JEE Advanced · Mathematics · 30. Vector Algebra
Consider the vectors
\(\vec{x}=\hat{i}+2 \hat{j}+3 \hat{k}, \quad \vec{y}=2 \hat{i}+3 \hat{j}+\hat{k}, \quad \text { and } \quad \vec{z}=3 \hat{i}+\hat{j}+2 \hat{k}\)
For two distinct positive real numbers \(\alpha\) and \(\beta\), define
\(\vec{X}=\alpha \vec{x}+\beta \vec{y}-\vec{z}, \quad \vec{Y}=\alpha \vec{y}+\beta \vec{z}-\vec{x}, \quad \text { and } \quad \vec{Z}=\alpha \vec{z}+\beta \vec{x}-\vec{y}\)
If the vectors \(\vec{X}, \vec{Y}\), and \(\vec{Z}\) lie in a plane, the value of \(\alpha+\beta-3\) is ______ .
- A -5
- B -2
- C -1
- D 5
Answer & Solution
Correct Answer
(B) -2
Step-by-step Solution
Detailed explanation
\(\Rightarrow\left|\begin{array}{ccc}\alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha\end{array}\right| \underbrace{\left|\begin{array}{ccc}1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2\end{array}\right|}_{-0}=0\)
\(\begin{aligned} & \Rightarrow\left(\alpha^3+\beta^3-1\right)-(-\alpha \beta-\alpha \beta-\alpha \beta)=0 \\ & \Rightarrow \alpha^3+\beta^3+3 \alpha \beta=1 \\ & \Rightarrow \alpha^3+\beta^3+(-1)^3=3(\alpha)(\beta)(-1) \\ & \Rightarrow \alpha+\beta-1=0\end{aligned}\)
So, \(\alpha+\beta-3=-2\)
\(\begin{aligned} & \Rightarrow\left(\alpha^3+\beta^3-1\right)-(-\alpha \beta-\alpha \beta-\alpha \beta)=0 \\ & \Rightarrow \alpha^3+\beta^3+3 \alpha \beta=1 \\ & \Rightarrow \alpha^3+\beta^3+(-1)^3=3(\alpha)(\beta)(-1) \\ & \Rightarrow \alpha+\beta-1=0\end{aligned}\)
So, \(\alpha+\beta-3=-2\)
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