Calculating Vector Distance in R^n Using Dot Product - Homework Solution

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SUMMARY

The discussion focuses on calculating the vector distance between two orthogonal unit vectors x and y in R^n, where ||x|| = ||y|| = 1 and x^T * y = 0. The solution requires using the equation ||x-y|| = sqrt((x-y)^T(x-y)), which simplifies to sqrt(x^Tx - x^Ty - y^Tx + y^Ty). Given that both x^Ty and y^Tx equal zero, the final result confirms that ||x-y|| = sqrt(2). This demonstrates the application of the dot product in determining distances in vector spaces.

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  • Basic principles of Euclidean geometry
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Homework Statement



Let x and y be vectos in R^n such that ||x|| = ||y|| = 1 and x^T * y = 0. Use Eq. (1) to show that ||x-y|| = sqrt(2).

Homework Equations



Eq. (1): ||x|| = sqrt(x^T * X)

The Attempt at a Solution



I could figure this out knowing that the dot product of the vectors is zero so the are perpendicular, the two sides of the triangle are 1 so the distance is sqrt(2) but this problem wants the answer to be found with Eq. (1) and I don't know how to do that and how to find ||x-y||.

Thanks for the help
 
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phrygian said:

Homework Statement



Let x and y be vectos in R^n such that ||x|| = ||y|| = 1 and x^T * y = 0. Use Eq. (1) to show that ||x-y|| = sqrt(2).

Homework Equations



Eq. (1): ||x|| = sqrt(x^T * X)

The Attempt at a Solution



I could figure this out knowing that the dot product of the vectors is zero so the are perpendicular, the two sides of the triangle are 1 so the distance is sqrt(2) but this problem wants the answer to be found with Eq. (1) and I don't know how to do that and how to find ||x-y||.

Thanks for the help
Your formula says that ||x- y||= [itex]\sqrt{(x- y)^T(x- y)}= \sqrt{x^Tx- x^Ty- y^Tx+ y^Ty}[/itex]. Both [itex]x^Ty[/itex] and [itex]y^Tx[/itex] are 0.
 

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