If the position vectors of the points A, B, C | vedclass

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(D) The position vectors are $\vec{OA} = i + j$,$\vec{OB} = i - j$,and $\vec{OC} = a i + b j + c k$.For the points $A, B, C$ to be collinear,the vecto

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$c = 0, a = 1$ and $b$ is an arbitrary scalar

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(D) The position vectors are $\vec{OA} = i + j$,$\vec{OB} = i - j$,and $\vec{OC} = a i + b j + c k$.For the points $A, B, C$ to be collinear,the vectors $\vec{AB}$ and $\vec{BC}$ must be parallel.Calculate $\vec{AB} = \vec{OB} - \vec{OA} = (i - j) - (i + j) = -2j$.Calculate $\vec{BC} = \vec{OC} - \vec{OB} = (a i + b j + c k) - (i - j) = (a - 1)i + (b + 1)j + ck$.Since $\vec{AB}$ and $\vec{BC}$ are collinear,there exists a scalar $k$ such that $\vec{AB} = k \vec{BC}$.$-2j = k((a - 1)i + (b + 1)j + ck)$.Comparing the coefficients of $i, j, k$ on both sides:$k(a - 1) = 0$$k(b + 1) = -2$$kc = 0$From $kc = 0$,since $k 
eq 0$ (otherwise $\vec{AB} = 0$,which is not possible),we must have $c = 0$.From $k(a - 1) = 0$,we get $a = 1$.From $k(b + 1) = -2$,$b$ can be any arbitrary scalar as long as $k$ is adjusted accordingly.Therefore,the condition for collinearity is $c = 0, a = 1$ and $b$ is an arbitrary scalar.

If the position vectors of the points $A, B, C$ are $i + j$,$i - j$,and $a i + b j + c k$ respectively,then the points $A, B, C$ are collinear if

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