Multivariable calculus: Exercises

04 Planes and linear spaces, part 2

Exercise 1: Point-slope equations in higher dimensions

Recall that the point-slope form of the equation of a hyperplane in \(\mathbb{R}^4\) is given by:

\[ w = m(x - x_0) + n(y - y_0) + p(z - z_0) + w_0 \]

where \((x_0, y_0, z_0, w_0)\) is a point on the plane, and \(m\), \(n\), \(p\), and \(q\) are the slopes in the positive \(x\)-, \(y\)-, \(z\)-, and \(w\)-directions, respectively.

Similarly, the point-slope form of the equation of a hyperplane in \(\mathbb{R}^5\) is given by:

\[ u = m(x - x_0) + n(y - y_0) + p(z - z_0) + q(w - w_0) + u_0 \]

where \((x_0, y_0, z_0, w_0, u_0)\) is a point on the hyperplane, and \(m\), \(n\), \(p\), \(q\), and \(r\) are the slopes in the positive \(x\)-, \(y\)-, \(z\)-, \(w\)-, and \(u\)-directions, respectively.

  1. Write the point-slope equation for a hyperplane in \(\mathbb{R}^4\) that passes through the point \((-1, 0, 4, 3)\) and has slopes \(2\) in the positive \(x\)-direction, \(-1\) in the positive \(y\)-direction, and \(3\) in the positive \(z\)-direction.

  2. Write the point-slope equation for a hyperplane in \(\mathbb{R}^5\) that passes through the point \((0, 1, -1, 2, 5)\) and has slopes \(1\) in the positive \(x\)-direction, \(2\) in the positive \(y\)-direction, \(-3\) in the positive \(z\)-direction, and \(4\) in the positive \(w\)-direction.

  3. Does the point \((2, 3, 4, 6)\) lie on the hyperplane from part (a)?

  4. Does the point \((1, 2, 0, 3, 8)\) lie on the hyperplane from part (b)?

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

  2. \(u = x + 2(y - 1) - 3(z + 1) + 4(w - 2) + 5\)

  3. Yes.

  4. No.

Exercise 2: Converting between equation forms in \(\mathbb{R}^4\)

Recall that the standard form of the equation of a hyperplane in \(\mathbb{R}^4\) is given by:

\[ a(x - x_0) + b(y - y_0) + c(z - z_0) + d(w - w_0) = 0 \]

where \((x_0, y_0, z_0, w_0)\) is a point on the plane, and \(a\), \(b\), \(c\), and \(d\) are the components of a normal vector to the hyperplane.

The affine form of the equation of a hyperplane in \(\mathbb{R}^4\) is given by:

\[ ax + by + cz + dw = e \]

where \(a\), \(b\), \(c\), and \(d\) are the components of a normal vector to the hyperplane, and \(e\) is a constant.

Consider the hyperplane in \(\mathbb{R}^4\) with point-slope equation: \[ w = 3(x-2) - 2(y+1) + 4(z-1) + 5 \]

  1. Identify a point on the hyperplane.

  2. Identify the slopes in the positive \(x\)-, \(y\)-, and \(z\)-directions.

  3. Write the equation in standard form.

  4. Identify a normal vector to the hyperplane.

  5. Write the affine equation for the hyperplane.

  1. \((2, -1, 1, 5)\)

  2. \(3\), \(-2\), \(4\)

  3. \(3(x-2) - 2(y+1) + 4(z-1) - (w - 5) = 0\)

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

  5. $ 3x - 2y + 4z - w = 7$

Exercise 3: Leveraging intuition from planes in \(\mathbb{R}^3\)

This next exercise is tricky. In order to solve it, you’ll need to leverage your intuition about planes in \(\mathbb{R}^3\) and apply it to hyperplanes in higher dimensions. Don’t give up and look at the solution too quickly.

  1. Consider the two hyperplanes in \(\mathbb{R}^5\) given by the equations: \[ 2x + 3y - z + w - 4u = 10 \] and \[ -4x - 6y + 2z - 2w + 8u = 5 \]

    1. Are these hyperplanes parallel? Explain.
    2. What is the distance between these two hyperplanes?

  2. Give the equation of the hyperplane in \(\mathbb{R}^5\) that passes through the point \((-1, 2, 0, 8, 5)\) and is parallel to the \(xzwu\)-coordinate hyperplane.

  3. You are standing at the point \((1, 0, -2, 3, 4)\) on the hyperplane with standard equation \[ 2(x-1) + 4y - 3(z+2) + 5(w-3) = 0. \] Identify a normal vector to the plane with positive \(x\)-component. Then, suppose you step 2 units along this normal vector. What are your new coordinates?

    1. Yes, these hyperplanes are parallel. Their normal vectors are \[ \begin{bmatrix} 2 \\ 3 \\ -1 \\ 1 \\ -4 \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} -4 \\ -6 \\ 2 \\ -2 \\ 8 \end{bmatrix}, \] which are scalar multiples of each other (the second is \(-2\) times the first) and hence parallel. But if their normal vectors are parallel, then the hyperplanes themselves must be parallel.

    2. A point on the first hyperplane is \((0, 0, -10, 0, 0)\) with position vector \[ \mathbf{r} = \begin{bmatrix} 0 \\ 0 \\ -10 \\ 0 \\ 0 \end{bmatrix}, \] while a normal vector to the same hyperplane is \[ \mathbf{n} = \begin{bmatrix} 2 \\ 3 \\ -1 \\ 1 \\ -4 \end{bmatrix}. \] If we can find a value of \(t\) for which the point with position vector \(\mathbf{r} + t\mathbf{n}\) lies on the second hyperplane, then the distance between the hyperplanes is given by \(\|t\mathbf{n}\|\). However, we have \[ \mathbf{r} + t\mathbf{n} = \begin{bmatrix} 2t \\ 3t \\ -10 - t \\ t \\ -4t \end{bmatrix}, \] and substituting the components of this vector into the second hyperplane’s equation gives \[ -4(2t) - 6(3t) + 2(-10 - t) - 2(t) + 8(-4t) = 5. \] Solving for \(t\) gives \(t = -25/62\), and thus the distance between the hyperplanes is \[ \|t\mathbf{n}\| = \left\| -\frac{25}{62} \begin{bmatrix} 2 \\ 3 \\ -1 \\ 1 \\ -4 \end{bmatrix} \right\| \approx 2.25. \]

  2. The \(xzwu\)-coordinate hyperplane is the hyperplane in \(\mathbb{R}^5\) given by the equation \(y = 0\). A hyperplane parallel to this hyperplane must also have the form \(y = c\) for some constant \(c\). Since the hyperplane must pass through the point \((-1, 2, 0, 8, 5)\), we must have \(y=2\).

  3. A normal vector to the plane with positive \(x\)-component is \[ \mathbf{n} = \begin{bmatrix} 2 \\ 4 \\ -3 \\ 5 \\ 0 \end{bmatrix}. \] In order to step 2 units along this normal vector, we first need to normalize it (i.e., make it a unit vector) and then scale it by 2. But the magnitude of \(\mathbf{n}\) is \[ \|\mathbf{n}\| = \sqrt{2^2 + 4^2 + (-3)^2 + 5^2 + 0^2} = 3 \sqrt{6}. \] So, the point that we end up at has position vector \[ \begin{bmatrix} 1 \\ 0 \\ -2 \\ 3 \\ 4 \end{bmatrix} + \frac{2}{3 \sqrt{6}} \begin{bmatrix} 2 \\ 4 \\ -3 \\ 5 \\ 0 \end{bmatrix} \approx \begin{bmatrix} 0.54 \\ 1.09 \\ -0.82 \\ 1.36 \\ 0 \end{bmatrix}. \]

Exercise 4: Level sets of linear functions

Recall from class that the level set of a function \(f: \mathbb{R}^2 \to \mathbb{R}\) is the set of points \((x, y)\) such that \(f(x, y) = c\) for some constant \(c\). Level sets of functions of two variables are curves in the \(xy\)-plane.

Similarly, the level set of a function \(g: \mathbb{R}^3 \to \mathbb{R}\) is the set of points \((x, y, z)\) such that \(g(x, y, z) = c\) for some constant \(c\). Level sets of functions of three variables are surfaces in \(\mathbb{R}^3\).

Consider the linear functions \(f: \mathbb{R}^2 \to \mathbb{R}\) and \(g: \mathbb{R}^3 \to \mathbb{R}\) given by: \[ z = f(x, y) = 3x - 2y + 6 \]

and

\[ w = g(x, y, z) = x + 2y - z + 4 \]

  1. Sketch the level sets of \(f\) for \(c = 0, 6, 12\) in the \(xy\)-plane.

  2. Sketch the level sets of \(g\) for \(c = 0, 4, 8\) in \(\mathbb{R}^3\).

  1. The level sets of \(f\) are given by the equations:
    • For \(c = 0\): \(3x - 2y + 6 = 0\) or \(y = \frac{3}{2}x + 3\)
    • For \(c = 6\): \(3x - 2y + 6 = 6\) or \(y = \frac{3}{2}x\)
    • For \(c = 12\): \(3x - 2y + 6 = 12\) or \(y = \frac{3}{2}x - 3\)

These are parallel lines in the \(xy\)-plane with slope \(\frac{3}{2}\).

  1. The level sets of \(g\) are given by the equations:
    • For \(c = 0\): \(x + 2y - z + 4 = 0\) or \(z = x + 2y + 4\)
    • For \(c = 4\): \(x + 2y - z + 4 = 4\) or \(z = x + 2y\)
    • For \(c = 8\): \(x + 2y - z + 4 = 8\) or \(z = x + 2y - 4\)

These are parallel planes in \(\mathbb{R}^3\) with normal vector \[ \mathbf{n} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}. \]

Exercise 5: Applied problem with hyperplanes

Though we have emphasize the geometric nature of hyperplanes in this section, they also have many applications in real-world problems. In this exercise, you’ll apply your understanding of hyperplanes to solve a problem in business.

A company’s profit, \(P\) (in thousands of dollars), depends on four factors:

  • \(x\) = advertising budget (in thousands of dollars)
  • \(y\) = number of sales representatives
  • \(z\) = product quality rating (on a scale of 0-10)
  • \(w\) = competitor’s price (in dollars)

The profit function is given by: \[ P = f(x, y, z, w) = 2x + 5y + 10z + 0.5w - 100 \]

  1. What is the base profit when all variables are zero?

  2. How much does profit increase for each additional thousand dollars spent on advertising?

  3. How much does profit increase for each additional sales representative?

  4. If the company spends \(\$50{,}000\) on advertising, employs \(15\) sales representatives, has a product quality rating of \(8\), and the competitor’s price is \(\$120\), what is the profit?

  5. The level set \(P = 0\) represents the break-even condition (zero profit). Write the equation of this level set. What geometric object does it represent in \(\mathbb{R}^4\)?

  1. \(-\$100{,}000\)

  2. \(\$2{,}000\) per additional thousand dollars spent on advertising.

  3. \(\$5{,}000\) per additional sales representative.

  4. \(\$215{,}000\)

  5. The equation is \[ 2x + 5y + 10z + 0.5w = 100, \] which is the affine form of a hyperplane in \(\mathbb{R}^4\).