We have, a. ortho (ab) = a.b - a.b/[a]² . /a/²
The /a/² cancel off each other, therefore,
a. ortho (ab) = a.b - a.b
a. ortho (ab) = 0
Therefore, the final result is 0.
Step-by-step explanation:
It should be noted that the orthogonal projection of a vector B onto A is given by a given formula known as:
proJ(ab) = a.b/[a]² . a equation (1)
Where [a] is the magnitude of the vector a,
therefore, [a]² = a.a equation (2)
Which means that if we have a dot product between a vector and the sum of vector, it can be distributed to each vector.
That is : a . (c - d + b )
a.b - a.d + a.b equation (3)
and since it is given that Ortho (ab) = b - proJ(ab)equation (4)
and in order to show that a two vector are orthogonal, we need to show that their dot product is zero.
Where : a.b = /a//b/ cos ( Ф)
and if the dot product is zero (0), this will imply that ,
cos (Ф) = 0
Now moving forward to the proof,,
The first thing we do is substitute equation (1) into equation (4)
That will give us : ortho (ab) = b - a.b/[a]² . a
and in order to show that vector a and orth(ab) are orthogonal, we need to show that their dot is zero
Therefore, a . ortho (ab) = a . b - a.b/[a]² . a equation (5)
Now, substituting equation (2) into (5),
We have, a. ortho (ab) = a.b - a.b/[a]² . /a/²
The /a/² cancel off each other, therefore,
a. ortho (ab) = a.b - a.b
a. ortho (ab) = 0
Therefore, the final result is 0.
