Question

Show that the vector orthab=b-projacb is orthogonal to a.

Answer 1

Answer:

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.