Question

The points with position vectors (120i + 6j), (80i - 16j), and (ai - 104j) are collinear if:​

Answer 1

Given:

A (120i + 6j) , B(80i - 16j)  and C (1i _ 104j) are 3 position vectors

To Find:

Value of a for which the points A , B and C are collinear.

Solution:

If 3 points are collinear , Area of triangle formed by joining three points is equal to 0.

Area of a triangle ABC = 0

AREA of triangle whose position coordinates are given.

$Area = 0.5 [{x_{1}( y_{2}- y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}( y_{1} - y_{2} ) }]\\x_{1} =120 x_{2}= 80 x_{3} = a y_{1} = 6 y_{2}= -16 y_{3}= -104\\\\\\Area = 0.5[120(-16 + 104) + 80(-104 - 6) + a(6 +16)\\Area = 0\\\\\\-1760 = a (22)\\a =-80$

The value of a for which the three points are collinear is -80 .