The points with position vectors 10,i + 3,j,1 | vedclass

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(C) Let the points be $A(10, 3)$,$B(12, -5)$,and $C(a, 11)$.Since the points are collinear,the vector $\overrightarrow{AB}$ must be parallel to the ve

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$8$

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(C) Let the points be $A(10, 3)$,$B(12, -5)$,and $C(a, 11)$.Since the points are collinear,the vector $\overrightarrow{AB}$ must be parallel to the vector $\overrightarrow{BC}$.$\overrightarrow{AB} = (12 - 10)i + (-5 - 3)j = 2i - 8j$.$\overrightarrow{BC} = (a - 12)i + (11 - (-5))j = (a - 12)i + 16j$.For collinearity,$\overrightarrow{AB} = k \overrightarrow{BC}$ for some scalar $k$.$2i - 8j = k[(a - 12)i + 16j]$.Comparing the coefficients of $j$: $-8 = 16k \Rightarrow k = -\frac{1}{2}$.Comparing the coefficients of $i$: $2 = k(a - 12)$.Substituting $k = -\frac{1}{2}$: $2 = -\frac{1}{2}(a - 12)$.$-4 = a - 12 \Rightarrow a = 8$.

The points with position vectors $10\,i + 3\,j$,$12\,i - 5\,j$,and $a\,i + 11\,j$ are collinear,if $a = $

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