Practice Exam 1

Vectors & the Geometry of Space

Questions10

Time90 min

Points100

SketchRequired for Q6, 8

1Vectors between points8 pts

Let \(P(1,-2,3)\) and \(Q(4,0,-1)\). Find \(\overrightarrow{PQ}\), its magnitude, the unit vector in its direction, and the midpoint of segment \(PQ\).

Solution

\[ \overrightarrow{PQ} = \langle 4-1,\ 0-(-2),\ -1-3\rangle = \langle 3, 2, -4\rangle,\qquad |\overrightarrow{PQ}| = \sqrt{9+4+16} = \sqrt{29}\approx 5.39 \] \[ \hat{\mathbf{u}} = \frac{1}{\sqrt{29}}\langle 3, 2, -4\rangle,\qquad M = \left(\tfrac{1+4}{2}, \tfrac{-2+0}{2}, \tfrac{3-1}{2}\right) = \left(\tfrac52, -1, 1\right) \]

Tip minus tail. \(\overrightarrow{QP}\) would be the negative. The midpoint is the average of the coordinates — equivalently \(P + \tfrac12\overrightarrow{PQ}\).

2Dot product8 pts

For \(\mathbf{a} = \langle 1, 2, 2\rangle\) and \(\mathbf{b} = \langle 3, 0, -4\rangle\), find the angle between them, the scalar projection \(\operatorname{comp}_{\mathbf{b}}\mathbf{a}\), and the vector projection \(\operatorname{proj}_{\mathbf{b}}\mathbf{a}\).

Solution

\[ \mathbf{a}\cdot\mathbf{b} = 3 + 0 - 8 = -5,\qquad |\mathbf{a}| = 3,\qquad |\mathbf{b}| = 5 \] \[ \cos\theta = \frac{-5}{15} = -\frac13\ \Rightarrow\ \theta = \arccos\!\left(-\tfrac13\right)\approx 109.5^\circ \] \[ \operatorname{comp}_{\mathbf{b}}\mathbf{a} = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|} = -1,\qquad \operatorname{proj}_{\mathbf{b}}\mathbf{a} = \frac{-5}{25}\langle 3,0,-4\rangle = \left\langle -\tfrac35, 0, \tfrac45\right\rangle \]

The negative dot product means an obtuse angle, and the projection points against \(\mathbf{b}\) — its length is \(|-1| = 1\), consistent with \(\operatorname{comp} = -1\). ✓

3Cross product & area10 pts

For \(\mathbf{a} = \langle 2, 1, -1\rangle\) and \(\mathbf{b} = \langle 1, -1, 3\rangle\): compute \(\mathbf{a}\times\mathbf{b}\), verify it is perpendicular to both, and find the area of the triangle with sides \(\mathbf{a}\) and \(\mathbf{b}\).

Solution

\[ \mathbf{a}\times\mathbf{b} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ 2&1&-1\\ 1&-1&3\end{vmatrix} = \mathbf{i}(3 - 1) - \mathbf{j}(6 + 1) + \mathbf{k}(-2 - 1) = \langle 2, -7, -3\rangle \] \[ \langle 2,-7,-3\rangle\cdot\langle 2,1,-1\rangle = 4 - 7 + 3 = 0\ ✓\qquad \langle 2,-7,-3\rangle\cdot\langle 1,-1,3\rangle = 2 + 7 - 9 = 0\ ✓ \] \[ \text{Triangle area} = \tfrac12|\mathbf{a}\times\mathbf{b}| = \tfrac12\sqrt{4 + 49 + 9} = \tfrac12\sqrt{62}\approx 3.94 \]

The perpendicularity check costs ten seconds and catches almost every sign error in the middle component. Do it every time.

4Lorentz force · EE10 pts

A proton (\(q = 1.6\times10^{-19}\) C) moves with \(\mathbf{v} = \langle 0,\ 3\times10^5,\ 0\rangle\) m/s through \(\mathbf{B} = \langle 0.2,\ 0,\ 0.1\rangle\) T. Find the force vector, its magnitude, and confirm it does no work on the proton.

Solution

\[ \mathbf{v}\times\mathbf{B} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ 0&3\times10^5&0\\ 0.2&0&0.1\end{vmatrix} = \mathbf{i}(3\times10^4 - 0) - \mathbf{j}(0 - 0) + \mathbf{k}(0 - 6\times10^4) = \langle 3\times10^4,\ 0,\ -6\times10^4\rangle \] \[ \mathbf{F} = q\,\mathbf{v}\times\mathbf{B} = \langle 4.8\times10^{-15},\ 0,\ -9.6\times10^{-15}\rangle\ \text{N},\qquad |\mathbf{F}| = 1.6\times10^{-19}\sqrt{(3\times10^4)^2 + (6\times10^4)^2} \approx 1.07\times10^{-14}\ \text{N} \]

No work: \(\mathbf{F}\cdot\mathbf{v} = 0 + 0 + 0 = 0\) — the force has no \(y\)-component and \(\mathbf{v}\) has only a \(y\)-component. The cross product is always perpendicular to \(\mathbf{v}\), so a magnetic field can only redirect a charge, never change its kinetic energy.

5Lines in space10 pts

(a) Find parametric and symmetric equations for the line \(L_1\) through \((1, 0, 2)\) and \((3, -1, 4)\). (b) Determine whether \(L_1\) intersects \(L_2:\ x = 2 + s,\ y = 1 + s,\ z = 3 - s\), is parallel to it, or is skew.

Solution

(a) Direction \(\mathbf{v} = \langle 2, -1, 2\rangle\).

\[ x = 1 + 2t,\quad y = -t,\quad z = 2 + 2t \qquad\Longleftrightarrow\qquad \frac{x-1}{2} = \frac{y}{-1} = \frac{z-2}{2} \]

(b) Not parallel: \(\langle 2,-1,2\rangle\) is not a multiple of \(\langle 1,1,-1\rangle\). Set the lines equal with different parameters:

\[ 1 + 2t = 2 + s,\qquad -t = 1 + s,\qquad 2 + 2t = 3 - s \]

From the second, \(s = -t - 1\). Substitute in the first: \(1 + 2t = 1 - t\), so \(t = 0\), \(s = -1\). Check the third: \(2 + 0 = 2\) but \(3 - (-1) = 4\). ✗ The lines do not meet: skew.

6Planes10 pts

Find the plane through \(P(1,1,0)\), \(Q(0,2,1)\), \(R(2,0,3)\). Then find the distance from the point \((3, 3, 5)\) to this plane.

Solution

\[ \overrightarrow{PQ} = \langle -1, 1, 1\rangle,\quad \overrightarrow{PR} = \langle 1, -1, 3\rangle,\quad \mathbf{n} = \overrightarrow{PQ}\times\overrightarrow{PR} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ -1&1&1\\ 1&-1&3\end{vmatrix} = \langle 3+1,\ 1+3,\ 1-1\rangle = \langle 4, 4, 0\rangle \] \[ 4(x-1) + 4(y-1) + 0 = 0\quad\Longrightarrow\quad x + y = 2 \]

Check \(R\): \(2 + 0 = 2\) ✓. The plane is vertical (no \(z\) in the equation) — \(z\) is free. Distance:

\[ D = \frac{|3 + 3 - 2|}{\sqrt{1 + 1}} = \frac{4}{\sqrt2} = 2\sqrt2\approx 2.83 \]

Using the simplified normal \(\langle 1,1,0\rangle\) in the distance formula is fine — normal vectors can be scaled freely, as long as numerator and denominator use the same one.

7Two planes10 pts

For the planes \(2x - y + z = 4\) and \(x + y - z = 1\): find the angle between them and parametric equations for their line of intersection.

Solution

Normals \(\mathbf{n}_1 = \langle 2,-1,1\rangle\), \(\mathbf{n}_2 = \langle 1,1,-1\rangle\). \(\mathbf{n}_1\cdot\mathbf{n}_2 = 2 - 1 - 1 = 0\): the planes are perpendicular, \(\theta = 90^\circ\).

\[ \mathbf{v} = \mathbf{n}_1\times\mathbf{n}_2 = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ 2&-1&1\\ 1&1&-1\end{vmatrix} = \langle 1 - 1,\ -(-2 - 1),\ 2 + 1\rangle = \langle 0, 3, 3\rangle \sim \langle 0, 1, 1\rangle \]

A point: set \(z = 0\): \(2x - y = 4\) and \(x + y = 1\). Adding, \(3x = 5\), \(x = \tfrac53\), \(y = -\tfrac23\).

\[ x = \tfrac53,\qquad y = -\tfrac23 + t,\qquad z = t \]

Check \(t = 1\): \((\tfrac53, \tfrac13, 1)\): first plane \(\tfrac{10}{3} - \tfrac13 + 1 = 4\) ✓; second \(\tfrac53 + \tfrac13 - 1 = 1\) ✓.

8Quadric surfaces10 pts

Identify each surface, name its axis or centre, and sketch it: (a) \(x^2 + 4y^2 - z^2 = 4\); (b) \(z = 4 - x^2 - y^2\); (c) \(x^2 - y^2 + z^2 = 0\); (d) \(x^2 + y^2 + z^2 - 2x + 4y = 4\); (e) \(y = z^2\).

Solution

(a) Divide by 4: \(\dfrac{x^2}{4} + y^2 - \dfrac{z^2}{4} = 1\). One minus sign, \(=1\): hyperboloid of one sheet around the \(z\)-axis. Waist at \(z = 0\) is the ellipse \(x^2/4 + y^2 = 1\).
(b) First power in \(z\), squares with the same sign: elliptic (circular) paraboloid, opening downward, vertex \((0,0,4)\).
(c) \(y^2 = x^2 + z^2\): all squares equal to zero: cone with axis along \(y\) (the variable on its own side).
(d) Complete the square: \((x-1)^2 + (y+2)^2 + z^2 = 4 + 1 + 4 = 9\): sphere, centre \((1,-2,0)\), radius 3.
(e) \(x\) is missing: parabolic cylinder. The parabola \(y = z^2\) in the \(yz\)-plane, extruded along \(x\).

9Coordinate systems12 pts

(a) Convert \((-\sqrt3,\ 1,\ -2)\) to cylindrical and spherical coordinates. (b) Identify the surface \(\rho = 4\cos\phi\). (c) Identify the surface \(r = 2\sin\theta\) (cylindrical).

Solution

(a) Cylindrical: \(r = \sqrt{3 + 1} = 2\). \(\tan\theta = 1/(-\sqrt3)\); the point has \(x < 0, y > 0\) (quadrant II), so \(\theta = 5\pi/6\), not \(-\pi/6\). \(z = -2\). Answer \((2,\ 5\pi/6,\ -2)\).

Spherical: \(\rho = \sqrt{3 + 1 + 4} = 2\sqrt2\). \(\cos\phi = z/\rho = -2/(2\sqrt2) = -1/\sqrt2\), so \(\phi = 3\pi/4\) (below the equator, as \(z < 0\) requires). Answer \((2\sqrt2,\ 5\pi/6,\ 3\pi/4)\). Check: \(r = \rho\sin\phi = 2\sqrt2\cdot\tfrac{1}{\sqrt2} = 2\) ✓.

(b) Multiply by \(\rho\): \(\rho^2 = 4\rho\cos\phi\), i.e. \(x^2 + y^2 + z^2 = 4z\). Complete the square: \(x^2 + y^2 + (z-2)^2 = 4\). A sphere of radius 2 centred at \((0,0,2)\) — it sits on the origin.

(c) Multiply by \(r\): \(r^2 = 2r\sin\theta\), i.e. \(x^2 + y^2 = 2y\), so \(x^2 + (y-1)^2 = 1\). A circular cylinder of radius 1 around the vertical line \(x = 0, y = 1\). No \(z\) appears, so it extends infinitely up and down.

10Torque & triple product · EE12 pts

(a) A force \(\mathbf{F} = \langle 0, 3, -1\rangle\) N acts on a motor coil at position \(\mathbf{r} = \langle 0.05, 0.02, 0\rangle\) m relative to the shaft. Find the torque vector and its magnitude. (b) Determine whether \(\mathbf{a} = \langle 1,2,0\rangle\), \(\mathbf{b} = \langle 0,1,3\rangle\), \(\mathbf{c} = \langle 2,0,1\rangle\) are coplanar, and if not, find the volume of the parallelepiped they span.

Solution

(a)

\[ \boldsymbol\tau = \mathbf{r}\times\mathbf{F} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ 0.05&0.02&0\\ 0&3&-1\end{vmatrix} = \mathbf{i}(-0.02 - 0) - \mathbf{j}(-0.05 - 0) + \mathbf{k}(0.15 - 0) = \langle -0.02,\ 0.05,\ 0.15\rangle\ \text{N}\cdot\text{m} \] \[ |\boldsymbol\tau| = \sqrt{0.0004 + 0.0025 + 0.0225} = \sqrt{0.0254}\approx 0.159\ \text{N}\cdot\text{m} \]

Most of the torque is about the \(z\)-axis — the \(\mathbf{k}\) component, \(0.15\), comes from the in-plane force \(3\,\mathbf{j}\) acting at the in-plane lever arm.

(b) Scalar triple product:

\[ \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) = \begin{vmatrix}1&2&0\\ 0&1&3\\ 2&0&1\end{vmatrix} = 1(1 - 0) - 2(0 - 6) + 0 = 1 + 12 = 13 \]

Nonzero, so not coplanar; the parallelepiped has volume \(13\). (Had it been zero, the three vectors would lie in one plane and span no volume.)

Scoring

Score	Where you are
90–100	Solid. On to vector functions.
75–89	Concepts are right, determinants are leaking signs. Slow down on the middle component of every cross product.
60–74	Setup errors — usually the wrong quadrant for \(\theta\), or one parameter used for two lines. Redo the misses from the sketch.
< 60	Re-read Sections 3–4 and spend twenty minutes in the vectors lab before continuing.

Where the points usually go

The middle component of the cross product (Q3, Q4, Q6, Q10); using the same parameter for both lines (Q5); putting \(\theta\) in the wrong quadrant (Q9); and forgetting to check that a plane passes through the third point (Q6). None of these are conceptual — they are all checkable in under a minute, and the solutions show the check each time.