Multivariable Calculus Quiz: Dot Product
2 questions · exam conditions
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Dot ProductQuestion 1 of 2

Two vectors u\mathbf{u} and v\mathbf{v} satisfy ∣∣u∣∣=3||\mathbf{u}|| = 3, ∣∣v∣∣=4||\mathbf{v}|| = 4, and ∣∣u+v∣∣=6||\mathbf{u} + \mathbf{v}|| = 6. What is the length of the projection of u−v\mathbf{u} - \mathbf{v} onto u+v\mathbf{u} + \mathbf{v}?

56\frac{5}{6}
76\frac{7}{6}
53\frac{5}{3}
73\frac{7}{3}
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Dot Product

Practice Dot Product in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Dot Product, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Two vectors u\mathbf{u} and v\mathbf{v} satisfy ∣∣u∣∣=3||\mathbf{u}|| = 3, ∣∣v∣∣=4||\mathbf{v}|| = 4, and ∣∣u+v∣∣=6||\mathbf{u} + \mathbf{v}|| = 6. What is the length of the projection of u−v\mathbf{u} - \mathbf{v} onto u+v\mathbf{u} + \mathbf{v}?

  1. 56\frac{5}{6}
  2. 76\frac{7}{6} (correct answer)
  3. 53\frac{5}{3}
  4. 73\frac{7}{3}
Explanation: First, we find u⋅v\mathbf{u} \cdot \mathbf{v} using ∣∣u+v∣∣2=∣∣u∣∣2+∣∣v∣∣2+2u⋅v||\mathbf{u} + \mathbf{v}||^2 = ||\mathbf{u}||^2 + ||\mathbf{v}||^2 + 2\mathbf{u} \cdot \mathbf{v}. We have 36=9+16+2u⋅v36 = 9 + 16 + 2\mathbf{u} \cdot \mathbf{v}, so u⋅v=112\mathbf{u} \cdot \mathbf{v} = \frac{11}{2}. The projection length is ∣(u−v)⋅(u+v)∣∣∣u+v∣∣\frac{|(\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v})|}{||\mathbf{u} + \mathbf{v}||}. We compute (u−v)⋅(u+v)=∣∣u∣∣2−∣∣v∣∣2=9−16=−7(\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = ||\mathbf{u}||^2 - ||\mathbf{v}||^2 = 9 - 16 = -7. Therefore, the projection length is ∣−7∣6=76\frac{|-7|}{6} = \frac{7}{6}. Option A uses the wrong dot product calculation, C and D incorrectly compute the denominator.

Question 2

Consider vectors p=⟨1,2,−1⟩\mathbf{p} = \langle 1, 2, -1 \rangle and q=⟨3,−1,2⟩\mathbf{q} = \langle 3, -1, 2 \rangle. If the angle between p\mathbf{p} and r\mathbf{r} is 60°60° and the angle between q\mathbf{q} and r\mathbf{r} is 120°120°, where ∣∣r∣∣=2||\mathbf{r}|| = 2, what is p⋅r+q⋅r\mathbf{p} \cdot \mathbf{r} + \mathbf{q} \cdot \mathbf{r}?

  1. 6−14\sqrt{6} - \sqrt{14} (correct answer)
  2. 26−2142\sqrt{6} - 2\sqrt{14}
  3. 14−6\sqrt{14} - \sqrt{6}
  4. 214−262\sqrt{14} - 2\sqrt{6}
Explanation: Using the dot product formula u⋅v=∣∣u∣∣∣∣v∣∣cos⁡θ\mathbf{u} \cdot \mathbf{v} = ||\mathbf{u}|| ||\mathbf{v}|| \cos \theta, we find: p⋅r=∣∣p∣∣∣∣r∣∣cos⁡60°=6⋅2⋅12=6\mathbf{p} \cdot \mathbf{r} = ||\mathbf{p}|| ||\mathbf{r}|| \cos 60° = \sqrt{6} \cdot 2 \cdot \frac{1}{2} = \sqrt{6} and q⋅r=∣∣q∣∣∣∣r∣∣cos⁡120°=14⋅2⋅(−12)=−14\mathbf{q} \cdot \mathbf{r} = ||\mathbf{q}|| ||\mathbf{r}|| \cos 120° = \sqrt{14} \cdot 2 \cdot (-\frac{1}{2}) = -\sqrt{14}. Therefore, p⋅r+q⋅r=6−14\mathbf{p} \cdot \mathbf{r} + \mathbf{q} \cdot \mathbf{r} = \sqrt{6} - \sqrt{14}. Here ∣∣p∣∣=1+4+1=6||\mathbf{p}|| = \sqrt{1 + 4 + 1} = \sqrt{6} and ∣∣q∣∣=9+1+4=14||\mathbf{q}|| = \sqrt{9 + 1 + 4} = \sqrt{14}. Option B incorrectly doubles both terms, C reverses the signs, and D both reverses signs and doubles.