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(C) Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.The

Vedclass · Accepted Answer

$2$

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(C) Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.The scalar triple product is given by the determinant:$\left|\begin{array}{ccc} -\lambda^2 & 1 & 1 \ 1 & -\lambda^2 & 1 \ 1 & 1 & -\lambda^2 \end{array}ight| = 0$Expanding the determinant along the first row:$-\lambda^2(\lambda^4 - 1) - 1(-\lambda^2 - 1) + 1(1 + \lambda^2) = 0$$-\lambda^6 + \lambda^2 + \lambda^2 + 1 + 1 + \lambda^2 = 0$$-\lambda^6 + 3\lambda^2 + 2 = 0$$\lambda^6 - 3\lambda^2 - 2 = 0$Let $x = \lambda^2$. Then the equation becomes $x^3 - 3x - 2 = 0$.By inspection,$x = -1$ is a root: $(-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0$.Dividing by $(x+1)$,we get $(x+1)(x^2 - x - 2) = 0$,which factors further to $(x+1)(x+1)(x-2) = 0$.So,$(x+1)^2(x-2) = 0$.Since $x = \lambda^2$,we have $(\lambda^2 + 1)^2(\lambda^2 - 2) = 0$.For real $\lambda$,$\lambda^2 + 1$ is always $\ge 1$,so $\lambda^2 + 1 
eq 0$.Thus,we must have $\lambda^2 - 2 = 0$,which gives $\lambda^2 = 2$.Therefore,$\lambda = \pm \sqrt{2}$.There are $2$ distinct real values of $\lambda$.

For how many distinct real values of $\lambda$ are the vectors $-\lambda^2 \hat{i} + \hat{j} + \hat{k}$,$\hat{i} - \lambda^2 \hat{j} + \hat{k}$,and $\hat{i} + \hat{j} - \lambda^2 \hat{k}$ coplanar?

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