Let vec{u} = ahat{i} + bhat{j} + chat{k},vec{ | vedclass

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(B) Given the vectors $\vec{u} = a\hat{i} + b\hat{j} + c\hat{k}$,$\vec{v} = b\hat{i} + c\hat{j} + a\hat{k}$,and $\vec{w} = c\hat{i} + a\hat{j} + b\hat

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coplanar

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(B) Given the vectors $\vec{u} = a\hat{i} + b\hat{j} + c\hat{k}$,$\vec{v} = b\hat{i} + c\hat{j} + a\hat{k}$,and $\vec{w} = c\hat{i} + a\hat{j} + b\hat{k}$.The scalar triple product $[\vec{u} \, \vec{v} \, \vec{w}]$ is given by the determinant:$\begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix} = 0$.Expanding the determinant,we get:$-(a^3 + b^3 + c^3 - 3abc) = 0$.This simplifies to:$-(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0$.Since $(a + b + c) 
eq 0$,we must have $(a^2 + b^2 + c^2 - ab - bc - ca) = 0$,which implies $\frac{1}{2}((a - b)^2 + (b - c)^2 + (c - a)^2) = 0$.This holds only if $a = b = c$.If $a = b = c$,then $\vec{u} = \vec{v} = \vec{w} = a(\hat{i} + \hat{j} + \hat{k})$.Since $\vec{w} = \lambda \vec{x} + \mu \vec{y}$,the vector $\vec{w}$ lies in the plane formed by $\vec{x}$ and $\vec{y}$.Since $\vec{u} = \vec{v} = \vec{w}$,all these vectors $\vec{u}, \vec{v}, \vec{w}$ lie in the plane spanned by $\vec{x}$ and $\vec{y}$.Therefore,the set of vectors $\vec{x}, \vec{y}, \vec{u}, \vec{v}, \vec{w}$ are coplanar.

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