# Solutions for HW 3

Problem 1.  (a) The domain of this function is $\mathbb{R}^2$ — there are no restrictions on the inputs since the function does not try to divide by zero, take an even root of a negative number or take a logarithm of a non-positive number.

(b)  The level set is a line in $\mathbb{R}^2$.  The standard equation for this line is $y = -2/3(x)$.

Problem 2.  (a) All of the vectors in this set lie on the line $y = -(2/3)x$ since any vector $\vec{w} = \langle x, y \rangle$ that makes this dot product zero must satisfy

$\langle 2, 3 \rangle \,\cdot\,\langle x, y \rangle = 0$

$2x + 3y = 0$

(b) This is the same picture that was drawn in question 1.

Problem 3.  (skipped for now)

Problem 4.  Given two vectors $\vec{w} = \langle w_1, w_2, w_3 \rangle$ and $\vec{v} = \langle v_1, v_2, v_3 \rangle$ in $\mathbb{R}^3$ where $\vec{w} = c\vec{v}$, it follows that

$w_1 = cv_1, w_2 = cv_2 \text{ and } w_3 = cv_3$.

When we compute the cross product we find

$\displaystyle \vec{v} \times \vec{w} = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{array} \right| =\left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ v_1 & v_2 & v_3 \\ cv_1 & cv_2 & cv_3 \end{array} \right|$

$= \vec{i}\left(v_2\,c\,v_3 - v_3\,c\,v_2\right) - \vec{j}\left(v_1\,c\,v_3 - v_3\,c\,v_1\right) + \vec{k}\left(v_1\,c\,v_2 - v_2\,c\,v_1\right) = \vec{0}$.

Problem 5.  Exercise 2. from 9.4 uses the vectors $\vec{x} = \langle 1, 1, 1 \rangle$ and $\vec{y} = \langle 0, 3, -2\rangle.$

(a) One computes that $\vec{x}\cdot\vec{y} = 3 - 2 = 1 \neq 0$ and so the vectors are not orthogonal.  The vectors are also not parallel since it is impossible to find a scalar $c \in \mathbb{R}$ so that $\vec{x} = c\vec{y}$ (equivalently, their cross product is not the zero vector).

(b) The cross product of these two vectors will be orthogonal to both of them.  That is, the vector

$\vec{x} \times \vec{y} = \langle -5, 2, 3 \rangle$

is perpendicular to both $\vec{x}$ and $\vec{y}$.  However, this is not (yet) a unit vector; to make this have unit length we form the new vector

$\displaystyle \vec{u} = \frac{\vec{x}\times\vec{y}}{\|\vec{x}\times\vec{y}\|} = \frac{1}{\sqrt{38}}\langle -5, 2, 3 \rangle$.

(c ) This part turned out to be more difficult than I intended, so don’t worry if you weren’t able to do it.  The main idea is that we want to find two scalars $c_1, c_2 \in \mathbb{R}$ and a vector $\vec{v}$ where

$\vec{y} = c_1\vec{x} + c_2\vec{v}$

where $\vec{v}\cdot\vec{x} = 0$.  There are lots of vectors to choose for $\vec{v}$ (indeed, an entire plane’s worth), but if you pick one arbitrarily and try to solve the above equation for the unknown constants $c_1$ and $c_2$, you can easily end up with three equations that are impossible to solve.

(d) The area of the parallelogram is given by

$\|\vec{x}\times\vec{y}\| = \sqrt{38}$.

Problem 6.  For problem 3a we have that $\vec{PQ} = \langle 2, 4, -5 \rangle$ and $\vec{PR} = \langle -1, -2, -1 \rangle$.

For problem 3b, we can compute the area of the triangle by computing

$\displaystyle \text{area } = \frac{1}{2}\| \vec{PQ} \times \vec{PR} \|$.

Problem 7.  Exercise 2 has several parts.  I’ll list some of my work and many of my answers for most of these.

(a) $\vec{v}_1 = \langle -2, 1, 3 \rangle$.

(b) The second line has parametric equations $x(t) = -4 - 2t, y(t) = 2 + t, z(t) = 17 + 5t$.

(c ) We can find where these two lines (possibly) intersect by setting the parametric equations for $x, y$ and $z$ equal to one another and solving.  This leads to three equations:

$x(t) = -4 - 2t = 4-2s = x(s)$

$y(t) = 2 + t = -2+s = y(s)$

$z(t) = 17+5t = 1+ 3s = z(s)$

We can use the first equation to find that $t = -4 + s$.  Plugging this into the second equation gives us $2 + (-4 + s) = -2 + s$ which is the same as $-2 + s = -2 + s$.  Since this equation is true for all possible values of $s$, we learn nothing from this substitution.  Finally, we plug what we do know (that $t = -4 + s$) into the third equation to find that $17 + 5(-4 + s) = 1 + 3s$.  This tells us that $s = 2$.  Using the fact that $t = -4 + s$ we learn that $t = -2$.  Therefore, these two lines intersect at the point $(0, 0,7)$

(d) The two lines intersect at the point $\vec{x}_0 = (0, 0, 7)$, and the angle the lines form can be detected by computing the angles their direction vectors make (note that we don’t actually need to know the intersection point to do this calculation).  In particular, we have that

$\vec{v}_1\cdot \vec{v}_2 = \langle -2, 1, 3 \rangle\,\cdot\,\langle-2, 1, 5\rangle = 4 + 1 + 15 = 20$.

We can also compute that $|\vec{v}_1| = \sqrt{4+1+9} = \sqrt{14}$ and $|\vec{v}_2| = \sqrt{4 + 1 + 25} = \sqrt{30}$.

By our formula for the dot product we then have

$\displaystyle \cos\theta = \cos^{-1}\,\frac{20}{\sqrt{14}\sqrt{30}}$.

(e) To find an equation for a plane we need a point and a normal.  Since these two lines intersect at th epoint $(0, 0, 7)$, we can use this as the given point $\vec{x}_0$.  For a normal vector, $\vec{n}$, we know that the direction vectors for each line each need to be perpendicular to $\vec{n}$.  This suggests using

$\vec{n} = \vec{v}_1 \times \vec{v}_2$.

Once this vector is computed we can then set up our (scalar) plane equation

$\left(\vec{x}-\vec{x}_0\right)\cdot\vec{n} = 0$.

Problem 8.  (Exercises 3a-3e in 9.5)

(a) For the first plane we can use $\vec{n}_1 = \langle 4, -5, 1 \rangle$.

(b) To find an equation for the second plane we need a point, $\vec{x}_0$, and a normal vector, $\vec{n}_2$.  We have our choice of three points for $\vec{x}_0$, and so we can use $\vec{x}_0 = \langle 1, 1, 1 \rangle$.  As discussed in class, we can construct a normal vector by setting

$\displaystyle \vec{n}_2 = \vec{PQ} \times \vec{PR}$

where $P = (1, 1, 1), Q = (0, 1, -1)$ and $R = (4, 2, -1)$.  In particular, we could compute

$\displaystyle \vec{n}_2 = \langle 1, 0, 2 \rangle \times \langle -3, -1, 2 \rangle$.

(c ) To find the angle between the planes we need to compute the angle between their normal vectors.  We would then find

$\displaystyle \theta = \cos^{-1} \frac{\vec{n}_1\cdot\vec{n}_2}{|\vec{n}_1|\,|\vec{n}_2|}$

(d) To find a parametric description of a line, we need a point $\vec{x}_0$ and a direction vector $\vec{v}$.  Since, as mentioned in the problem, the direction vector must be perpendicular to both normals we can find $\vec{v}$ by setting

$\displaystyle \vec{v} = \vec{n}_1 \times \vec{n}_2$.

Problem 9.  (skipped)

Problem 10.  Since the plane is to be parallel to the given one, we can take as its normal vector the same one used in the given plane.  That is we can use $\vec{n} = \langle 2, 3, 5 \rangle$.  We want our plane to pass through $\vec{x}_0 = (1, 1, 1)$.  This gives us the following equation:

$\displaystyle \left(\vec{x}-\vec{x}_0\right)\cdot\vec{n} = 0$

$\displaystyle \langle x-1, y-1, z-1 \rangle \cdot \langle 2, 3, 5 \rangle = 0$

$\displaystyle 2(x-1) + 3(y-1) + 5(z-1) = 0$

$\displaystyle 2x + 3y + 5z -10 = 0$

Problem 11.  It is not hard to convince oneself that the plane $z = 1$ touches the unit sphere at the north pole $(0, 0, 1)$. Here is a not-so-terrible picture of both the plane and the sphere: