$ , define $\vec{X}=\alpha \vec{x}+\beta \vec{y}-\vec{z}$ , $\vec{Y}=\alpha \vec{y}+\beta \vec{z}-\vec{x}$ ,and $\vec{Z}=\alpha \vec{z}+\beta \vec{x}-\vec{y}$ . If the vectors $\vec{X}, \vec{Y}$ , and $\vec{Z}$ lie in a plane, the value of $\alpha+\beta-3$ is $\qquad$

Answer: -2

Numerical answer — enter this value.

Step-by-step solution

$\quad[\vec{x} \vec{y} \vec{z}]=0$

$\Rightarrow\left|\begin{array}{ccc}\alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha\end{array}\right| \underbrace{\left|\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2\end{array}\right|}_{\neq 0}=0$

$\Rightarrow\left(\alpha^{3}+\beta^{3}-1\right)-(-\alpha \beta-\alpha \beta-\alpha \beta)=0$

$\Rightarrow \alpha^{3}+\beta^{3}+3 \alpha \beta=1$

$\Rightarrow \alpha^{3}+\beta^{3}+(-1)^{3}=3(\alpha)(\beta)(-1)$

$\Rightarrow \alpha+\beta-1=0$ So, $\alpha+\beta-3=-2$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2025
Paper: Paper 2
Subject: Mathematics
Chapter: Vector Algebra
Topic: Algebra of Vectors