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Unformatted text preview: A nicer way to express this: Note that since d = (d/a)d when a = 0, d −bd/a = d a Thus, with d arbitrary, the vectors of the form of the vector −b . a Notice that the prescription all of the multiples of the vector a b −b a −b a −b a −bd/a d . are precisely all of the multiples also gives us the set of vectors orthogonal to for any vector a in IR2 , deﬁne a⊥ = Then, we can answer: when a = 0 as long as b = 0. Therefore, . (∗) • If a = 0, then every vector in IR2 is orthogonal to a. Otherwise, the set of all vectors that are orthogonal to a is the set of all multiples of a⊥ . Later on, we will see that the deﬁnition (∗) is quite useful. The symbol ⊥ is called the perp symbol, and one reads a⊥ as a perp.
4.9 Let x and y be two non–zero vectors in IRn . Suppose that |x + y|2 = |x|2 + |y|2 . What is the angle between these two vectors? SOLUTION Using the dot product to compute the length of x + y, |x + y|2 = |x|2 + |y|2 + x · y + y · x = |x|2 + |y|2 + 2x · y . Therefore, the condition |x + y|2 = |x|2 + |y|2 implies that x·y =0 . That is, x and y are orthognal, and so the angle between them is π/2.
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