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(C) For the vectors to be coplanar,their scalar triple product must be zero:$\begin{vmatrix} \lambda & -1 & 1 \ 1 & 2 & \mu \ 3 & -4 & 5 \end{vmatri

Vedclass · Accepted Answer

$2290$

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(C) For the vectors to be coplanar,their scalar triple product must be zero:$\begin{vmatrix} \lambda & -1 & 1 \ 1 & 2 & \mu \ 3 & -4 & 5 \end{vmatrix} = 0$Expanding the determinant:$\lambda(10 + 4\mu) - (-1)(5 - 3\mu) + 1(-4 - 6) = 0$$10\lambda + 4\lambda\mu + 5 - 3\mu - 10 = 0$$10\lambda + 4\lambda\mu - 3\mu - 5 = 0$Given $\lambda - \mu = 5$,so $\lambda = \mu + 5$. Substituting this into the equation:$10(\mu + 5) + 4(\mu + 5)\mu - 3\mu - 5 = 0$$10\mu + 50 + 4\mu^2 + 20\mu - 3\mu - 5 = 0$$4\mu^2 + 27\mu + 45 = 0$$(4\mu + 15)(\mu + 3) = 0$So,$\mu_1 = -15/4$ and $\mu_2 = -3$.Corresponding $\lambda$ values are $\lambda_1 = -15/4 + 5 = 5/4$ and $\lambda_2 = -3 + 5 = 2$.The set $S = \{(5/4, -15/4), (2, -3)\}$.Now,calculate $\sum_{(\lambda, \mu) \in S} 80(\lambda^2 + \mu^2)$:$= 80[((5/4)^2 + (-15/4)^2) + (2^2 + (-3)^2)]$$= 80[(25/16 + 225/16) + (4 + 9)]$$= 80[250/16 + 13] = 80[15.625 + 13] = 80[28.625] = 2290$.

Let $S$ be the set of all $(\lambda, \mu)$ for which the vectors $\lambda \hat{i} - \hat{j} + \hat{k}$,$\hat{i} + 2\hat{j} + \mu \hat{k}$ and $3\hat{i} - 4\hat{j} + 5\hat{k}$,where $\lambda - \mu = 5$,are coplanar,then $\sum_{(\lambda, \mu) \in S} 80(\lambda^2 + \mu^2)$ is equal to :

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