12.1: Three-Dimensional Coordinate Systems #

Problem 3: Which of the points $A(-4, 0, -1)$, $B(3, 1, -5)$, and $C(2, 4, 6)$ is closest to the $yz$-plane? Which point lies in the $xz$-plane?

To find the distance between a point and a plane, we find the distance between that point and it’s projection to the plane in question. For example, $A$’s projection to the $yz$-plane is $(0, 0, -1)$ as the projection removes the $x$ term. Note now that the distance between these two is very simple, it is

$$ \sqrt{(-4 - 0)^2 + (0 - 0)^2 + (-1 + 1)^2} = \sqrt{16 + 0 + 0} = 4 $$

Doing similar calculations tells us that $C$ is closest to the $yz$-plane and $A$ is closest to the $xz$-plane.

Problem 9: Find the lengths of the sides of the triangle $\emph{PQR}$ where

$$ P(3, -2, -3), \ Q(7, 0, 1), \ R(1, 2, 1) $$

Is it a right triangle? Is it an isosceles triangle?

Again, we do distance calculations to find the lengths of the sides, for example the length of $\emph{PQ}$ is

$$ \sqrt{(3 - 7)^2 + (-2 - 0)^2 + (-3 - 1)^2} = \sqrt{16 + 4 + 16} = 6 $$

The lengths of $\emph{PR}$ and $\emph{QR}$ is $6$ and $\sqrt{40}$ respectively. This is an isosceles triangle, but is it not a right triangle as $6^2 + \sqrt{40}^2 > 6^2$.

Problem 11: Determine whether the points lie on a straight line

$(a) \ A(2, 4, 2), \ B(3, 7, -2), \ C(1, 3, 3)$

$(b) \ A(0, -5, 5), \ B(1, -2, 4), \ C(3, 4, 2)$

The set of points for the line from two points is

$$ \overline{AB} = \{ A + c(B - A) : c \in \R \} $$

This is in set builder notation, which without going into full detail, is the collection of all points in 3-dimensional space that are of the form $A + c(A - B)$. So $C$ is on the line $\overline{AB}$ if there exists a $c$ such that $C = A + c(A - B)$. For $(a)$ there does not exist such a $c$, note that there could only exist one $c$, $c = -1$, since the $x$ coordinate of

$$ A + (-1)(A - B) = (2, 4, 2) + (-1)((3, 7, -2) - (2, 4, 2)) $$

$$ = (2, 4, 2) + (-1, -3, 4) = (1, 1, 6) $$

matches $C$’s $x$ coordinate, $1$. But as we can tell the other coordinates are no good. It is now your turn to find a $c$ for $(b)$!

Problem 15: Find an equation of the sphere that passes through the point $(4, 3, -1)$ and has center $(3, 8, 1)$.

We need to find the radius of the sphere we want, which is computed with the distance function,

$$ \sqrt{(4 - 3)^2 + (3 - 8)^2 + (-1 - 1)^2} = \sqrt{1 + 25 + 4} = \sqrt{30} $$

Now it is formulaic, and our sphere we want is

$$ (x - 4)^2 + (y - 3)^2 + (z + 1)^2 = \sqrt{30}^2 = 30 $$

Note that if you plugged in the point we wanted the sphere to contain, we would use the same calculations as above!

Problem 19: Show that the equation below represents a sphere; find its center and radius.

$$ 2x^2 + 2y^2 + 2z^2 = 8x - 24z + 1 $$

This almost looks like the canonical equation of a sphere we want, so lets pull the “junk”, $8x - 24z$, to the other side, getting,

$$ 2x^2 - 8x + 2y^2 + 2z^2 + 24z = 1 $$

We want to factor out the $2$, then complete the square for both the $x$ and $z$ terms,

$$ 1 = 2x^2 - 8x + 2y^2 + 2z^2 + 24z = 2(x^2 - 4x + y^2 + z^2 + 12z) $$

$$ = 2(x^2 - 4x + 4 - 4 + y^2 + z^2 + 12z + 36 - 36) $$

$$ = 2((x - 2)^2 + y^2 + (z + 6)^2) - 8 - 72 $$

We’re almost there, moving over the $-80$ and dividing by two we have,

$$ (x - 2)^2 + y^2 + (z + 6)^2 = 81/2 $$

Problem 35: Describe in words the region of $\R^3$ represented by the equation or inequality, (try drawing it!)

$$ 1 \leq x^2 + y^2 + z^2 \leq 5 $$

It is your job to draw this sucker, but in words, its everything that is in-between the sphere of radius $1$ centered at the origin and the sphere of radius $\sqrt{5}$ at the origin. It looks like the skin of an inflated basketball, if the skin were way thicker.

12.2: Vectors #

Problem 25: Find a unit vector that has the same direction as the vector $v$,

$(a) \ v = (6, -2)$

$(b) \ v = -5i + 3j - k$

$(c) \ v = 8i - j + 4k$

Making a vector a unit vector is formulaic, but to explain the geometry of what we are doing, we are scaling by the reciprocal of the length of the vector. Since the length of a vector is a function that commutes with scalars (this means that if $f$ is the length function, $f(c v) = c f(v)$, additionally show that this property actually holds for all $c$ and $v$), we will always end up with a vector that is pointing in the same direction but is unit in length. For $(a)$ the length is $\sqrt{36 + 4} = \sqrt{40}$ so our desired unit vector is $1/\sqrt{40} \cdot (6, -2)$

Problem 29: If $v$ lies in the first quadrant and makes an angle $\pi/3$ with the positive $x$-axis and $|v| = 4$, find $v$ in component form.

If it is in the first quadrant and makes an angle $\pi/3$ with the positive $x$-axis, then it is $(1, 0)$ rotated by $\pi/3$ counter-clockwise and is scaled by $4$. Rotating $(1, 0)$ by $\pi/3$ is simple you can find it on those unit circle charts, otherwise we could use a rotational transformation, which is just a matrix. Using the former (since this is very nice angle) we see that our desired vector is $(1/2, \sqrt{3}/2)$, and by scaling by $4$ we have, $(2, 2\sqrt{3})$.

Problem 39: A boatman wants to cross a canal that is $3$ km wide and wants to land at a point $2$ km upstream from his starting point. The current in the canal flows at $3.5$ km/h and the speed of his boat is $13$ km/h.

$(a) $ In what direction should he steer?

$(b) $ How long will the trip take?

We want our vector, $(x, y)$ to point in the same direction as $(3, 2)$, after the force of the river affects the boat. We have 2 requirements,

$$ \sqrt{x^2 + y^2} = 13 $$

and that $x = 3c$ and $y - 3.5 = 2c$, thus we get the equation,

$$ 13 = \sqrt{(3c)^2 + (2c + 3.5)^2} \sqrt{13c^2 + 14c + 12.25} $$

Solving for $c$ we get $2$ values, $c = -4.05238$, $2.97546$, obviously the negative value is erroneous, so our final vector is

$$ (3 \cdot 2.97546, 2 \cdot 2.97546 + 3.5) = (8.92638, 9.45092) $$

Then also means that the boat ride takes $3/8.92638$ hours.

12.3: The Dot Product #

Problem 4: Find $a \cdot b$, where $a = (6, -2, 3)$, $b = (2, 4, -1)$.

Finding the dot product when you are given the components of 2 vectors is formulaic,

$$ (a_1, a_2, a_3) \cdot (b_1, b_2, b_3) = a_1b_1 + a_2b_2 + a_3b_3 $$

So in our case we have,

$$ (6, -2, 3) \cdot (2, 4, -1) = 12 + (-8) + (-3) = 1 $$

Problem 10: Find $a \cdot b$, where $|a| = 80$, $|b| = 50$, and where the angle between $a$ and $b$ is $3\pi/4$.

We use a handy formula that relates the dot product, lengths, and the angle between them

$$ c \cdot d = |c||d| \cos(\theta) $$

Usually the intension of the formula is to find the angle, but it works perfectly for our needs! In this case, $\cos(3/4 \pi) = -\sqrt{2}/2$ thus,

$$ |a||b| \cos(3/4 \pi) = 400 \frac{-\sqrt{2}}{2} = -200\sqrt{2} = a \cdot b $$

Problem 27: Find both unit vectors orthogonal to both $i + j$ and $i + k$.

The reason why we are asking for both is because negating the vector also gives a unit vector that is orthogonal to our vectors. If you looked ahead you might just say, “uhh, Nels… just use the cross product dummy” and you would be perfectly correct! Keeping that in mind, lets do this problem without the cross product, noting how it informs us why the cross product works the way it does!

So lets looks at all vectors $(a, b, c)$ that satisfy,

$$ 0 = (a, b, c) \cdot (i + j) = (a, b, c) \cdot (1, 1, 0) = a + b $$

and,

$$ 0 = (a, b, c) \cdot (i + k) = (a, b, c) \cdot (1, 0, 1) = a + c $$

This is a set of linear equations that we want to solve, we get that $b = -a$ and $c = -a$ so all of the vectors that are mutually orthogonal to our asked vectors fit the form $(a, -a, -a)$. The length of these vectors is $\sqrt{a^2 + a^2 + a^2} = a \sqrt{3}$, so our unit vector is

$$ \frac{1}{a \sqrt{3}}(a, -a, -a) = \left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right) $$

As said before, negating this gives us another desired unit vector,

$$ \left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) $$

Note that we started with 3 free variables, and then with 2 linear equations, our parameters reduced to one, then the unital condition removed our solution space from the whole real line, to 2 variables!

Problem 53: Use a scalar projection to show that the distance from a point $(x_1, y_1)$ to the line $ax + by + c = 0$ is

$$ \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} $$

Use this formula to find the distance from the point $(-2, 3)$ to the line $3x - 4y + 5 = 0$.

IOU on this one.

12.4: The Cross Product #

Problem 9: Find the vectors, not with determinants, but by using properties of cross product.

$(a) \ (i \times j) \times k$

$(b) \ k \times (i - 2j)$

For the first one, note that the cross product of $i$ and $j$ is $k$, as they all lie on the axises. Then we have that $k \times k = 0$.

For the second problem, since the cross product distributes, and is compatible with scalars, we have,

$$ k \times (i - 2j) = (k \times i) - 2 (k \times j) = j + 2i $$

Problem 19: Find two unit vectors orthogonal $(3, 2, 1)$ and $(-1, 1, 0)$.

We compute the cross product,

$$ (3, 2, 1) \times (-1, 1, 0) = \det \! \begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix} i - \det \! \begin{pmatrix} 3 & 1 \\ -1 & 0 \end{pmatrix} j + \det \! \begin{pmatrix} 3 & 2 \\ -1 & 1 \end{pmatrix} k = -i -j + 5k $$

Now the length of this vector is $\sqrt{27}$ so multiplying by $\pm 1/\sqrt{27}$ will give us both unit vectors.

Problem 27: Find the area of the following parallelogram,

$A = (-3, 0)$, $B = (-1, 3)$, $C = (5, 2)$, $D = (3, -1)$.

Finding the magnitude of cross product of 2 vectors from the same point gives us what is desired, but to do that we must embed our vectors into $\R^3$ by adding the $z$-axis. So $A_z = (-3, 0, 0)$, $B_z = (-1, 3, 0)$ and so on. So our volume follows the computation,

$$ |(A_z - B_z) \times (C_z - B_z)| = |(-2, -3, 0) \times (6, -1, 0)| = |(0, 0, 20)| = 20 $$

After some communication with a student it might not be clear what the geometry of the scenario is, so here is a desmos graph of the scenario, move $t$ around to see how the parallelogram shifts by $B$.

Problem 29: Find a nonzero vector orthogonal to the plane through the points $P$, $Q$, and $R$, find the area of triangle $\emph{PQR}$.

$P = (1, 0, 1)$, $Q = (-2, 1, 3)$, $R = (4, 2, 5)$.

Do the same process as in 27, the vector produced in the cross product carries the data of the area of the parallelogram, and a vector orthonormal to it, so use that information to solve this problem!

12.5: Equations of Lines and Planes #

Problem 2: Find a vector equation and parametric equations for the line through the point $P = (6, -5, 2)$ and parallel to the vector $r = (1, 3, -2/3)$.

This is a plug-in-play problem, our vector equation is $P + t r$ and thus our parametric equation is $(6 + t, -5 + 3t, 2 + -2/3t)$.

Problem 13: Is the line through the points $(-4, -6, 1)$ and $(-2, 0, -3)$ parallel to the line through $(10, 18, 4)$ and $(5, 3, 14)$?

Lets find the vectors that generate the line,

$$ (-4, -6, 1) - (-2, 0, -3) = (-2, -6, 4) $$

and,

$$ (10, 18, 4) - (5, 3, 14) = (5, 15, -10) $$

Since $-5/2 (-2, -6, 4) = (5, 15, -10)$, they are parallel!

Problem 23: Find an equation for the plane through the origin and perpendicular to the vector $(1, -2, 5)$

This is all of the points $(x, y, z)$ that are perpendicular to the vector above, which is when these vectors have dot product $0$,

$$ 0 = (1, -2, 5) \cdot (x, y, z) = x - 2y + 5z $$