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

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(C) Three vectors $\vec{u}, \vec{v}, \vec{w}$ are coplanar if their scalar triple product $[\vec{u}, \vec{v}, \vec{w}] = 0$. Given vectors are $\vec{u

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all except two values of $\lambda$

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(C) Three vectors $\vec{u}, \vec{v}, \vec{w}$ are coplanar if their scalar triple product $[\vec{u}, \vec{v}, \vec{w}] = 0$. Given vectors are $\vec{u} = \vec{a}+2 \vec{b}+3 \vec{c}$, $\vec{v} = 0\vec{a} + \lambda \vec{b}+4 \vec{c}$, and $\vec{w} = 0\vec{a} + 0\vec{b} + (2 \lambda-1) \vec{c}$. The scalar triple product is given by the determinant of the coefficients of $\vec{a}, \vec{b}, \vec{c}$: $\left|\begin{array}{ccc} 1 & 2 & 3 \ 0 & \lambda & 4 \ 0 & 0 & 2 \lambda-1 \end{array}ight| = 0$. Expanding along the first column: $1 	imes [\lambda(2 \lambda-1) - 0] = 0$. $\lambda(2 \lambda-1) = 0$. This gives $\lambda = 0$ or $\lambda = \frac{1}{2}$. Since the vectors are non-coplanar when the determinant is non-zero, the vectors are non-coplanar for all real values of $\lambda$ except $\lambda = 0$ and $\lambda = \frac{1}{2}$.

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

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