Multivariable Calculus Exam 1 Study Guide: Step-by-Step Guidance

Q1a. Is $\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}$ always true, sometimes true, or never true?

Background

Topic: Dot Product Properties

This question tests your understanding of the symmetry property of the dot product in vector algebra.

Key Terms and Formulas:

• Dot product: $\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3$

• Commutativity: $\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}$

Step-by-Step Guidance

1. Recall the definition of the dot product for two vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^3$.

2. Write out the dot product in terms of components: $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$.

3. Compare $\mathbf{u} \cdot \mathbf{v}$ and $\mathbf{v} \cdot \mathbf{u}$ by substituting the components.

4. Think about whether the order of multiplication affects the result for real numbers.

Try solving on your own before revealing the answer!

Final Answer: Always True (AT)

The dot product is commutative because $\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 = v_1u_1 + v_2u_2 + v_3u_3 = \mathbf{v} \cdot \mathbf{u}$.

This holds for all vectors in $\mathbb{R}^3$.

Q1b. If $\mathbf{u} \cdot \mathbf{v} = 0$, does it follow that $\mathbf{u} = 0$ or $\mathbf{v} = 0$? Is this always, sometimes, or never true?

Background

Topic: Orthogonality and Dot Product

This question tests your understanding of what it means for two vectors to be orthogonal and whether the zero dot product implies one vector is zero.

Key Terms and Formulas:

• Orthogonal vectors: $\mathbf{u} \cdot \mathbf{v} = 0$

• Zero vector: $\mathbf{u} = 0$ or $\mathbf{v} = 0$

Step-by-Step Guidance

1. Recall the definition of the dot product and what it means for it to be zero.

2. Consider whether two nonzero vectors can have a dot product of zero.

3. Think about examples of orthogonal vectors that are not zero vectors.

4. Try to construct a counterexample or proof based on your reasoning.

Try solving on your own before revealing the answer!

Final Answer: Sometimes True (ST)

If $\mathbf{u} = 0$ or $\mathbf{v} = 0$, then $\mathbf{u} \cdot \mathbf{v} = 0$. However, two nonzero vectors can also have a dot product of zero if they are orthogonal (e.g., $\mathbf{u} = (1,0,0)$ and $\mathbf{v} = (0,1,0)$).

So, it is sometimes true: provide examples for both cases.

Q1c. Is the cross product of two unit vectors always a unit vector?

Background

Topic: Cross Product Properties

This question tests your understanding of the magnitude of the cross product and its relation to the original vectors.

Key Terms and Formulas:

• Unit vector: $|\mathbf{u}| = 1$

• Cross product magnitude: $|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}| |\mathbf{v}| \sin \theta$

Step-by-Step Guidance

1. Recall the formula for the magnitude of the cross product.

2. Substitute $|\mathbf{u}| = 1$ and $|\mathbf{v}| = 1$ into the formula.

3. Consider the value of $\sin \theta$ for different angles between the vectors.

4. Think about cases where $\sin \theta = 1$ and $\sin \theta = 0$.

Try solving on your own before revealing the answer!

Final Answer: Sometimes True (ST)

The cross product of two unit vectors is a unit vector only if they are perpendicular ($\theta = 90^\circ$), since $|\mathbf{u} \times \mathbf{v}| = \sin \theta$. If they are parallel, the cross product is zero. So, it is sometimes true.

Q1d. Is $\mathbf{v} \times \mathbf{v} = 0$ always, sometimes, or never true?

Background

Topic: Cross Product Properties

This question tests your understanding of the cross product of a vector with itself.

Key Terms and Formulas:

• Cross product: $\mathbf{v} \times \mathbf{v}$

• Magnitude: $|\mathbf{v} \times \mathbf{v}| = |\mathbf{v}| |\mathbf{v}| \sin 0 = 0$

Step-by-Step Guidance

1. Recall the formula for the cross product and its magnitude.

2. Consider the angle between a vector and itself.

3. Calculate $\sin 0$ and see what happens to the magnitude.

4. Think about whether this result holds for all vectors.

Try solving on your own before revealing the answer!

Final Answer: Always True (AT)

The cross product of any vector with itself is always zero because the angle between them is zero and $\sin 0 = 0$.

Q2a. Given $\mathbf{a} = \mathbf{i} + \mathbf{j} - 2\mathbf{k}$ and $\mathbf{b} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k}$, calculate $\mathbf{c} = \mathbf{a} + \mathbf{b}$.

Background

Topic: Vector Addition

This question tests your ability to add vectors component-wise.

Key Terms and Formulas:

• Vector addition: $\mathbf{a} + \mathbf{b} = (a_1 + b_1)\mathbf{i} + (a_2 + b_2)\mathbf{j} + (a_3 + b_3)\mathbf{k}$

Step-by-Step Guidance

1. Write $\mathbf{a}$ and $\mathbf{b}$ in component form: $\mathbf{a} = (1, 1, -2)$, $\mathbf{b} = (3, -2, 1)$.

2. Add corresponding components: $c_1 = a_1 + b_1$, $c_2 = a_2 + b_2$, $c_3 = a_3 + b_3$.

3. Express $\mathbf{c}$ in terms of $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$.

Try solving on your own before revealing the answer!

Final Answer: $\mathbf{c} = 4\mathbf{i} - \mathbf{j} - \mathbf{k}$

Adding the components: $1+3=4$, $1+(-2)=-1$, $-2+1=-1$.

Q2b. Determine the angle between $\mathbf{a}$ and $\mathbf{c}$.

Background

Topic: Angle Between Vectors

This question tests your ability to use the dot product to find the angle between two vectors.

Key Terms and Formulas:

• Dot product: $\mathbf{a} \cdot \mathbf{c} = |\mathbf{a}| |\mathbf{c}| \cos \theta$

• Magnitude: $|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$

• Angle: $\theta = \arccos \left( \frac{\mathbf{a} \cdot \mathbf{c}}{|\mathbf{a}| |\mathbf{c}|} \right)$

Step-by-Step Guidance

1. Write $\mathbf{a}$ and $\mathbf{c}$ in component form.

2. Compute the dot product $\mathbf{a} \cdot \mathbf{c}$.

3. Find the magnitudes $|\mathbf{a}|$ and $|\mathbf{c}|$.

4. Set up the formula for $\theta$ using the values you calculated.

Try solving on your own before revealing the answer!

Final Answer: $\theta \approx 54.7^\circ$

Using the formula $\theta = \arccos \left( \frac{\mathbf{a} \cdot \mathbf{c}}{|\mathbf{a}| |\mathbf{c}|} \right)$ and plugging in the computed values gives approximately $54.7^\circ$.

Q2c. Determine the angle between $\mathbf{b}$ and $\mathbf{c}$.

Background

Topic: Angle Between Vectors

• This question tests your ability to use the dot product to find the angle between two vectors.

  Key Terms and Formulas:

  • Dot product: $\mathbf{b} \cdot \mathbf{c} = |\mathbf{b}| |\mathbf{c}| \cos \theta$

  • Magnitude: $|\mathbf{b}| = \sqrt{b_1^2 + b_2^2 + b_3^2}$

• Angle: $\theta = \arccos \left( \frac{\mathbf{b} \cdot \mathbf{c}}{|\mathbf{b}| |\mathbf{c}|} \right)$

Step-by-Step Guidance

1. Write $\mathbf{b}$ and $\mathbf{c}$ in component form.

2. Compute the dot product $\mathbf{b} \cdot \mathbf{c}$.

3. Find the magnitudes $|\mathbf{b}|$ and $|\mathbf{c}|$.

4. Set up the formula for $\theta$ using the values you calculated.

Try solving on your own before revealing the answer!

Final Answer: $\theta \approx 60.0^\circ$

Using the formula $\theta = \arccos \left( \frac{\mathbf{b} \cdot \mathbf{c}}{|\mathbf{b}| |\mathbf{c}|} \right)$ and plugging in the computed values gives approximately $60.0^\circ$.

Q3. Find an equation of the plane containing the points $A(1, 0, 0)$, $B(2, 0, -1)$, and $C(1, 4, 3)$.

Background

Topic: Equation of a Plane in $\mathbb{R}^3$

This question tests your ability to find the equation of a plane given three points in space.

Key Terms and Formulas:

• General equation of a plane: $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$

• Normal vector: $\mathbf{n} = \mathbf{AB} \times \mathbf{AC}$

Step-by-Step Guidance

1. Find vectors $\mathbf{AB}$ and $\mathbf{AC}$ using the given points.

2. Compute the cross product $\mathbf{AB} \times \mathbf{AC}$ to get the normal vector $\mathbf{n}$.

3. Use the normal vector and one of the points (e.g., $A$) to write the equation of the plane.

4. Set up the equation in the form $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.

Try solving on your own before revealing the answer!

Final Answer: $4x + 2y + 4z = 4$

The normal vector is $(4, 2, 4)$, and using point $A(1, 0, 0)$, the equation simplifies to $4x + 2y + 4z = 4$.

Q4. Find a parametric equation for the line through the point $(-2, 2, 4)$ that is perpendicular to the plane $2x - y + 5z = 12$.

Background

Topic: Parametric Equations of Lines and Planes

This question tests your ability to write a parametric equation for a line given a point and a direction vector.

Key Terms and Formulas:

• Normal vector to the plane: $\mathbf{n} = (2, -1, 5)$

• Parametric equation: $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{n}$

Step-by-Step Guidance

1. Identify the point through which the line passes: $(-2, 2, 4)$.

2. Find the normal vector to the plane, which gives the direction of the line.

3. Write the parametric equations for $x$, $y$, and $z$ using the point and direction vector.

4. Set up the equations in the form $x = x_0 + 2t$, $y = y_0 - t$, $z = z_0 + 5t$.

Try solving on your own before revealing the answer!

Final Answer:

$x = -2 + 2t$, $y = 2 - t$, $z = 4 + 5t$

This line passes through $(-2, 2, 4)$ and is perpendicular to the given plane.

Q5a. Write an equation for a hyperboloid of two sheets with its axis on the y-axis.

Background

Topic: Quadric Surfaces

This question tests your knowledge of the standard equations for quadric surfaces, specifically the hyperboloid of two sheets.

Key Terms and Formulas:

• Hyperboloid of two sheets: $\frac{y^2}{b^2} - \frac{x^2}{a^2} - \frac{z^2}{c^2} = -1$

Step-by-Step Guidance

1. Recall the general form for a hyperboloid of two sheets.

2. Identify which variable is squared positively (the axis of the surface).

3. Write the equation with $y$ as the axis variable.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{y^2}{b^2} - \frac{x^2}{a^2} - \frac{z^2}{c^2} = -1$

This equation describes a hyperboloid of two sheets with its axis along the y-axis.

Q5b. Sketch the $xy$-, $yz$-, and $xz$-traces for the hyperboloid of two sheets.

Background

Topic: Traces of Quadric Surfaces

This question tests your ability to visualize and sketch traces (cross-sections) of a quadric surface in different coordinate planes.

Key Terms and Formulas:

• Trace: The intersection of the surface with a coordinate plane (e.g., $z=0$ for $xy$-trace).

Step-by-Step Guidance

1. Set $z=0$ to find the $xy$-trace: $\frac{y^2}{b^2} - \frac{x^2}{a^2} = -1$.

2. Set $x=0$ to find the $yz$-trace: $\frac{y^2}{b^2} - \frac{z^2}{c^2} = -1$.

3. Set $y=0$ to find the $xz$-trace: $-\frac{x^2}{a^2} - \frac{z^2}{c^2} = -1$.

4. Analyze the resulting equations to determine the shapes (hyperbolas or ellipses).

Try solving on your own before revealing the answer!

Final Answer:

The $xy$- and $yz$-traces are hyperbolas, while the $xz$-trace is an ellipse (for $y=0$).

Sketch each trace accordingly in its respective plane.

Q5c. Combine the traces into a 3D sketch of the quadric surface (label axes).

Background

Topic: 3D Visualization of Quadric Surfaces

This question tests your ability to synthesize traces into a full 3D sketch and label axes.

Key Terms and Formulas:

• 3D sketch: Combine traces to visualize the surface.

Step-by-Step Guidance

1. Draw the $xy$-, $yz$-, and $xz$-traces in their respective planes.

2. Label the $x$, $y$, and $z$ axes clearly.

3. Combine the traces to show the two-sheet structure of the hyperboloid.

4. Indicate the axis of symmetry (y-axis) in your sketch.

Try solving on your own before revealing the answer!

Final Answer:

The 3D sketch should show two separate sheets opening along the y-axis, with hyperbolic traces in $xy$ and $yz$ planes and elliptical traces in $xz$ plane. Axes should be labeled $x$, $y$, and $z$.