The sum of the distinct real values of mu,for | vedclass

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(B) For three vectors to be coplanar,their scalar triple product must be zero.$\begin{vmatrix} \mu & 1 & 1 \ 1 & \mu & 1 \ 1 & 1 & \mu \end{vmatrix}

Vedclass · Accepted Answer

$-1$

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(B) For three vectors to be coplanar,their scalar triple product must be zero.$\begin{vmatrix} \mu & 1 & 1 \ 1 & \mu & 1 \ 1 & 1 & \mu \end{vmatrix} = 0$Expanding the determinant along the first row:$\mu(\mu^2 - 1) - 1(\mu - 1) + 1(1 - \mu) = 0$$\mu(\mu - 1)(\mu + 1) - 1(\mu - 1) - 1(\mu - 1) = 0$$(\mu - 1) [\mu(\mu + 1) - 1 - 1] = 0$$(\mu - 1) (\mu^2 + \mu - 2) = 0$$(\mu - 1) (\mu + 2) (\mu - 1) = 0$$(\mu - 1)^2 (\mu + 2) = 0$The distinct real values of $\mu$ are $\mu = 1$ and $\mu = -2$.The sum of these distinct values is $1 + (-2) = -1$.

The sum of the distinct real values of $\mu$,for which the vectors $\mu \hat{i}+\hat{j}+\hat{k}$,$\hat{i}+\mu \hat{j}+\hat{k}$,and $\hat{i}+\hat{j}+\mu \hat{k}$ are coplanar is

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