Given that vector A i3j B 4i 6j C3i 6j D2i 6j Show that

Given that vector A = i+3j, B = 4i - 6j, C=-3i - 6j, D=2i - 6j.. Show that vector A, B and C are collinear. Find the value of |AB|.

Solution

given vectors A= i +3j , B=4i - 6j , C= 3i -6j , D=2i - 6j

Given points a, b and c form the line segments ab, bc and ac. If ab + bc = ac then the three points are collinear.

The line segments can be translated to vectors ab, bc and ac where the magnitude of the vectors are equal to the length of the respective line segments mentioned.

so now AB = B - A = (4i - 6j) - (i + 3j)

= (4i - i) -6j-3j

= 3i - 9j

now |AB| is magnitude of AB = sqrt(3^2 + 9^2 )

= sqrt( 9 +81)

= sqrt(90) --->ANSWER for b) part

now BC = C - B = ( 3i -6j ) - (4i - 6j)

= 3i-4i -6j+6j

= -i

now |BC| = sqrt(1^2) = 1

now AC = C - A = ( 3i -6j ) - ( i +3j)

= (3i-i) -6j-3j

= 2i -9j

now |AC| = sqrt(2^2 + 9^2) = sqrt(4+81) = sqrt(85)

now check for |AB|+|BC| = |AC| if this is true then the vectors are collinear.

sqrt(90)+1 = sqrt(85)

this is not true

soA,B,C are not collinear.

and |AB|= sqrt(90)