Given vectors $u$ and $v$, with $| u |=10$ and $| v |=12$, and the angle between the vectors is $53^{\circ}$, what is $u \cdot v$?
Answer (2)
The dot product of vectors u and v can be calculated using the formula u ⋅ v = ∣ u ∣∣ v ∣ cos ( θ ) . Substituting the given values, we find the dot product is approximately 72.22 . This value highlights the relationship between the magnitudes of the vectors and the angle between them.
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Recall the dot product formula: u ⋅ v = ∣ u ∣∣ v ∣ cos ( θ ) .
Substitute the given magnitudes ∣ u ∣ = 10 , ∣ v ∣ = 12 , and angle θ = 5 3 ∘ into the formula.
Calculate cos ( 5 3 ∘ ) ≈ 0.6018 .
Compute the dot product: u ⋅ v = 10 × 12 × 0.6018 = 72.216 ≈ 72.22 .
The dot product of vectors u and v is 72.22 .
Explanation
Problem Analysis We are given two vectors, u and v , with magnitudes ∣ u ∣ = 10 and ∣ v ∣ = 12 . The angle between them is θ = 5 3 ∘ . Our goal is to find the dot product u ⋅ v .
Dot Product Formula The dot product of two vectors is defined as: u ⋅ v = ∣ u ∣∣ v ∣ cos ( θ ) where ∣ u ∣ and ∣ v ∣ are the magnitudes of the vectors, and θ is the angle between them.
Finding Cosine of the Angle We are given ∣ u ∣ = 10 , ∣ v ∣ = 12 , and θ = 5 3 ∘ . We need to find cos ( 5 3 ∘ ) .
Using a calculator, we find that cos ( 5 3 ∘ ) ≈ 0.6018 .
Calculating the Dot Product Now, we substitute the given values into the dot product formula: u ⋅ v = ( 10 ) ( 12 ) cos ( 5 3 ∘ ) u ⋅ v = ( 10 ) ( 12 ) ( 0.6018 ) u ⋅ v = 120 × 0.6018 u ⋅ v = 72.216 Rounding to two decimal places, we get u ⋅ v ≈ 72.22 .
Final Answer Therefore, the dot product of the two vectors is approximately 72.22.
Examples
The dot product is incredibly useful in physics and engineering. For example, if you're calculating the work done by a force, you can use the dot product of the force vector and the displacement vector. Imagine pushing a box across the floor; the work done is maximized when you push in the direction of motion and is zero when you push perpendicular to the motion. The dot product captures this relationship perfectly, allowing engineers to design efficient systems and physicists to understand energy transfer.
