Linear Algebra: Basic dot product property

[Idealistic]

Let a and b be vectors. If a*a = 2 and a*b = 2, then is it true that a = b? The "*" represents the dot product.

 

[Deleted member 4993]

Let a and b be vectors. If a*a = 2 and a*b = 2, then is it true that a = b? The "*" represents the dot product.

The short answer is NO.

Can you provide an example where \(\displaystyle a \ne b\)?

 

[Idealistic]

The short answer is NO.

Can you provide an example where \(\displaystyle a \ne b\)?

so if a*a = 2 and a*b = 2, then

a*a - a*b = 0 and

a*(a - b) = 0

does this mean either:

(1) a and b are different but the vector a - b has to be orthogonal to a

(2) a and b are equal?

 

[Deleted member 4993]

so if a*a = 2 and a*b = 2, then

a*a - a*b = 0 and

a*(a - b) = 0

does this mean either:

(1) a and b are different but the vector a - b has to be orthogonal to a

(2) a and b are equal?

Suppose vector a is <a_1, a₂> and vector b is <1/a₁ , 1/a₂>.

Then

what is a*b?

 

[Idealistic]

Suppose vector a is <a_1, a₂> and vector b is <1/a₁ , 1/a₂>.

Then

what is a*b?

It'll be two, but im having trouble understanding of the relationship between the magnitude/direction of vector a, the magnitude is sqrt(2), and the magnitude/direction of b.

All the possible choices for <a1, a2> lie on the circle with radius sqrt(2), what are all of the possible choices for b? Is there a relationship? (e.g. is b always parallel or a scalar multiple of a, or is a - b always orthogonal to a)?

 

[Idealistic]

It'll be two, but im having trouble understanding of the relationship between the magnitude/direction of vector a, the magnitude is sqrt(2), and the magnitude/direction of b.

All the possible choice for <a1, a2> lie on the circle with radius sqrt(2), what are all of the possible choices for b? Is there a relationship? (e.g. is b always parallel or a scalar multiple of a, or is a - b always orthogonal to a)?

I get it. If a and b arn't equal, but the satisfy the assumed property a*a = a*b, then a - b will be orthogonal to a by design.

Even if we choose a = <a₁, a₂> and b = <1/a₁, 1/a₂>, if a and b arn't equal a*(a - b) will still equal zero.

since a - b = <(a₁² - 1)/a₁, (a₂² - 1)/a₂>

a*b = a₁² - 1 + a₂² - 1 = |a|^2 - 2 = 0 since the magnitude of a is given as sqrt(2).