Show that proj(b⃗a⃗) =[(a⃗⋅b⃗)/(b⃗⋅b⃗)](b⃗)
To show that the projection of vector b onto vector a is equal to [(a · b) / (b · b)] times vector b, we can use the definition of projection and some vector algebra.
The projection of vector b onto vector a, denoted as proj(b,a), is defined as a vector that lies on vector a and has the same direction as vector a. Mathematically, we can express this as:
proj(b,a) = |proj(b,a)| * unit vector of a
where |proj(b,a)| denotes the magnitude (or length) of the projection of b onto a.
To find the magnitude of the projection of b onto a, we can use the dot product. The dot product of two vectors a and b is defined as:
a · b = |a| * |b| * cos(theta)
where |a| and |b| represent the magnitudes of vectors a and b, respectively, and theta is the angle between them.
Now, let's find the magnitude of the projection of b onto a.
|proj(b,a)| = |b| * cos(theta)
The cosine of the angle theta can be written as:
cos(theta) = (b · a) / (|b| * |a|)
Substituting this value into the previous expression, we get:
|proj(b,a)| = |b| * (b · a) / (|b| * |a|)
Simplifying this further, we have:
|proj(b,a)| = (b · a) / |a|
Now, we want to express this in terms of vector b. Since the magnitude of vector a is present in the numerator, we can rewrite |a| as |b| * |a| / |b|. This gives us:
|proj(b,a)| = (b · a) / (|b| * |a| / |b|)
Simplifying, we obtain:
|proj(b,a)| = (b · a) / |b|
Finally, the projection of b onto a is given by:
proj(b,a) = (b · a) / |b| * unit vector of b
Using the definition of unit vector (a vector with magnitude 1), we can express this as:
proj(b,a) = [(b · a) / (|b| * |b|)] * b
By canceling out the magnitudes of vector b, we arrive at the desired result:
proj(b,a) = [(a · b) / (b · b)] * b