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Topic 1

Vectors, Dot Product & Cross Product

Two products do almost all the work. The dot product a⋅b\mathbf a\cdot\mathbf b is a number: it gives angles, perpendicular tests and . The cross product a×b\mathbf a\times\mathbf b is a vector perpendicular to both: its length gives areas, and with a third vector it gives volumes.

6 kinds of question. Open the one you are stuck on.

1

Find the center and radius of a sphere by completing the square

e.g. “Describe the set of all points (x,y,z)(x, y, z) in R3\mathbb R^3 that satisfy x2+y2+z2=…x^2 + y^2 + z^2 = \dots”
  1. 1
    Collect the terms so x2x^2, y2y^2, z2z^2 each have coefficient 11

    From a distance condition, square both distances first. Divide through if all three squares share a coefficient.

  2. 2
    Complete the square in each variable

    x2+bx=(x+b2)2−b24x^2 + bx = \left(x + \tfrac{b}{2}\right)^2 - \tfrac{b^2}{4}. Every constant you subtract on the left moves to the right side.

  3. 3
    Read off the center and r2r^2

    (x+4)2(x + 4)^2 means the center's xx-coordinate is −4-4.

  4. 4
    Name the set: the sphere, its inside, or its outside

    == is the sphere, << the points inside it, >> the points outside it.

Remember: Group, complete, read, name.

See it in a problem3 worked examples

Describe the set of points (x,y,z)(x, y, z) with x2+y2+z2−6x+2z=6x^2 + y^2 + z^2 - 6x + 2z = 6.

  1. 1
    Collect the terms so x2x^2, y2y^2, z2z^2 each have coefficient 11

    (x2−6x)+y2+(z2+2z)=6(x^2 - 6x) + y^2 + (z^2 + 2z) = 6.

  2. 2
    Complete the square in each variable

    (x−3)2−9+y2+(z+1)2−1=6(x - 3)^2 - 9 + y^2 + (z + 1)^2 - 1 = 6, so (x−3)2+y2+(z+1)2=16(x - 3)^2 + y^2 + (z + 1)^2 = 16.

  3. 3
    Read off the center and r2r^2

    Center (3,0,−1)(3, 0, -1), r2=16r^2 = 16, so r=4r = 4.

  4. 4
    Name the set: the sphere, its inside, or its outside

    An equation (==), so it is the sphere itself.

Answer:

The sphere of radius 44 centered at (3,0,−1)(3, 0, -1).

Common mistakes4
  • Center signs: (x−3)2+(y+5)2+z2(x - 3)^2 + (y + 5)^2 + z^2 gives center (3,−5,0)(3, -5, 0).
  • Reporting r2r^2 as the radius: ⋯=16\dots = 16 means r=4r = 4.
  • Completing the square on the left but not adding the same constants on the right.
  • Given the endpoints of a diameter? The center is their midpoint and the radius is half the distance between them.
2

Find a unit vector parallel or perpendicular to a vector

e.g. “Find a unit vector in the same direction as a\mathbf a”
  1. 1
    Find the length ∣a∣|\mathbf a|

    ∣⟨a1,a2,a3⟩∣=a12+a22+a32|\langle a_1, a_2, a_3\rangle| = \sqrt{a_1^2 + a_2^2 + a_3^2}.

  2. 2
    Divide by it: a^=a∣a∣\hat{\mathbf a} = \dfrac{\mathbf a}{|\mathbf a|}

    That is the in the same direction. "All unit vectors parallel" means ±a^\pm\hat{\mathbf a}.

  3. 3
    For length LL, scale: ±L a^\pm L\,\hat{\mathbf a}
  4. 4
    For perpendicular: dot product zero, then divide by the length

    In 2D, ⟨a1,a2⟩\langle a_1, a_2\rangle is perpendicular to ⟨a2,−a1⟩\langle a_2, -a_1\rangle. In 3D there are infinitely many: set one component to 00 and solve.

Remember: Divide by the length; plus or minus for all of them.

See it in a problem2 worked examples

Let a=⟨5,−12⟩\mathbf a = \langle 5, -12\rangle. Find the unit vector in the direction of a\mathbf a, all vectors parallel to a\mathbf a of length 2626, and all unit vectors perpendicular to a\mathbf a.

  1. 1
    Find the length ∣a∣|\mathbf a|

    ∣a∣=25+144=13|\mathbf a| = \sqrt{25 + 144} = 13.

  2. 2
    Divide by it: a^=a∣a∣\hat{\mathbf a} = \dfrac{\mathbf a}{|\mathbf a|}

    a^=113⟨5,−12⟩\hat{\mathbf a} = \tfrac{1}{13}\langle 5, -12\rangle.

  3. 3
    For length LL, scale: ±L a^\pm L\,\hat{\mathbf a}

    ±26⋅113⟨5,−12⟩=±⟨10,−24⟩\pm 26 \cdot \tfrac{1}{13}\langle 5, -12\rangle = \pm\langle 10, -24\rangle.

  4. 4
    For perpendicular: dot product zero, then divide by the length

    ⟨12,5⟩\langle 12, 5\rangle works: 60−60=060 - 60 = 0. Its length is 1313, so the unit vectors are ±113⟨12,5⟩\pm\tfrac{1}{13}\langle 12, 5\rangle.

Answer:

113⟨5,−12⟩\tfrac{1}{13}\langle 5, -12\rangle;  ±⟨10,−24⟩\ \pm\langle 10, -24\rangle;  ±113⟨12,5⟩\ \pm\tfrac{1}{13}\langle 12, 5\rangle

Common mistakes3
  • Forgetting the −- sign when the question says all parallel vectors: there are two.
  • Dividing by a1+a2+a3a_1 + a_2 + a_3 instead of the length a12+a22+a32\sqrt{a_1^2 + a_2^2 + a_3^2}.
  • In 3D, a perpendicular unit vector is not unique: any one that works gets the marks, unless the question asks for all.
3

Find the angle between two vectors using the dot product

e.g. “Determine the angle between the vectors a\mathbf a and b\mathbf b”
  1. 1
    Compute the dot product a⋅b\mathbf a\cdot\mathbf b

    Multiply matching components and add. For an angle at a vertex, both vectors must start at that vertex.

  2. 2
    Compute both lengths
  3. 3
    Divide: cos⁡θ=a⋅b∣a∣ ∣b∣\cos\theta = \dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a|\,|\mathbf b|}
  4. 4
    Take θ=arccos⁡(⋅)\theta = \arccos(\cdot)

    A negative cosine means an obtuse angle. Give exact angles where you can (π4\tfrac{\pi}{4}), otherwise degrees to one decimal.

Remember: Dot, lengths, divide, arccos.

See it in a problem3 worked examples

Find the angle between a=⟨2,−1,2⟩\mathbf a = \langle 2, -1, 2\rangle and b=⟨1,1,4⟩\mathbf b = \langle 1, 1, 4\rangle.

  1. 1
    Compute the dot product a⋅b\mathbf a\cdot\mathbf b

    a⋅b=2−1+8=9\mathbf a\cdot\mathbf b = 2 - 1 + 8 = 9.

  2. 2
    Compute both lengths

    ∣a∣=9=3|\mathbf a| = \sqrt{9} = 3, ∣b∣=18=32|\mathbf b| = \sqrt{18} = 3\sqrt 2.

  3. 3
    Divide: cos⁡θ=a⋅b∣a∣ ∣b∣\cos\theta = \dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a|\,|\mathbf b|}

    cos⁡θ=992=12\cos\theta = \dfrac{9}{9\sqrt 2} = \dfrac{1}{\sqrt 2}.

  4. 4
    Take θ=arccos⁡(⋅)\theta = \arccos(\cdot)

    θ=π4\theta = \dfrac{\pi}{4}.

Answer:

θ=π4=45∘\theta = \dfrac{\pi}{4} = 45^\circ

Common mistakes3
  • For the angle at AA in a triangle, use AB→\overrightarrow{AB} and AC→\overrightarrow{AC}, not AB→\overrightarrow{AB} and BC→\overrightarrow{BC}.
  • Writing a⋅b\mathbf a\cdot\mathbf b as a vector: it is a single number.
  • Mixing degrees and radians in the final answer.
4

Check whether two vectors are perpendicular or parallel

e.g. “Determine whether the given pair of vectors is perpendicular”
  1. 1
    Build the vectors you need

    Between points, head minus tail: PQ→=Q−P\overrightarrow{PQ} = Q - P.

  2. 2
    Write the condition

    Perpendicular: a⋅b=0\mathbf a\cdot\mathbf b = 0. Parallel: a=c b\mathbf a = c\,\mathbf b for one constant cc (or a×b=0\mathbf a\times\mathbf b = \mathbf 0).

  3. 3
    Solve, keeping every solution

    A quadratic in the unknown can give two values.

  4. 4
    Answer the question in words

Remember: Perpendicular: dot is zero. Parallel: one is a multiple of the other.

See it in a problem4 worked examples

Find all yy for which ⟨y,2,−3⟩\langle y, 2, -3\rangle and ⟨y,y,1⟩\langle y, y, 1\rangle are perpendicular.

  1. 1
    Build the vectors you need

    Already given: ⟨y,2,−3⟩\langle y, 2, -3\rangle and ⟨y,y,1⟩\langle y, y, 1\rangle.

  2. 2
    Write the condition

    y2+2y−3=0y^2 + 2y - 3 = 0.

  3. 3
    Solve, keeping every solution

    (y+3)(y−1)=0(y + 3)(y - 1) = 0, so y=1y = 1 or y=−3y = -3.

  4. 4
    Answer the question in words

    They are perpendicular exactly when y=1y = 1 or y=−3y = -3.

Answer:

y=1y = 1 or y=−3y = -3

Common mistakes3
  • Parallel needs the same cc in every component: ⟨2,4,5⟩\langle 2, 4, 5\rangle is not parallel to ⟨1,2,3⟩\langle 1, 2, 3\rangle.
  • Stopping at one root of a quadratic like y2+2y−3=0y^2 + 2y - 3 = 0.
  • points: build both vectors from the same point.
5

Find the projection of one vector onto another

e.g. “Find the projection of b\mathbf b on a\mathbf a”
  1. 1
    Compute a⋅b\mathbf a\cdot\mathbf b and ∣a∣2|\mathbf a|^2

    a\mathbf a is the vector you project onto.

  2. 2
    Component, a number: a⋅b∣a∣\dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a|}
  3. 3
    Projection, a vector: a⋅b∣a∣2 a\dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a|^2}\,\mathbf a
  4. 4
    Perpendicular part: b−projab\mathbf b - \text{proj}_{\mathbf a}\mathbf b

    Check: its dot product with a\mathbf a is 00.

Remember: Divide by a's length for the number, by its square for the vector.

See it in a problem3 worked examples

Let a=⟨1,2,−2⟩\mathbf a = \langle 1, 2, -2\rangle and b=⟨6,3,−3⟩\mathbf b = \langle 6, 3, -3\rangle. Find the component of b\mathbf b in the direction of a\mathbf a, the projection of b\mathbf b on a\mathbf a, and the projection of b\mathbf b perpendicular to a\mathbf a.

  1. 1
    Compute a⋅b\mathbf a\cdot\mathbf b and ∣a∣2|\mathbf a|^2

    a⋅b=6+6+6=18\mathbf a\cdot\mathbf b = 6 + 6 + 6 = 18, ∣a∣2=9|\mathbf a|^2 = 9.

  2. 2
    Component, a number: a⋅b∣a∣\dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a|}

    183=6\dfrac{18}{3} = 6.

  3. 3
    Projection, a vector: a⋅b∣a∣2 a\dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a|^2}\,\mathbf a

    189⟨1,2,−2⟩=⟨2,4,−4⟩\dfrac{18}{9}\langle 1, 2, -2\rangle = \langle 2, 4, -4\rangle.

  4. 4
    Perpendicular part: b−projab\mathbf b - \text{proj}_{\mathbf a}\mathbf b

    ⟨6,3,−3⟩−⟨2,4,−4⟩=⟨4,−1,1⟩\langle 6, 3, -3\rangle - \langle 2, 4, -4\rangle = \langle 4, -1, 1\rangle. Check: 4−2−2=04 - 2 - 2 = 0.

Answer:

Component 66; projection ⟨2,4,−4⟩\langle 2, 4, -4\rangle; perpendicular part ⟨4,−1,1⟩\langle 4, -1, 1\rangle

Common mistakes3
  • Dividing by ∣b∣2|\mathbf b|^2: the projection onto a\mathbf a divides by ∣a∣2|\mathbf a|^2 and points along a\mathbf a.
  • Using ∣a∣|\mathbf a| instead of ∣a∣2|\mathbf a|^2 in the vector projection.
  • A negative component is fine: it means the angle between the vectors is obtuse.
6

Find the area of a triangle or volume of a parallelepiped with the cross product

e.g. “Calculate the area of the triangle with vertices …”
  1. 1
    Get the edge vectors from one shared corner
  2. 2
    Compute the cross product

    a×b=⟨a2b3−a3b2, a3b1−a1b3, a1b2−a2b1⟩\mathbf a\times\mathbf b = \langle a_2b_3 - a_3b_2,\ a_3b_1 - a_1b_3,\ a_1b_2 - a_2b_1\rangle. In 2D, add a 00 third component.

  3. 3
    Take its length, or dot it with the third edge

    Parallelogram area: ∣a×b∣|\mathbf a\times\mathbf b|. Volume: a⋅(b×c)\mathbf a\cdot(\mathbf b\times\mathbf c).

  4. 4
    Halve for a triangle; drop the sign for a volume

Remember: Cross for area, cross then dot for volume.

See it in a problem3 worked examples

Find the area of the triangle with vertices A=(1,0,1)A = (1, 0, 1), B=(2,2,1)B = (2, 2, 1) and C=(0,1,3)C = (0, 1, 3).

  1. 1
    Get the edge vectors from one shared corner

    AB→=⟨1,2,0⟩\overrightarrow{AB} = \langle 1, 2, 0\rangle, AC→=⟨−1,1,2⟩\overrightarrow{AC} = \langle -1, 1, 2\rangle.

  2. 2
    Compute the cross product

    AB→×AC→=⟨4−0, 0−2, 1+2⟩=⟨4,−2,3⟩\overrightarrow{AB}\times\overrightarrow{AC} = \langle 4 - 0,\ 0 - 2,\ 1 + 2\rangle = \langle 4, -2, 3\rangle.

  3. 3
    Take its length, or dot it with the third edge

    ∣⟨4,−2,3⟩∣=29|\langle 4, -2, 3\rangle| = \sqrt{29}.

  4. 4
    Halve for a triangle; drop the sign for a volume

    Triangle: 1229\tfrac{1}{2}\sqrt{29}.

Answer:

292\dfrac{\sqrt{29}}{2}

Common mistakes3
  • The middle component is a3b1−a1b3a_3b_1 - a_1b_3. Its order is flipped compared with the other two.
  • Forgetting the one half for a triangle.
  • A negative : the volume is its absolute value.

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