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Question: Answered & Verified by Expert
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
MathematicsVector AlgebraJEE AdvancedJEE Advanced 2007 (Paper 1)
Options:
  • A
    zero
  • B
    one
  • C
    two
  • D
    three
Solution:
2484 Upvotes Verified Answer
The correct answer is:
two
$$
\begin{aligned}
& \left|\begin{array}{ccc}
-\lambda^2 & 1 & 1 \\
1 & -\lambda^2 & 1 \\
1 & 1 & -\lambda^2
\end{array}\right|=0 \Rightarrow \lambda^6-3 \lambda^2-2=0 \\
& \Rightarrow \quad\left(1+\lambda^2\right)^2\left(\lambda^2-2\right)=0 \Rightarrow \lambda=\pm \sqrt{2} \\
&
\end{aligned}
$$

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