If vec{a}, vec{b}, vec{c} are non-coplanar ve | vedclass

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(C) Let $\vec{\alpha} = \vec{a} + 2\vec{b} + 3\vec{c}$,$\vec{\beta} = \lambda\vec{b} + 4\vec{c}$,and $\vec{\gamma} = (2\lambda - 1)\vec{c}$.The vector

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For all values of $\lambda$ except two values

Vedclass · Answer

(C) Let $\vec{\alpha} = \vec{a} + 2\vec{b} + 3\vec{c}$,$\vec{\beta} = \lambda\vec{b} + 4\vec{c}$,and $\vec{\gamma} = (2\lambda - 1)\vec{c}$.The vectors are non-coplanar if their scalar triple product is non-zero,i.e.,$[\vec{\alpha}, \vec{\beta}, \vec{\gamma}] 
eq 0$.$[\vec{\alpha}, \vec{\beta}, \vec{\gamma}] = \begin{vmatrix} 1 & 2 & 3 \ 0 & \lambda & 4 \ 0 & 0 & (2\lambda - 1) \end{vmatrix} [\vec{a}, \vec{b}, \vec{c}]$$= 1 \cdot \lambda \cdot (2\lambda - 1) [\vec{a}, \vec{b}, \vec{c}] = \lambda(2\lambda - 1) [\vec{a}, \vec{b}, \vec{c}]$Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,$[\vec{a}, \vec{b}, \vec{c}] 
eq 0$.Thus,the vectors are non-coplanar if $\lambda(2\lambda - 1) 
eq 0$.This implies $\lambda 
eq 0$ and $\lambda 
eq \frac{1}{2}$.Therefore,the vectors are non-coplanar for all values of $\lambda$ except two values,$0$ and $\frac{1}{2}$.

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and $\lambda$ is a real number,then for what value of $\lambda$ are the vectors $\vec{a} + 2\vec{b} + 3\vec{c}$,$\lambda\vec{b} + 4\vec{c}$,and $(2\lambda - 1)\vec{c}$ non-coplanar?

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