2. Use the properties of the dot product to show that (⃗b·⃗c)a−(⃗a·⃗c)⃗b is perpendicular to ⃗c.
Must be shown for arbitrary vectors.
Im sorry, I'm really stuck on this. I know that is a vector is perpendicular to another their dot product is 0. Can anyone steer me in the right direction?
To show that (⃗b·⃗c)a - (⃗a·⃗c)⃗b is perpendicular to ⃗c, we need to prove that their dot product is zero.
Let's start by calculating the dot product of (⃗b·⃗c)a - (⃗a·⃗c)⃗b and ⃗c:
(⃗b·⃗c)a - (⃗a·⃗c)⃗b) · ⃗c
Expanding the dot product, we have:
(⃗b·⃗c)a · ⃗c - (⃗a·⃗c)⃗b · ⃗c
Using the properties of the dot product, we can factor out constants:
⃗b·⃗c · (a · ⃗c) - ⃗a·⃗c · (⃗b · ⃗c)
Notice that ⃗b · ⃗c is a scalar, so we can interchange the order:
⃗b·⃗c · (a · ⃗c) - ⃗a·⃗c · (⃗c · ⃗b)
Since the dot product is commutative, we can further simplify:
⃗b·⃗c · (a · ⃗c) - ⃗a·⃗c · (⃗b · ⃗c)
Again, ⃗c · ⃗b is a scalar, so we can rearrange:
⃗b·⃗c · (a · ⃗c) - (⃗a·⃗c) · (⃗b · ⃗c)
Since scalar multiplication is commutative, we can write:
(a · ⃗c) · (⃗b·⃗c) - (⃗a·⃗c) · (⃗b · ⃗c)
Now it's clear that we have two dot products subtracted from each other. By the dot product property, we know that a · ⃗c and ⃗a · ⃗c are scalar quantities, so we can interchange their positions:
(a · ⃗c) · (⃗b·⃗c) - (⃗b·⃗c) · (⃗a·⃗c)
Now, let's factor out the common factor (⃗b·⃗c):
(⃗b·⃗c)[(a · ⃗c) - (⃗a·⃗c)]
Finally, we can see that (a · ⃗c) - (⃗a·⃗c) is simply the difference between two scalar quantities, hence it's still a scalar. Therefore, we can rewrite this as:
(⃗b·⃗c) · (scalar)
Since the dot product of a scalar and a vector is just a scalar multiplied by each component of the vector, the dot product of (⃗b·⃗c) and any vector results in a scalar. Consequently, (⃗b·⃗c) · (scalar) equals zero, implying that (⃗b·⃗c)a - (⃗a·⃗c)⃗b is perpendicular to ⃗c.
Therefore, we have shown that (⃗b·⃗c)a - (⃗a·⃗c)⃗b is perpendicular to ⃗c for arbitrary vectors ⃗a, ⃗b, and ⃗c.