Vectors and their Products

Chapter 6 · Mathematics · 3 questions in the paper · 11 min read

Vectors and their Products

What the syllabus asks for 2
  • Vectors in plane and space, algebra of vectors, linear combination of vectors, linearly dependent and independent sets of vectors
  • Product of two vectors, scalar and vector product of two vectors, scalar triple product

Vectors carry about 3 questions (our estimate).

The booklet's model question — the projection of \vec b=2\hat i+3\hat j-2\hat k on \vec a=\hat i+2\hat j+3\hat k — is typical: a dot product, a cross product, or a scalar triple product, computed from components.

The whole chapter reduces to four formulas used fluently.

What the paper asks from this chapter
  • Components, magnitude, unit vector, section formula, collinearity.
  • Dot product: \vec a\cdot\vec b=ab\cos\theta; projection of \vec b on \vec a is \dfrac{\vec a\cdot\vec b}{|\vec a|}.
  • Cross product: |\vec a\times\vec b|=ab\sin\theta, area of a triangle \frac{1}{2}|\vec a\times\vec b|.
  • Scalar triple product: volume, coplanarity ([\vec a\,\vec b\,\vec c]=0).
  • Linear dependence: two vectors are dependent iff parallel; three iff coplanar.

6.1 Vectors in plane and space, algebra, linear combination, dependence

A vector has magnitude and direction. For \vec a=a_1\hat i+a_2\hat j+a_3\hat k: |\vec a|=\sqrt{a_1^2+a_2^2+a_3^2} and the unit vector is \hat a=\vec a/|\vec a|.

Direction cosines \frac{a_1}{|\vec a|},\frac{a_2}{|\vec a|},\frac{a_3}{|\vec a|}. The vector \overrightarrow{AB}=\vec b-\vec a (head minus tail).

  • Triangle law: \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}; the parallelogram law; |\vec a\pm\vec b|^2=a^2+b^2\pm2\vec a\cdot\vec b.
  • Section formula: point dividing AB in the ratio m:n internally is \dfrac{m\vec b+n\vec a}{m+n} (externally \dfrac{m\vec b-n\vec a}{m-n}); midpoint \frac{\vec a+\vec b}{2}; centroid \frac{\vec a+\vec b+\vec c}{3}.
  • Collinear points A,B,C: \overrightarrow{AC}=k\,\overrightarrow{AB}.
  • Linear combination: \vec r=x\vec a+y\vec b+z\vec c. A set is linearly dependent if some non-trivial combination is zero: two vectors iff parallel; three iff coplanar ([\vec a\,\vec b\,\vec c]=0); any four in space always. Three non-coplanar vectors form a basis of space.

6.2 Product of two vectors, scalar and vector product, scalar triple product

Scalar (dot) product

\vec a\cdot\vec b=ab\cos\theta=a_1b_1+a_2b_2+a_3b_3; commutative and distributive. \hat i\cdot\hat i=1, \hat i\cdot\hat j=0. \vec a\cdot\vec a=|\vec a|^2.

Perpendicular iff \vec a\cdot\vec b=0. \cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}. Projection of \vec b on \vec a: \dfrac{\vec a\cdot\vec b}{|\vec a|} (a scalar); the vector component is that times \hat a.

Work =\vec F\cdot\vec d. (\vec a+\vec b)\cdot(\vec a-\vec b)=a^2-b^2.

Vector (cross) product

\vec a\times\vec b=ab\sin\theta\,\hat n (right-hand rule), anticommutative; \hat i\times\hat j=\hat k, \hat j\times\hat k=\hat i, \hat k\times\hat i=\hat j.

\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k \\ a_1&a_2&a_3 \\ b_1&b_2&b_3\end{vmatrix}.

|\vec a\times\vec b|^2=a^2b^2-(\vec a\cdot\vec b)^2 (Lagrange). Parallel iff \vec a\times\vec b=\vec 0. Area of a parallelogram with sides \vec a,\vec b is |\vec a\times\vec b|; of the triangle, half.

A unit vector perpendicular to both: \pm\dfrac{\vec a\times\vec b}{|\vec a\times\vec b|}. Moment of a force: \vec r\times\vec F.

Triple products

[\vec a\,\vec b\,\vec c]=\vec a\cdot(\vec b\times\vec c)= the 3×3 determinant of components; cyclic invariant, changes sign when two vectors swap.

Volume of the parallelepiped =|[\vec a\,\vec b\,\vec c]|; tetrahedron =\frac{1}{6}|[\vec a\,\vec b\,\vec c]|.

Coplanar iff [\vec a\,\vec b\,\vec c]=0. Vector triple product: \vec a\times(\vec b\times\vec c)=\vec b(\vec a\cdot\vec c)-\vec c(\vec a\cdot\vec b) ("BAC minus CAB").

Common mistake

The dot product is a scalar, the cross product a vector. \vec a\times(\vec b\times\vec c)\ne(\vec a\times\vec b)\times\vec c. The projection can be negative.

photograph
(a) Tug of war: each team pulls with a force that has a size and a direction
diagram
(b) Resultant of 120 N, 100 N and 80 N at 120 degrees spacing
Plate 6.1. Adding vectors. (a) In a tug of war the rope tension is a vector: the pull has a size and a direction, and the result depends on both. (b) Three ropes tied to a ring pull with 120 N at 0 degrees, 100 N at 120 degrees and 80 N at 240 degrees. Adding components gives Rx = 30 N and Ry = 17.32 N, so the resultant has magnitude 34.64 N at 30 degrees. Drawn head to tail, the three forces and the resultant form a closed polygon. A fourth force of 34.64 N at 210 degrees would give equilibrium. Any three vectors in a plane are linearly dependent.Photo credits: (a) Joy Agyepong, CC BY-SA 4.0.

photograph
(a) A click torque wrench: turning effect equals r cross F
diagram
(b) Torque of a 150 N push on a 0.40 m handle vs angle
Plate 6.2. The vector product. (a) A torque wrench measures the turning effect of a force, which is the vector product tau = r x F, with magnitude r F sin(theta). (b) For a handle length r = 0.40 m and a force F = 150 N at 60 degrees to the handle, the cross product of the vectors r = (0.40, 0, 0) and F = 150 (cos 60, sin 60, 0) is (0, 0, 51.96), so the torque is 51.96 N m along the axis of the bolt. The check r . tau = 0 and F . tau = 0 confirms that the result is perpendicular to both vectors. The torque is greatest at 90 degrees (60 N m) and zero at 0 and 180 degrees.Photo credits: (a) No machine-readable author provided. Heron, CC BY-SA 3.0.

photograph
(a) A road-gradient warning sign: a car on a slope
diagram
(b) Weight and displacement on a 20 per cent grade
Plate 6.3. The dot product as work. (a) A steep-gradient sign warns drivers that gravity has a large component along the road. (b) Take a 1,200 kg car on a 20 per cent grade (rise 1 for every 5 along the ground), so the angle is 11.31 degrees. The weight is W = (0, -11,760) N. Along 100 m of road the displacement is d = (98.06, 19.61) m, and W . d = -230,633 J, which equals -m g h for a rise of 19.61 m. The part of the weight pulling along the road is m g sin(theta) = 2306 N, and this is the force the brakes must resist.Photo credits: (a) Bidgee, CC BY-SA 2.5 au.

Formula and fact sheet

Topic Result
angle \cos\theta=\frac{\vec a\cdot\vec b}{ab}
projection of \vec b on \vec a \frac{\vec a\cdot\vec b}{|\vec a|}
triangle area \frac{1}{2}|\vec a\times\vec b|
Lagrange |\vec a\times\vec b|^2=a^2b^2-(\vec a\cdot\vec b)^2
volume |[\vec a\,\vec b\,\vec c]|; tetrahedron \frac{1}{6} of it
triple product \vec a\times(\vec b\times\vec c)=\vec b(\vec a\cdot\vec c)-\vec c(\vec a\cdot\vec b)

Chapter summary

  • Head minus tail gives \overrightarrow{AB}; the section formula weights the opposite end.
  • Dot product zero means perpendicular; cross product zero means parallel; triple product zero means coplanar.
  • Projection is \frac{\vec a\cdot\vec b}{|\vec a|}, a signed scalar.
  • Areas come from |\vec a\times\vec b|, volumes from the scalar triple product.

Practice questions

Set A — Recall (10 questions)

  1. For \vec a=2\hat i+3\hat j+\hat k and \vec b=\hat i-\hat j+2\hat k, \vec a\cdot\vec b equalsa7b5c1d-1
  2. The magnitude of 3\hat i+4\hat j-12\hat k isa13b19c11d5
  3. \hat i\times\hat j equalsa0b-\hat kc1d\hat k
  4. The vectors 2\hat i+\lambda\hat j+\hat k and \hat i-2\hat j+3\hat k are perpendicular for \lambda=a2b5/2c3/2d1
  5. The area of the parallelogram with adjacent sides \hat i+\hat j and \hat j+\hat k isa2b\sqrt{2}c\sqrt{3}d3
  6. If [\vec a\,\vec b\,\vec c]=0, the three vectors areacoplanarbparallelcmutually perpendiculardcollinear
  7. The angle between \hat i+\hat j and \hat j isa30^\circb60^\circc90^\circd45^\circ
  8. The midpoint of A(2,4,6) and B(4,0,2) isa(6,4,8)b(3,2,4)c(1,2,2)d(3,4,2)
  9. The unit vector along 3\hat i-4\hat j isa\dfrac{3\hat i-4\hat j}{25}b3\hat i-4\hat jc\dfrac{3\hat i+4\hat j}{5}d\dfrac{3\hat i-4\hat j}{5}
  10. If |\vec a|=5 and |\vec b|=3, then (\vec a+\vec b)\cdot(\vec a-\vec b) isa16b34c4d25
Answers and explanations — Set A

1. (c) — 2-3+2=1.

2. (a) — \sqrt{9+16+144}=13.

3. (d) — right-hand rule: \hat i\times\hat j=\hat k (\hat j\times\hat i=-\hat k).

4. (b) — 2-2\lambda+3=0\Rightarrow\lambda=\frac{5}{2}.

5. (c) — \vec a\times\vec b=(1,-1,1), magnitude \sqrt{3}.

6. (a) — a zero scalar triple product means the vectors lie in one plane.

7. (d) — \cos\theta=\frac{1}{\sqrt{2}}.

8. (b) — average of the coordinates.

9. (d) — divide by the magnitude 5.

10. (a) — |\vec a|^2-|\vec b|^2=25-9=16.

Set B — Understanding (7 questions)

  1. The projection of \vec b=2\hat i+3\hat j-2\hat k on \vec a=\hat i+2\hat j+3\hat k isa2/\sqrt{17}b2/\sqrt{14}c2/7d-2/7
  2. If |\vec a|=3, |\vec b|=4 and the angle between them is 30^\circ, then |\vec a\times\vec b| isa12b6\sqrt{3}c6d3
  3. The vectors 2\hat i-\hat j+\hat k, \hat i+2\hat j-3\hat k and 3\hat i+\lambda\hat j+5\hat k are coplanar for \lambda=a4b-1c1d-4
  4. The area of the triangle with vertices (1,0,0), (0,1,0), (0,0,1) isa\frac{\sqrt{3}}{2}b\sqrt{3}c\frac{1}{2}d\frac{3}{2}
  5. The volume of the parallelepiped with edges \hat i+\hat j, \hat j+\hat k, \hat k+\hat i isa1b4c2d3
  6. A unit vector perpendicular to both \hat i+\hat j and \hat j+\hat k isa\frac{\hat i+\hat j+\hat k}{\sqrt{3}}b\frac{\hat i-\hat j+\hat k}{\sqrt{3}}c\frac{\hat i-\hat j-\hat k}{3}d\frac{\hat i+\hat j-\hat k}{\sqrt{3}}
  7. The value of \hat i\times(\hat j\times\hat i) isa\hat jb0c-\hat jd\hat k
Answers and explanations — Set B

1. (b) — \vec a\cdot\vec b=2+6-6=2, |\vec a|=\sqrt{14}. The projection of \vec b on \vec a divides by |\vec a|.

2. (c) — 3\cdot4\sin30^\circ=6.

3. (d) — the determinant is 28+7\lambda=0.

4. (a) — \overrightarrow{AB}\times\overrightarrow{AC}=(1,1,1); half of \sqrt{3}.

5. (c) — \begin{vmatrix}1&1&0 \\ 0&1&1 \\ 1&0&1\end{vmatrix}=1+1=2.

6. (b) — (\hat i+\hat j)\times(\hat j+\hat k)=\hat i-\hat j+\hat k, magnitude \sqrt{3}.

7. (a) — \hat j\times\hat i=-\hat k, and \hat i\times(-\hat k)=\hat j. Check with BAC–CAB: \hat j(\hat i\cdot\hat i)-\hat i(\hat i\cdot\hat j)=\hat j.

Set C — Application (6 questions)

  1. The points (1,1,1), (2,3,4) and (3,5,\lambda) are collinear when \lambda=a6b7c8d5
  2. The work done by \vec F=2\hat i+3\hat j+\hat k in moving a particle from (1,1,1) to (3,2,4) isa10b12c8d9
  3. If \vec a,\vec b are unit vectors at 60^\circ to each other, |\vec a+\vec b| isa\sqrt{2}b2c1d\sqrt{3}
  4. If |\vec a|=2, |\vec b|=3 and \vec a\cdot\vec b=3, then |\vec a\times\vec b| isa3b6c9d3\sqrt{3}
  5. The vectors 2\hat i+3\hat j and \hat i+\lambda\hat j are linearly dependent for \lambda=a2/3b3/2c6d1/6
  6. The volume of the tetrahedron with vertices O, (1,0,0), (0,2,0), (0,0,3) isa1b6c3d2
Answers and explanations — Set C

1. (b) — \overrightarrow{AB}=(1,2,3), \overrightarrow{AC}=(2,4,\lambda-1) must be 2\overrightarrow{AB}: \lambda-1=6.

2. (a) — displacement (2,1,3): 4+3+3=10.

3. (d) — |\vec a+\vec b|^2=1+1+2\cdot\frac{1}{2}=3.

4. (d) — |\vec a\times\vec b|^2=36-9=27.

5. (b) — parallel: \frac{2}{1}=\frac{3}{\lambda}.

6. (a) — \frac{1}{6}|1\cdot2\cdot3|=1.

Group D — Model answers for long questions (10 marks each, for deeper revision)

  1. For \mathbf{a}=(1,2,3) and \mathbf{b}=(2,-1,1) find the scalar product, angle, vector product, parallelogram area and projection. [10]
  2. Use the scalar triple product to test coplanarity and to find volumes; find the work done by a force and the torque of a force. [10]

Model answer D1: Scalar and vector products (10 marks)

Introduction. The scalar (dot) product is \mathbf{a}\cdot\mathbf{b}=\lvert\mathbf{a}\rvert\lvert\mathbf{b}\rvert\cos\theta=a_1b_1+a_2b_2+a_3b_3; the vector (cross) product \mathbf{a}\times\mathbf{b} has magnitude \lvert\mathbf{a}\rvert\lvert\mathbf{b}\rvert\sin\theta and is perpendicular to both vectors.

Solution.

Quantity Working Value
\mathbf{a}\cdot\mathbf{b} 2-2+3 3
\lvert\mathbf{a}\rvert,\ \lvert\mathbf{b}\rvert \sqrt{14},\ \sqrt6 3.742,\ 2.449
Angle \theta \cos\theta=\frac{3}{\sqrt{84}}=0.327 70.9^\circ
\mathbf{a}\times\mathbf{b} (2\cdot1-3(-1),\ 3\cdot2-1\cdot1,\ 1(-1)-2\cdot2) (5,5,-5)
Area of parallelogram \lvert\mathbf{a}\times\mathbf{b}\rvert=\sqrt{75} 8.66
Projection of \mathbf{a} on \mathbf{b} \frac{\mathbf{a}\cdot\mathbf{b}}{\lvert\mathbf{b}\rvert}=\frac{3}{\sqrt6} 1.22

Checks. (\mathbf{a}\times\mathbf{b})\cdot\mathbf{a}=5+10-15=0 and (\mathbf{a}\times\mathbf{b})\cdot\mathbf{b}=10-5-5=0, so the result is perpendicular to both. The area of the triangle with these sides is half: 4.33.

Properties.

  1. \mathbf{a}\cdot\mathbf{b}=\mathbf{b}\cdot\mathbf{a}, but \mathbf{a}\times\mathbf{b}=-\mathbf{b}\times\mathbf{a}.
  2. Two non-zero vectors are perpendicular if \mathbf{a}\cdot\mathbf{b}=0 and parallel if \mathbf{a}\times\mathbf{b}=\mathbf{0}.
  3. \hat i\cdot\hat i=1, \hat i\times\hat j=\hat k.
  4. Section formula: the point dividing \mathbf{a} and \mathbf{b} in the ratio m:n is \frac{m\mathbf{b}+n\mathbf{a}}{m+n}.

Conclusion. The dot product measures how much two vectors point together and the cross product measures the area they span.

Model answer D2: Triple product, work and torque (10 marks)

Introduction. The scalar triple product [\mathbf{a}\ \mathbf{b}\ \mathbf{c}]=\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) equals the determinant of the three vectors. Its absolute value is the volume of the parallelepiped.

1. Coplanarity. Three vectors are coplanar if [\mathbf{a}\ \mathbf{b}\ \mathbf{c}]=0. For \mathbf{a}=(1,2,3), \mathbf{b}=(2,-1,1), \mathbf{c}=(1,1,2):

\begin{vmatrix}1&2&3\\2&-1&1\\1&1&2\end{vmatrix}=1(-2-1)-2(4-1)+3(2+1)=-3-6+9=0.

So they are coplanar: indeed \mathbf{c}=\frac35\mathbf{a}+\frac15\mathbf{b}, since \frac35+\frac25=1, \frac65-\frac15=1 and \frac95+\frac15=2.

2. Volume. For \mathbf{c}=(3,1,0): the determinant =1(0-1)-2(0-3)+3(2+3)=-1+6+15=20. The parallelepiped has volume 20 and the tetrahedron has volume \frac{20}{6}=3.33.

3. Work done. W=\mathbf{F}\cdot\mathbf{d}. For \mathbf{F}=(2,3,1)\ \text{N} and \mathbf{d}=(3,-1,2)\ \text{m}: W=6-3+2=5\ \text{J}.

4. Torque. \tau=\mathbf{r}\times\mathbf{F}. For \mathbf{r}=(2,1,0) and \mathbf{F}=(1,-1,3): \tau=(1\cdot3-0,\ 0-2\cdot3,\ 2(-1)-1\cdot1)=(3,-6,-3), with magnitude \sqrt{9+36+9}=\sqrt{54}=7.35\ \text{N m}.

Quantity Formula Value
Coplanarity test [\mathbf{a}\ \mathbf{b}\ \mathbf{c}]=0 0 (coplanar)
Parallelepiped volume \lvert[\mathbf{a}\ \mathbf{b}\ \mathbf{c}]\rvert 20
Work \mathbf{F}\cdot\mathbf{d} 5\ \text{J}
Torque \mathbf{r}\times\mathbf{F} 7.35\ \text{N m}

Identity. \mathbf{a}\times(\mathbf{b}\times\mathbf{c})=\mathbf{b}(\mathbf{a}\cdot\mathbf{c})-\mathbf{c}(\mathbf{a}\cdot\mathbf{b}).

Conclusion. The triple product gives volume and a coplanarity test, and the two products model work and torque in physics.