JEE Advanced · Mathematics · 30. Vector Algebra
The number of distinct real values of \(\lambda\), for which the vectors \(-\lambda^2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\), \(\hat{\mathbf{i}}-\lambda^2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\lambda^2 \hat{\mathbf{k}}\) are coplanar, is
- A
zero
- B
one
- C
two
- D
three
Answer & Solution
Correct Answer
(C)
two
Step-by-step Solution
Detailed explanation
\[
\begin{aligned}
& \left|\begin{array}{ccc}
-\lambda^2 & 1 & 1 \\
1 & -\lambda^2 & 1 \\
1 & 1 & -\lambda^2
\end{array}\right|=0 \Rightarrow \lambda^6-3 \lambda^2-2=0 \\
& \Rightarrow \quad\left(1+\lambda^2\right)^2\left(\lambda^2-2\right)=0 \Rightarrow \lambda=\pm \sqrt{2} \\
&
\end{aligned}
\]
\begin{aligned}
& \left|\begin{array}{ccc}
-\lambda^2 & 1 & 1 \\
1 & -\lambda^2 & 1 \\
1 & 1 & -\lambda^2
\end{array}\right|=0 \Rightarrow \lambda^6-3 \lambda^2-2=0 \\
& \Rightarrow \quad\left(1+\lambda^2\right)^2\left(\lambda^2-2\right)=0 \Rightarrow \lambda=\pm \sqrt{2} \\
&
\end{aligned}
\]
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