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Examples for The scalar product

Example 1

i · i = 1

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Solution

Since the length (or also called the magnitude) of i is 1 and the angle h here is 0 we get
i · i = |i||i|cos0 = 1.

Example 2

i · j = 0

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Solution

Since the angle h between i and j is 90° we get
i · j = |i||j|cos90 = 0.

Example 3

Given vectors u and v such that |u| = 1, |v| = 3 and the angle between u and v is 45°. Calculate u · v.

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Solution

By definition
u · v = |u||v|cosh = (1)(3)(1/ V~ 2) = 3/V ~ 2.

Example 4

Given u = i - 2j + 2k and v = 4i - 3k. Calculate u · v.

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Solution

Using the formula for the scalar product in terms of components we get
u · v = (1)(4) + (- 2)(0) + (2)(- 3) = -2.