Using Inner Product Properties to Solve Vector Problems

  • #1
Blackbear38
1
0
Summary:: I need to solve a problem for an assignment but just couldn't find the right approach. I fail to eliminate b or c to get only the magnitude of a.

Let a, b and c be unit vectors such that a⋅b=1/4, b⋅c=1/7 and a⋅c=1/8. Evaluate (write in the exact form):
- ||4a||
- 3a.5b
- a.(b-c)
- (a+b+c).(a-b)

What I first did was ab.ac = a^(2).bc then substitute values of ab, ac, and bc, but I cannot confirm that this is the correct approach. Hence, I found:

ab.ac = a^(2).bc
(1/4)(1/8)=a^(2)(1/7)
a^(2) = 7/32
Hence, ||a|| = sqrt(14)/8

I really hope that this doubt can be clarified for all the parts of my question. Thanks!

[Moderator's note: moved from a technical forum.]
 
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  • #2
What you need to know is

a is a unit vector, so what is its magnitude?
scalars and vectors are commutative.
The scalar product is commutative and distributive.

The rest is given.
 
  • #3
Blackbear38 said:
Summary:: I need to solve a problem for an assignment but just couldn't find the right approach. I fail to eliminate b or c to get only the magnitude of a.

Let a, b and c be unit vectors such that a⋅b=1/4, b⋅c=1/7 and a⋅c=1/8. Evaluate (write in the exact form):
- ||4a||
- 3a.5b
- a.(b-c)
- (a+b+c).(a-b)

What I first did was ab.ac = a^(2).bc
I have no idea why you did this, and if '.' means "dot product" the above makes no sense.
##a \cdot b## is a number (given) and ##a \cdot c## is also a number (also given). The dot product is defined for vectors, but not plain old numbers.
Blackbear38 said:
Then substitute values of ab, ac, and bc, but I cannot confirm that this is the correct approach.
It's not.
Blackbear38 said:
Hence, I found:

ab.ac = a^(2).bc
(1/4)(1/8)=a^(2)(1/7)
a^(2) = 7/32
Hence, ||a|| = sqrt(14)/8

I really hope that this doubt can be clarified for all the parts of my question. Thanks!

[Moderator's note: moved from a technical forum.]
Following up on @gleem's comments, you need to be looking at the properties of the dot product, such as ##ku \cdot v = k u \cdot v## and ##u \cdot (v +w) = u \cdot v + u \cdot w##, etc. This is a very easy set of problems if you know these properties, plus the fact that a, b, and c are all unit vectors.
 
  • #4
Sorry for this question but are you sure you know what a unit vector is?
 
  • #5
You need to rely more on the basic properties of the inner product. The inner product is linear in both its arguments for real scalars. That should give you almost all the answers.
 
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