Answer:
[tex]\dfrac{425}{289}[/tex]
Step-by-step explanation:
Given vectors a =(1, 4) and b = (2, 3) we are to find orthab*orthab =
[tex]orth_ab[/tex] [tex]= b - proj_ab[/tex]
First we need to get [tex]proj_ab[/tex]
[tex]proj_ab = \dfrac{a.b}{|a|^2} * a[/tex]
a.b = (1, 4).(2, 3)
a.b = 1(2)+4(3)
a.b = 2+12
a.b = 14
|a| is the modulus of vector a
|a| = √1²+4²
|a| = √1+16
|a| = √17
square both sides
|a|² = (√17)²
|a|² = 17
[tex]proj_ab = \dfrac{14}{17} . (1,4)\\proj_ab = (\dfrac{14}{17}, \dfrac{56}{17})\\[/tex]
[tex]orth_ab = (2, 3) - (\frac{14}{17}, \frac{56}{17})\\orth_ab = (2- \frac{14}{17}, 3 - \frac{56}{17})\\orth_ab = (\frac{20}{17}, \frac{-5}{17})\\orth_ab.orth_ab = (\frac{20}{17}, \frac{-5}{17}).(\frac{20}{17}, \frac{-5}{17})\\\\orth_ab.orth_ab = (\frac{20}{17}.\frac{20}{17})+(\frac{-5}{17},.\frac{-5}{17})\\\\orth_ab.orth_ab = (\frac{400}{289}+(\frac{25}{289})\\\\orth_ab.orth_ab = \frac{425}{289}[/tex]
