If the vectors vec{a} + lambda vec{b} + 3vec{ | vedclass

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(D) Three vectors $\vec{u}, \vec{v}, \vec{w}$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{u}, \vec{v}, \vec{w}] = 0$.L

Vedclass · Accepted Answer

$-2$

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(D) Three vectors $\vec{u}, \vec{v}, \vec{w}$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{u}, \vec{v}, \vec{w}] = 0$.Let $\vec{u} = \vec{a} + \lambda \vec{b} + 3\vec{c}$,$\vec{v} = -2\vec{a} + 3\vec{b} - 4\vec{c}$,and $\vec{w} = \vec{a} - 3\vec{b} + 5\vec{c}$.The condition $[\vec{u}, \vec{v}, \vec{w}] = 0$ implies the determinant of the coefficients of $\vec{a}, \vec{b}, \vec{c}$ must be zero:$\begin{vmatrix} 1 & \lambda & 3 \ -2 & 3 & -4 \ 1 & -3 & 5 \end{vmatrix} = 0$Expanding the determinant along the first row:$1(3 	imes 5 - (-4) 	imes (-3)) - \lambda((-2) 	imes 5 - (-4) 	imes 1) + 3((-2) 	imes (-3) - 3 	imes 1) = 0$$1(15 - 12) - \lambda(-10 + 4) + 3(6 - 3) = 0$$1(3) - \lambda(-6) + 3(3) = 0$$3 + 6\lambda + 9 = 0$$6\lambda + 12 = 0$$6\lambda = -12$$\lambda = -2$.

If the vectors $\vec{a} + \lambda \vec{b} + 3\vec{c}$,$-2\vec{a} + 3\vec{b} - 4\vec{c}$,and $\vec{a} - 3\vec{b} + 5\vec{c}$ are coplanar,and $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,find the value of $\lambda$.

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