Question

A and B are vectors in the same plane. If A= 8i+6j. Find the vector B such that |A+B| = 14.14 units and A is perpendicular to B.

Answer 1

A is perpendicular to B
so
A dot B = 0
AxBx + AyBy = 0
8 Bx + 6 By = 0
and
A + B = (Ax + Bx)i+ (Ay+By)j
|A+B|^2 = 14.14^2 = (Ax+Bx)^2+(Ay+By)^2
so
(Ax+Bx)^2+(Ay+By)^2 = 100
(8+Bx)^2 +(6+By)^2 = 100
but we know
By =-(8/6)Bx = -(4/3) Bx
plug and chug.

Answer 2

Sorru for the inconvenience... but I didnt get this concept
|A+B|^2 = 14 ??!!

Answer 3

|A+B|^2 = 14.14^2 = 200 not 100

I did not use calculator, knew sqrt 2 = 1.414

Answer 4

Oh .. the rest of the steps are the same then ?

Answer 5

yes