The number of distinct values of $\lambda$ for which the vectors $\lambda^2\hat{i}+\hat{j}+\hat{k}$, $\hat{i}+\lambda^2\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+\lambda^2\hat{k}$ are coplanar is

✅ Correct Answer: 2

Explanation

For coplanar vectors, determinant equals zero $|\lambda^2\ 1\ 1;\ 1\ \lambda^2\ 1;\ 1\ 1\ \lambda^2|=0$ Simplifying gives $(1-\lambda^2)^2=0$ $\lambda=\pm1$ Number of distinct values is $2$

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The number of distinct values of $\lambda$ for which the vectors $\lambda^2\hat{i}+\hat{j}+\hat{k}$, $\hat{i}+\lambda^2\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+\lambda^2\hat{k}$ are coplanar is | ACME Academy