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A First Course in Linear Algebra, 2017a

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4.7. The Dot Product 169<br />

Example 4.34: Us<strong>in</strong>g Geometric Description to F<strong>in</strong>d a Dot Product<br />

Let ⃗u,⃗v be vectors with ‖⃗u‖ = 3 and ‖⃗v‖ = 4. Suppose the angle between ⃗u and⃗v is π/3. F<strong>in</strong>d⃗u•⃗v.<br />

Solution. From the geometric description of the dot product <strong>in</strong> Proposition 4.31<br />

⃗u •⃗v =(3)(4)cos(π/3)=3 × 4 × 1/2 = 6<br />

Two nonzero vectors are said to be perpendicular, sometimes also called orthogonal, if the <strong>in</strong>cluded<br />

angle is π/2radians(90 ◦ ).<br />

Consider the follow<strong>in</strong>g proposition.<br />

Proposition 4.35: Perpendicular Vectors<br />

Let ⃗u and⃗v be nonzero vectors <strong>in</strong> R n . Then, ⃗u and⃗v are said to be perpendicular exactly when<br />

⃗u •⃗v = 0<br />

♠<br />

Proof. This follows directly from Proposition 4.31. <strong>First</strong> if the dot product of two nonzero vectors is equal<br />

to 0, this tells us that cosθ = 0 (this is where we need nonzero vectors). Thus θ = π/2 and the vectors are<br />

perpendicular.<br />

If on the other hand ⃗v is perpendicular to ⃗u, then the <strong>in</strong>cluded angle is π/2 radians. Hence cosθ = 0<br />

and ⃗u •⃗v = 0.<br />

♠<br />

Consider the follow<strong>in</strong>g example.<br />

Example 4.36: Determ<strong>in</strong>e if Two Vectors are Perpendicular<br />

Determ<strong>in</strong>e whether the two vectors,<br />

⎡ ⎤ ⎡ ⎤<br />

2 1<br />

⃗u = ⎣ 1 ⎦,⃗v = ⎣ 3 ⎦<br />

−1 5<br />

are perpendicular.<br />

Solution. In order to determ<strong>in</strong>e if these two vectors are perpendicular, we compute the dot product. This<br />

is given by<br />

⃗u •⃗v =(2)(1)+(1)(3)+(−1)(5)=0<br />

Therefore, by Proposition 4.35 these two vectors are perpendicular.<br />

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