Multivariable calculus: Exercises

03 Planes and linear spaces, part 1

Exercise 1: Point-slope equations for planes

Recall that the point-slope equation for a plane through a point \((x_0,y_0,z_0)\) with slope \(m\) in the positive \(x\)-direction and slope \(n\) in the positive \(y\)-direction is

\[ z = m(x-x_0) + n(y-y_0) + z_0. \]

  1. Find the point-slope equation for a plane through the point \((2, -1, 4)\) with slope \(3\) in the positive \(x\)-direction and slope \(-2\) in the positive \(y\)-direction.

  2. Find the point-slope equation for a plane through the origin with slope \(1\) in the positive \(x\)-direction and slope \(5\) in the positive \(y\)-direction.

  3. The point-slope equation of a plane is \[ z = -2(x+3) + 4(y-1) + 7. \] Through what point does this plane pass? What are the slopes in the \(+x\) and \(+y\) directions?

  4. The point-slope equation of a plane is \[ z = 5y - 2. \] Through what point does this plane pass? What are the slopes in the \(+x\) and \(+y\) directions?

  5. Find the point-slope equation for a plane through the point \((1, 1, 1)\) that is parallel to the \(xy\)-plane.

  1. \(z = 3(x-2) - 2(y+1) + 4\)

  2. \(z = x + 5y\)

  3. The plane passes through \((-3, 1, 7)\). The slope in the \(+x\) direction is \(-2\), and the slope in the \(+y\) direction is \(4\).

  4. The plane passes through \((0, 0, -2)\), has slope \(0\) in the \(+x\) direction, and slope \(5\) in the \(+y\) direction.

  5. A plane parallel to the \(xy\)-plane has slope \(0\) in both the \(+x\) and \(+y\) directions. So the equation is \(z = 0(x-1) + 0(y-1) + 1\), or simply \(z = 1\).

Exercise 2: Normal vectors to planes

Recall that a vector \(\mathbf{n}\) is called a normal vector to a plane if it is orthogonal to all vectors that lie in the plane. For a plane with point-slope equation

\[ z = m(x-x_0) + n(y-y_0) + z_0, \]

a normal vector is given by \(\mathbf{n} = \begin{bmatrix} -m \\ -n \\ 1 \end{bmatrix}\).

  1. Find a normal vector to the plane \(z = 3(x-2) - 2(y+1) + 4\).

  2. Find a normal vector to the plane \(z = x + 5y\).

  3. Find a normal vector to the plane \(z = -4x + 2y - 1\).

  4. Find a normal vector to the plane \(z = 7\).

  5. Find a normal vector to the plane \(y = 3\).

  6. Is \(\mathbf{n} = \begin{bmatrix} -6 \\ 4 \\ 2 \end{bmatrix}\) also a normal vector to the plane \(z = 3(x-2) - 2(y+1) + 4\)? Explain.

  1. \(\mathbf{n} = \begin{bmatrix} -3 \\ 2 \\ 1 \end{bmatrix}\)

  2. \(\mathbf{n} = \begin{bmatrix} -1 \\ -5 \\ 1 \end{bmatrix}\)

  3. \(\mathbf{n} = \begin{bmatrix} 4 \\ -2 \\ 1 \end{bmatrix}\)

  4. \(\mathbf{n} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\).

  5. \(\mathbf{n} = \mathbf{e}_2\).

  6. Yes. Notice that \(\begin{bmatrix} -6 \\ 4 \\ 2 \end{bmatrix} = 2\begin{bmatrix} -3 \\ 2 \\ 1 \end{bmatrix}\). Any scalar multiple (except zero) of a normal vector is also a normal vector.

Exercise 3: Vector equations of planes

Recall that the vector equation of a plane with normal vector \(\mathbf{n}\) passing through the point with position vector \(\mathbf{r}_0\) is

\[ \langle \mathbf{n}, \mathbf{r}-\mathbf{r}_0 \rangle = 0, \]

where \(\mathbf{r} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\) is a variable position vector in \(\mathbb{R}^3\).

  1. Find the point-slope equation of the plane with normal vector \(\mathbf{n} = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}\) passing through the point \((1, 0, -2)\).

  2. Find the point-slope equation of the plane with normal vector \(\mathbf{n} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\) passing through the origin.

  1. \(z = -\frac{2}{3}(x-1) + \frac{1}{3}(y-0) - 2\)

  2. \(z = -x - y\)

Exercise 4: Finding planes from geometric information

  1. Find a point-slope equation of the plane passing through the points \((1, 0, 2)\), \((2, 1, 3)\), and \((0, 1, 1)\).

  2. Find a point-slope equation of the plane that contains the point \((0, 1, 1)\) and the \(x\)-axis.

  3. Find a plane that passes through the point \((2, -1, 3)\) and is perpendicular to the vector \[ \mathbf{v} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}. \]

  4. Find the point-slope equation of the plane that is parallel to the \(xz\)-plane and passes through the point \((1, 3, 2)\), if possible. If no such equation exists, explain why, and which equation you would use instead.

  1. Let \[ \mathbf{r}_1 = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}, \quad \mathbf{r}_2 = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}, \quad \mathbf{r}_3 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \] be position vectors for the three points. Two vectors in the plane are \[ \mathbf{v}_1 = \mathbf{r}_2 - \mathbf{r}_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad \mathbf{v}_2 = \mathbf{r}_3 - \mathbf{r}_1 = \begin{bmatrix} -1 \\ 1 \\ -1 \end{bmatrix}. \] We need a vector \[ \mathbf{n} = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \] orthogonal to both: \[ \langle \mathbf{n}, \mathbf{v}_1 \rangle = a + b + c = 0, \quad \langle \mathbf{n}, \mathbf{v}_2 \rangle = -a + b - c = 0. \] From the second equation, \(b = a + c\). Substituting into the first: \(a + (a+c) + c = 0\), so \(2a + 2c = 0\), giving \(c = -a\). Then \(b = a + (-a) = 0\). Choosing \(a = 1\), we get the normal vector \[ \mathbf{n} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}. \] Then, a point-slope equation of the plane is \[ z = (x-1) + 2. \]

  2. A normal vector to the plane is \[ \mathbf{n} = \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}. \] Using this, we get a point-slope equation \[ z = (y-1) + 1 \]

  3. If the plane is perpendicular to \(\mathbf{v}\), then \(\mathbf{v}\) is a normal vector. So \[ \mathbf{n} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}. \] Then a point-slope equation of the plane is \[ z = (x-2) + 2(y+1) + 3 \]

  4. A plane parallel to the \(xz\)-plane has normal vector \[ \mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}. \] But this can’t be written in the form \[ \mathbf{n} = \begin{bmatrix} -m \\ -n \\ 1 \end{bmatrix}, \] so no point-slope equation exists for this plane. Instead, we use the standard form: \[ y-3 = 0. \]

Exercise 5: Temperature distribution

The temperature \(T\) (in degrees Celsius) at a point \((x, y, z)\) in a room is modeled by a linear function. Measurements show that:

  • At the point \((0, 0, 0)\), the temperature is \(20°C\).
  • As you move 1 meter in the positive \(x\)-direction, the temperature increases by \(2°C\) per meter.
  • As you move 1 meter in the positive \(y\)-direction, the temperature decreases by \(1°C\) per meter.
  • The temperature does not change as you move in the \(z\)-direction.

Because of the third bullet point, the temperature is actually a function of the form \(T = f(x,y)\), independent of \(z\).

  1. Find a formula for \(T = f(x, y)\).

  2. What is the temperature at the point \((3, 2, 5)\)?

  3. Find all points where the temperature is exactly \(25°C\).

  4. Does the set of solutions in part (c) form a plane? If so, find a normal vector.

  1. Since temperature changes linearly with position and increases at \(2°C\)/meter in the \(+x\) direction and decreases at \(1°C\)/meter in the \(+y\) direction, and doesn’t change with \(z\), we have \[ T = f(x,y)= 2x - y + 20. \] (Since \(f(0,0,0) = 20\).)

  2. \(T = f(3, 2) = 24°C\).

  3. Setting \(T = 25\) and solving \(f(x,y)=25\), we get: \[ \begin{align*} 2x - y + 20 = 25\\ 2x - y = 5. \end{align*} \]

  4. Yes, the set of solutions forms a plane. Note that the equation can be rewritten in the standard form \[ 2(x-0) - (y+5)=0. \] Thus, a normal vector is \[ \mathbf{n} = \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}. \]

Exercise 6: Standard equations of planes

Recall that a standard equation of a plane in \(\mathbb{R}^3\) passing through the point \((x_0,y_0,z_0)\) with normal vector \[ \mathbf{n} = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \] is given by \[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0. \]

  1. Find a standard equation of the plane with normal vector \[ \mathbf{n} = \begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix} \] passing through the point \((1, 2, -1)\).

  2. A standard equation of a plane is \[ 3(x-1) - 2(y+2) + 4z = 0. \] Find a normal vector to this plane and identify a point on the plane.

  3. Expand the standard equation from part (a) to obtain an equation of the form \(ax + by + cz = d\).

  4. Consider the plane with point-slope equation \(z = 2(x-1) - 3(y+1) + 4\). Write this in standard form.

  1. Using the formula with \(a = 2\), \(b = -3\), \(c = 1\) and the point \((1, 2, -1)\): \[ 2(x-1) - 3(y-2) + (z+1) = 0. \]

  2. The normal vector is \[ \mathbf{n} = \begin{bmatrix} 3 \\ -2 \\ 4 \end{bmatrix}. \] One point on the plane is \((1, -2, 0)\).

  3. \(2x - 3y + z = -5\)

  4. \(2(x-1) - 3(y+1) - (z-4) = 0\)

Exercise 7: Affine equations of planes

Recall that an affine equation for a plane is given by

\[ ax + by + cz = d, \]

where \(a\), \(b\), \(c\), and \(d\) are constants (and not all of \(a\), \(b\) and \(c\) are \(0\)).

  1. Rewrite the point-slope equation \[ z = 2(x-1) - 3(y+1) + 4 \] as an affine equation.
  2. Rewrite the affine equation \[ 3x - 2y +4z = 5 \] as both a point-slope equation. What is the slope in the \(+x\)-direction? What is the slope in the \(+y\)-direction?
  3. Rewrite the affine equation \[ 2x + 4y = 8 \] as a point-slope equation, if possible. If it isn’t possible, explain why.
  4. Find a normal vector to the plane with affine equation \[ 2x - y + 3z = 7. \]

  1. \(2x - 3y - z = 1\)

  2. Equation: \(z = - \frac{3}{4}x + \frac{1}{2}y + \frac{5}{4}\). Slope in \(+x\)-direction: \(-\frac{3}{4}\). Slope in \(+y\)-direction: \(\frac{1}{2}\).

  3. \(2x + 4y = 8\) cannot be rewritten as a point-slope equation because it does not define \(z\) as a function of \(x\) and \(y\).

  4. A normal vector to the plane is \[ \mathbf{n} = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}. \]

Exercise 8: More practice with planes

  1. Suppose you are standing at the point \((1, 2, 3)\) on the plane with equation \[ z = 2(x-1) + 3(y-2) + 3. \]

    1. If you step \(2\) units in the \(+x\)-direction, what point are you at?
    2. If you step 1 unit in the \(-x\)-direction, what point are you at?
    3. If you step 3 units in the \(-x\)-direction and \(4\) in the \(-y\)-direction, what point are you at?

  2. Suppose you are standing at \((1, -2, 0)\) on the plane with equation \[ 2(x-1) -4 (y+2) +2z = 0. \]

    1. If you step \(1\) unit in the \(-x\)-direction, what point are you at?
    2. If you step \(2\) units in the \(+y\) direction, what point are you at?

  3. Consider the plane with equation \[ 2x-3y + 3z = 5. \] This plane has a unique normal vector that points in the positive \(+z\)-direction (i.e., the \(z\)-component of the normal vector is positive). If you begin at the point \((1,0,1)\) and step \(2\) units along this vector, what point are you at?

  4. Consider the plane with equation \[ 2x + 3y + 4z = 1. \] What is the point on this plane closest to the point \((2,3,-1)\)?

  1. \((3, 2, 7)\), \((0, 2, 1)\), \((-2, -2, -15)\)

  2. \((0, -2, 1)\), \((1, 0, 4)\)

  3. A normal vector with positive \(z\)-component is \[ \mathbf{n} = \begin{bmatrix} 2 \\ -3 \\ 3 \end{bmatrix} \] Now, we want to step \(2\) units along this vector, and for this, it will be helpful to first normalize \(\mathbf{n}\) to get a unit normal vector: \[ \hat{\mathbf{n}} = \frac{\mathbf{n}}{\|\mathbf{n}\|} = \frac{1}{\sqrt{22}} \begin{bmatrix} 2 \\ -3 \\ 3 \end{bmatrix}. \] So, if we begin at the point \((1,0,1)\) and step \(2\) units along \(\hat{\mathbf{n}}\), the position vector of our new point will be \[ \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + \frac{2}{\sqrt{22}} \begin{bmatrix} 2 \\ -3 \\ 3 \end{bmatrix} = \begin{bmatrix} 1 + \frac{4}{\sqrt{22}} \\ 0 - \frac{6}{\sqrt{22}} \\ 1 + \frac{6}{\sqrt{22}} \end{bmatrix} \approx \begin{bmatrix} 1.85 \\ -1.28 \\ 2.28 \end{bmatrix}. \] Thus, the new point is approximately \((1.85, -1.28, 2.28)\).

  4. Let \(\mathbf{r}\) be the position vector of the point \((2,3,-1)\), and let \(\mathbf{r}_0\) be the position vector of the point \((x_0,y_0,z_0)\) on the plane closest to \((2,3,-1)\). Then it follows that the difference \(\mathbf{r}-\mathbf{r_0}\) is orthogonal to the plane. But a normal vector to the plane is \[ \mathbf{n} = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}, \] so the difference \(\mathbf{r}-\mathbf{r_0}\) must be parallel to \(\mathbf{n}\), which means there is a scalar \(t\) such that \(\mathbf{r}-\mathbf{r_0} = t \mathbf{n}\). Then we have \(\mathbf{r}_0 = \mathbf{r} - t \mathbf{n}\): \[ \begin{bmatrix} x_0 \\ y_0 \\ z_0 \end{bmatrix} = \begin{bmatrix} 2 - 2t \\ 3 - 3t \\ -1 - 4t \end{bmatrix}. \] But since \((x_0,y_0,z_0)\) lies on the plane, it must be a solution to the equation of the plane: \[ 2(2 - 2t) + 3(3 - 3t) + 4(-1 - 4t) = 1. \] Solving for \(t\), we get \(t = 8/29\). Therefore, the point on the plane closest to \((2,3,-1)\) has position vector \[ \mathbf{r_0} = \mathbf{r} - \frac{8}{29}\mathbf{n} = \begin{bmatrix} \frac{42}{29} \\ \frac{63}{29} \\ -\frac{61}{29} \end{bmatrix} \approx \begin{bmatrix} 1.45 \\ 2.17 \\ -2.10 \end{bmatrix}. \] Thus, the point on the plane closest to \((2,3,-1)\) is approximately \((1.45, 2.17, -2.10)\).