4 Vector Algebra
4.1 Scalars and Vectors
A scalar is a quantity that is specified by a number only (and the units the quantity is measured in). Examples include temperature, energy, time, density. A vector is a quantity that has magnitude and direction. Examples include displacement, velocity, acceleration. Vectors are denoted with bold typeface or an underline.
4.1.1 Addition and subtraction of vectors
Vector sum: \[ {\mathbf a} + {\mathbf b} = {\mathbf b} + {\mathbf a} = {\mathbf c} \] where \(\vec{c}\) is called the resultant vector. Vectors can be added in any order: \[ ({\mathbf a}+ {\mathbf b}) + {\mathbf c} = {\mathbf a}+ ({\mathbf b} + {\bf c}) \]
\[ {\mathbf a} - {\mathbf b} = {\mathbf a} + (-{\mathbf b}) \] where \(-{\mathbf b}\) has the same magnitude but opposite direction as \({\mathbf b}\).
\[ {\mathbf b} - {\mathbf b} = {\mathbf 0} \] where \({\bf 0}\) is the zero vector. It has magnitude zero and no direction.
4.1.2 Multiplication of a vector by a scalar
Note: This is NOT the same as the scalar product.
Multiplication of a vector by a scalar is distributive, associative, and commutative and results in a vector with the same direction but proportional magnitude.
A point P divides a line segment AB into the ratio \(\lambda:\mu\). Find the position vector of the point P.
It is helpful to draw this - please take a look in the Riley textbook to check your diagram. I drew my diagram such that the origin is at \(O\) and the tail of \({\bf a}\) and \({\bf b}\) originate at \(O\). The line segment \({\bf AB}\) then points from the tip of \({\bf a}\) at point \(A\) to the tip of \({\bf b}\) at point \(B\). The point \(P\) is somewhere along the line segment AB. There are two ways to get to \(P\) either \(O\) to \(A\) to \(P\) or \(O\) to \(B\) to \(P\). We will choose the former. \[ \begin{aligned} {\bf AB} &= {\bf b} - {\bf a}\\ {\bf OP} &= {\bf p} = {\bf a} + \frac{\lambda}{\lambda + \mu} {\bf AB}\\ &= {\bf a} + \frac{\lambda}{\lambda + \mu} ({\bf b} - {\bf a})\\ &= \frac{\mu}{\lambda + \mu} {\bf a} + \frac{\lambda}{\lambda + \mu} {\bf b} \end{aligned} \]
4.1.3 Basis vectors and components
Given any three different vectors \({\bf e_1}\), \({\bf e_2}\), \({\bf e_3}\) which do not lie in a plane it is possible in 3D space to write any vector in terms of them and scalar multipliers: \[ {\bf a} = a_1 {\bf e_1} + a_2 {\bf e_2} + a_3 {\bf e_3}. \] In this case \({\bf e_1}\), \({\bf e_2}\), \({\bf e_3}\) form a basis. The scalar multipliers \(a_1, a_2, a_3\) can have any value and are called components of \({\bf a}\).
We generally use basis vectors that are mutually perpendicular. The number of basis vectors is equal to the number of dimensions of the space and no one basis vector can be written as the linear combination of the other two. In Cartesian coordinates we will use the basis \({\hat {\imath}}, {\hat {\jmath}}, {\hat k}\).
\[ \begin{aligned} {\bf a} &= a_x \hat {\imath} + a_y \hat {\jmath} + a_z \hat k\\ \hat {\imath} &= (1, 0, 0)\\ \hat {\imath} &= (0, 1, 0)\\ \hat {\imath} &= (0, 0, 1)\\ {\bf a} + {\bf b} &= (a_x +b_x) \hat {\imath} + (a_y +b_y) \hat {\jmath} + (a_z +b_z) \hat k\\ {\bf a} - {\bf b} &= (a_x - b_x) \hat {\imath} + (a_y - b_y) \hat {\jmath} + (a_z - b_z) \hat k \end{aligned} \]
Two particles have velocities \(v_1 = \hat \imath + 3 \hat \jmath + 6 \hat k\) and \(v_2 = \hat \imath - 2 \hat k\). Find the velocity \({\mathbf u}\) of the second particle relative to the first.
4.2 Magnitude of a Vector
The magnitude of a vector can be written as \(\left| \mathbf{a} \right|\) or \(a\) in either case the magnitude equals the sqaure root of the sum of the squares of the vector components.
\[ \begin{aligned} \mathbf{a} = -3 \hat{\imath} - 8 \hat{k}\\ a = \sqrt{a_x^2 + a_y^2 +a_z^2}\\ a = \sqrt{73} \end{aligned} \]
The magnitude of a vector is the scalar quantity which for displacement is the “length” for force it is the “strength” for velocity it is the “speed”.
A vector whose magnitude equals unity is a unit vector: \[ \begin{aligned} \mathbf{\hat{a}} = \frac{\mathbf a}{\left| \mathbf{a} \right|}\\ \mathbf{a} = \lambda \mathbf{\hat{a}} \end{aligned} \] where \(\mathbf{a}\) has a magnitude \(\lambda\) and a direction \(\mathbf{\hat{a}}\).
4.3 Multiplication of Vectors
When we multiply vectors together there are two choices of operation: we can use a scalar product (dot product) or a vector product (cross product). The scalar product of two vectors produces a scalar (a number with no direction). A vector product of two vectors produces another vector.
4.3.1 Scalar product
A scalar product is written as \[ \mathbf{a} \cdot \mathbf{b} \equiv a\,b\,\cos\theta \quad 0\le\theta\le \pi \] where \(\theta\) is the angle between the two vectors placed tail to tail or head to head.
\(\mathbf{a} \cdot \mathbf{b}\) is the magnitude of \(\mathbf{a}\) multiplied by the projection of \(\mathbf{b}\) onto \(\mathbf{a}\). This means that if \(\mathbf{a} \cdot \mathbf{b} = 0\) \(\mathbf{a}\) is perpendicular to \(\mathbf{b}\).
\[ \begin{aligned} \hat{\imath} \cdot \hat{\imath} = \hat{\jmath} \cdot \hat{\jmath} = \hat{k} \cdot \hat{k} = 1\\ \hat{\imath} \cdot \hat{\jmath} = \hat{\imath} \cdot \hat{k} = \hat{\jmath} \cdot \hat{k} = 0 \end{aligned} \]
Work is a common example of scalar product in physics \(W = \mathbf{F} \cdot \mathbf{r}\). If the force is applied perpendicular to the displacement no work is done.
Properties of scalar product: \[ \begin{aligned} \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}\\ \mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} \end{aligned} \]
Non-coplanar points \(A, B, C, D\) are placed such that \(\mathbf{AD}\) is perpendicular to \(\mathbf{BC}\) and \(\mathbf{BD}\) is perpendicular to \(\mathbf{AC}\). Show that \(\mathbf{CD}\) is perpendicular to \(\mathbf{AB}\).
Given that the Cartesian basis vectors are perpendicular to each other \(\mathbf{a} \cdot \mathbf{b} = (a_x \hat{\imath} + a_y \hat{\jmath} + a_z\hat{k})\cdot(b_x \hat{\imath} + b_y \hat{\jmath} + b_z\hat{k}) = a_x\,b_x + a_y\,b_y + a_z\,b_z\).
Find the angle between \(\mathbf{a} = \hat{\imath} + 2\hat{\jmath} + 3\hat{k}\) and \(\mathbf{b} = 2\hat{\imath} + 3\hat{\jmath} + 4\hat{k}\).
4.3.2 Vector Product
The vector product or cross product is written as \(\mathbf{a} \times \mathbf{b}\) and produces a vector with magnitude \(\left| \mathbf{a} \times \mathbf{b} \right| = a \, b \, \sin \theta\) and a direction perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\). The direction is found by rotating \(\mathbf{a}\) into \(\mathbf{b}\) through the smallest angle (use right hand rule).
The vector product is distributive in addition but unlike the scalar product it is anti-commutative and non-associative. \[ \begin{aligned} (\mathbf{a} + \mathbf{b}) \times \mathbf{c} &= (\mathbf{a} \times \mathbf{b}) + (\mathbf{b} \times \mathbf{c})\\ \mathbf{b} \times \mathbf{a} &= -\mathbf{a} \times \mathbf{b} \\ (\mathbf{a} \times \mathbf{b}) \times \mathbf{c} &\neq \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \end{aligned} \]
If \(\mathbf{a} \times \mathbf{b} = \mathbf{0}\) then \(\mathbf{a}\) is either parallel or antiparallel to \(\mathbf{b}\) assuming neither is \(\mathbf{0}\). This means that the vector product of a vector with itself is the zero vector.
Show that if \(\mathbf{a} = \mathbf{b} + \lambda \mathbf{c}\) for some scalar \(\lambda\) then \(\mathbf{a}\times\mathbf{c} = \mathbf{b} \times \mathbf{c}\).
The cross product is used to find the angular momentum and torque in mechanics but the magnitude of the cross product can also be used to find the area of a parallelogram with sides \(\mathbf{a}\) and \(\mathbf{b}\).
Since the Cartesian basis vectors are all mutually perpendicular to each other the cross product of each of them with themselves is the zero vector.
One can find the resultant vector of the cross product by using the determinant method: \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} \]
The resultant vector will be perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\).
Find the area of the parallelogram with sides \(\mathbf{a} = \hat{\imath} + 2\hat{\jmath} + 3\hat{k}\) and \(\mathbf{b} = 3\hat{\imath} + 4\hat{\jmath} + 5\hat{k}\).