The points with position vectors 60,i + 3,j,4 | vedclass

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(A) Let the points be $A(60, 3)$,$B(40, -8)$,and $C(a, -52)$.Since the points are collinear,the vector $\overrightarrow{AB}$ must be parallel to the v

Vedclass · Accepted Answer

$-40$

Vedclass · Answer

(A) Let the points be $A(60, 3)$,$B(40, -8)$,and $C(a, -52)$.Since the points are collinear,the vector $\overrightarrow{AB}$ must be parallel to the vector $\overrightarrow{BC}$.$\overrightarrow{AB} = (40 - 60)i + (-8 - 3)j = -20i - 11j$.$\overrightarrow{BC} = (a - 40)i + (-52 - (-8))j = (a - 40)i - 44j$.For collinearity,$\overrightarrow{AB} = k\overrightarrow{BC}$ for some scalar $k$.$-20i - 11j = k((a - 40)i - 44j)$.Comparing the coefficients of $j$: $-11 = -44k$,which gives $k = \frac{11}{44} = \frac{1}{4}$.Comparing the coefficients of $i$: $-20 = k(a - 40)$.Substituting $k = \frac{1}{4}$: $-20 = \frac{1}{4}(a - 40)$.$-80 = a - 40$.$a = -80 + 40 = -40$.

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

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