The points with position vectors 60i + 3j,40i | vedclass

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Question

(A) Let the points be $A(60, 3)$,$B(40, -8)$,and $C(a, -52)$.For the points to be collinear,the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}

Vedclass · Accepted Answer

$-40$

Vedclass · Answer

(A) Let the points be $A(60, 3)$,$B(40, -8)$,and $C(a, -52)$.For the points to be collinear,the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ must be parallel,i.e.,$\overrightarrow{AB} = k(\overrightarrow{BC})$ for some scalar $k$.First,calculate $\overrightarrow{AB} = (40 - 60)i + (-8 - 3)j = -20i - 11j$.Next,calculate $\overrightarrow{BC} = (a - 40)i + (-52 - (-8))j = (a - 40)i - 44j$.Setting $\overrightarrow{AB} = k(\overrightarrow{BC})$,we get $-20i - 11j = k((a - 40)i - 44j)$.Comparing the coefficients of $j$,we have $-11 = -44k$,which gives $k = \frac{1}{4}$.Comparing the coefficients of $i$,we have $-20 = k(a - 40)$.Substituting $k = \frac{1}{4}$,we get $-20 = \frac{1}{4}(a - 40)$.$-80 = a - 40$,which implies $a = -40$.

The points with position vectors $60i + 3j$,$40i - 8j$,and $ai - 52j$ are collinear if $a = \dots$

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