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.
I have tried to assign each vector (b,a,c) arbitrary identities with unit vectors i,j,k, but I'm really stuck. Any help would be greatly appreciated.
their are 360 legs in total and 6x the amount of birds than cats.How many birds and cats are their?
Oh...I know theres a multiplicity involved, and the difference of those two expressions has to equal 0, because their difference dot the c vector has to be 0, so they can be perpendicular, correct? Sorry, I'm not sure how what you said relates to this problem when I have to use arbitrary vectors...
To show that (⃗b·⃗c)a−(⃗a·⃗c)⃗b is perpendicular to ⃗c, we can use the following properties of the dot product:
1. Distributive property: ⃗a·(⃗b+⃗c) = ⃗a·⃗b + ⃗a·⃗c
2. Commutative property: ⃗a·⃗b = ⃗b·⃗a
3. Associative property: (k⃗a)·⃗b = k(⃗a·⃗b)
Let's start the proof:
First, let's calculate the dot products involved:
⃗b·⃗c = b1c1 + b2c2 + b3c3 (where b1, b2, b3, c1, c2, c3 are the components of the vectors ⃗b and ⃗c)
⃗a·⃗c = a1c1 + a2c2 + a3c3 (where a1, a2, a3, c1, c2, c3 are the components of the vectors ⃗a and ⃗c)
Now, let's compute the expression (⃗b·⃗c)a−(⃗a·⃗c)⃗b:
(⃗b·⃗c)a = (b1c1 + b2c2 + b3c3)a = b1c1a + b2c2a + b3c3a
(⃗a·⃗c)⃗b = (a1c1 + a2c2 + a3c3)⃗b = a1c1⃗b + a2c2⃗b + a3c3⃗b
Combining these two expressions:
(⃗b·⃗c)a−(⃗a·⃗c)⃗b = (b1c1a + b2c2a + b3c3a) − (a1c1⃗b + a2c2⃗b + a3c3⃗b)
We can now rewrite this expression using the distributive property:
= b1c1a − a1c1⃗b + b2c2a − a2c2⃗b + b3c3a − a3c3⃗b
Next, we can rearrange the terms by using the commutative property:
= (b1a − a1⃗b)c1 + (b2a − a2⃗b)c2 + (b3a − a3⃗b)c3
Now, let's compare this expression with the dot product of ⃗c and the vector ⃗v = (b1a − a1⃗b, b2a − a2⃗b, b3a − a3⃗b):
⃗c·⃗v = c1(b1a − a1⃗b) + c2(b2a − a2⃗b) + c3(b3a − a3⃗b)
= c1b1a − c1a1⃗b + c2b2a − c2a2⃗b + c3b3a − c3a3⃗b
We can now see that ⃗c·⃗v is equal to (⃗b·⃗c)a−(⃗a·⃗c)⃗b.
Finally, we can conclude that ⃗c·⃗v = (⃗b·⃗c)a−(⃗a·⃗c)⃗b, which means that ⃗c and the vector ⃗v are parallel since their dot product is zero.
Therefore, we have shown that (⃗b·⃗c)a−(⃗a·⃗c)⃗b is perpendicular to ⃗c, as desired.
I hope this explanation helps you understand how to prove the perpendicularity of vectors using the properties of the dot product.