ACTIVITY 5.6
PERPENDICULARITY OF VECTORS AND SCALAR PRODUCT

According to the definition of scalar product, that it's·=||||cos(), if we have two vectors and not null and perpendicular, his scalar product is zero (because they form an angle of  90º and cos90º = 0).

Reciprocity, if the scalar product of two not null vectors and is zero, the cosine of the angle that they form has to be zero (because they aren't zero not  || neither||), the formed angle is 90º, and vectors are perpendicular.

This fact let us to give the next definition:

 Two vectors are perpendicular (or orthogonal)
when and only when their scalar product is zero

Symbolically, if we indicate the perpendicularity with :      · = 0

It is interesting to observe two things:
1) Remembering how do you calculate the scalar product from his components, the condition of perpendicularity in two given vectors to his components is
                                        = (a₁,a₂)   = (b₁,b₂)        · = a₁b₁+ a₂b₂ = 0
2) We have extended the definition of perpendicularity also to null vectors, and according to this new definition, null vector is perpendicular to any other vector.

INTERACTIVE ACTIVITY

1) Using  ·= (a₁,a₂)·(b₁,b₂) = a₁b₁+ a₂b₂  say which of the next pairs of vectors has scalar product zero and, therefore, are perpendicular. Check it in the applet of the right side.

a)  (4,1) and (1,4)	b)  (4,1) and (-1,4)
c)  (-1,2) and (4,2)	d)  (3,2) and (1,-3)
e)  (6,2) and (-1,3)	f)  (5,2) and (0,0)
g)  (3,2) and (-3,-2)	h)  (6,-3) and (2,-1)
i)  (2,1) and (4,5)	j)  (3,0) and (0,-2)

2) Draw vector = (-2,3) and try to draw two perpendicular vectors to. Which are his components?

SOLUTION

END OF ACTIVITY 5.6
PERPENDICULARITY OF VECTORS AND SCALAR PRODUCT