26.8 The Vector Product

The scalar product of the vector \(\underline {a}\) and \(\underline {b}\) is \(\, \underline {a}\cdot \underline {b} = \left |\underline {a}\right | \left |\underline {b}\right |\cos \theta \), where \(\theta \) is the angle between the vectors \(\underline {a}\) and \(\underline {b}\).

The vectors (or cross) product of the vectors \(\underline {a}\) and \(\underline {b}\) is defined as \[\underline {a}\times \underline {b} = \left |\underline {a}\right | \left |\underline {b}\right |\sin \theta \,(\hat {\textbf {n}})\]

where \(\theta \) is the angle between \(\underline {a}\) and \(\underline {b}\) and \(\hat {\textbf {n}}\) is a unit vector perpendicular to both \(\underline {a}\) and \(\underline {b}\). The directions of \(\hat {\textbf {n}}\) is that in which a right handed cork screws would move when turned from \(\underline {a}\) to \(\underline {b}\).

If the turn is in the opposite sense, that is from \(\underline {b}\) to \(\underline {a}\), the corkscrew moves in the direction of \(-\hat {\textbf {n}}\).

\begin {align*} \text {So}\quad \underline {b}\times \underline {a} & = \left |\underline {b}\right | \left |\underline {a}\right |\sin \theta \,(-\hat {\textbf {n}}) \\ & = - \left |\underline {a}\right | \left |\underline {b}\right |\sin \theta \,(\hat {\textbf {n}})\\\\ \implies \quad \underline {b}\times \underline {a} & = - \underline {a}\times \underline {b} \\ \end {align*}

So the cross product is not commutative: in general \(\underline {a}\times \underline {b}\neq \underline {b}\times \underline {a}\). It is anti-commutative, the two products being equal in magnitude and opposite in direction.

By definition, \begin {align*} \textbf {i} \times \textbf {j} & = -\left (\textbf {j} \times \textbf {i}\right ) = \textbf {k}\\ \textbf {j} \times \textbf {k} & = -\left (\textbf {k} \times \textbf {j}\right ) = \textbf {i}\\ \textbf {k} \times \textbf {i} & = -\left (\textbf {i} \times \textbf {k}\right ) = \textbf {j} \end {align*}

\[\textbf {i} \times \textbf {i} = \textbf {j} \times \textbf {j} = \textbf {k} \times \textbf {k} = \textbf {0}\]

Let \(\, \underline {a} = a_1 \textbf {i} + a_2\textbf {j} + a_3\textbf {k}\,\) and \(\, \underline {b} = b_1 \textbf {i} + b_2\textbf {j} + b_3\textbf {k}\,\)

\begin {align*} \underline {a}\times \underline {b} & = \left ( a_1 \textbf {i} + a_2\textbf {j} + a_3\textbf {k}\right )\times \left (b_1 \textbf {i} + b_2\textbf {j} + b_3\textbf {k}\right )\\ & = a_1b_2\textbf {k} + a_1b_3\left (-\textbf {j}\right ) + a_2b_1\left (-\textbf {k}\right ) + a_2b_3\textbf {i} + a_3b_1\textbf {j} + a_3b_2\left (-\textbf {i} \right )\\ & = a_1b_2\textbf {k} - a_1b_3\textbf {j} - a_2b_1\textbf {k} + a_2b_3\textbf {i} + a_3b_1\textbf {j} - a_3b_2\textbf {i}\\ & = \left (a_2b_3 - a_3b_2\right )\textbf {i} - \left (a_3b_1 - a_1b_3\right )\textbf {j} + \left (a_1b_2 - a_2b_1\right )\textbf {k} \end {align*}

\begin {align*} \underline {a}\times \underline {b} & = \begin {vmatrix} \textbf {i} & \textbf {j} & \textbf {k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ \end {vmatrix}\\\\ & = \textbf {i}\begin {vmatrix} a_2 & a_3\\ b_2 & b_3\\ \end {vmatrix} - \textbf {j}\begin {vmatrix} a_1 & a_3\\ b_1 & b_3\\ \end {vmatrix} + \textbf {k}\begin {vmatrix} a_1 & a_2\\ b_1 & b_2\\ \end {vmatrix}\\ \end {align*}

So,

1.

\(\,\underline {a} \times \underline {b} = |\underline {a}||\underline {b}|\,\sin \theta \left (\hat {\textbf {n}}\right )\quad \) formula

2.

\(\,\underline {a} \times \underline {b} = \left ( a_1 \textbf {i} + a_2\textbf {j} + a_3\textbf {k}\right )\times \left (b_1 \textbf {i} + b_2\textbf {j} + b_3\textbf {k}\right )\quad \) Directly

\(\underline {a}\times \underline {b} = \begin {vmatrix} \textbf {i} & \textbf {j} & \textbf {k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ \end {vmatrix}\quad \) Appropriate determinate

 

\(*\quad \) If \(\,\underline {a}\times \underline {b} = 0,\,\) then \(\,\underline {a}\times \underline {b} = \left |\underline {a}\right | \left |\underline {b}\right |\sin \theta \,(\hat {\textbf {n}})\,\) either \(\, \left |\underline {a}\right | = 0 \implies \underline {a}= \underline {0}\,\) or \(\, \left |\underline {b}\right | = 0\implies \underline {b} = \underline {0}\)

\[\sin \theta = 0 \quad \implies \quad \theta = 0, \quad \pi \]

Then \(\underline {a}\) and \(\underline {b}\) are in the same direction or in the same sense or opposite sense. If \(\sin \theta = 0\) then \(\underline {a}\) and \(\underline {b}\) are parallel.

Questions

1.

(a)

Given the vectors \(\quad \underline {u} = 2\alpha \textbf {i} + 4\textbf {j} - 3\textbf {k}\,\) and \(\quad \underline {v} = 4\textbf {i} + \beta \textbf {j} + 3\textbf {k}\,\), find

i.

the values of \(\alpha \) and \(\beta \) if the vectors are parallel

ii.

a relation between \(\alpha \) and \(\beta \) if the vectors are perpendicular.

(b)

Find the area of the parallelogram \(ABCD\) given that its vertices are \(\, A(2,1,-1),\\ \quad B(1,-7,3),\quad C(-2,5,1)\,\) and \(\, D(-1, 13, -3)\).

2.

Given that \(\quad \underline {a} = 4\textbf {i} - 3\textbf {j} - \textbf {k}\,m\), \(\quad \underline {b} = \textbf {i} + 2\textbf {j} + 3\textbf {k}\,\) and \(\quad \underline {c} = -5\textbf {i} - 3\textbf {j} + 5\textbf {k}\,\).

Find the following:

(a)

\(\, 2\underline {a}\cdot \underline {b}\)

(b)

\(\, \underline {a}\times 3\underline {b}\)

(c)

\(\, \underline {a}\cdot \left (\underline {b}\times \underline {c}\right )\)

(d)

Find the area of the triangle \(ABC\) whose vertices are \(\, A(2,-3,4),\quad B(0,1,2)\,\) and \(\, C(-1,2,0)\)

3.

(a)

If the point \(P\) has position vector \(\quad 2\textbf {i} - 4\textbf {j} + 5\textbf {k}\,\) and \(\quad \overrightarrow {PQ} = 3\textbf {i} + 6\textbf {j} - 2\textbf {k},\,\) find the position vector of the point \(Q\).

(b)

Find the values of \(a\) and \(b\) for which the vector \(\quad 2\textbf {i} + \textbf {j} + 3\textbf {k}\,\) is perpendicular to both \(\, 3\textbf {i}- b\textbf {k}\quad \) and \(\quad a\textbf {i} + 2 \textbf {j} + 2b\textbf {k}\,\).

(c)

Given that \(\quad \underline {c} = 3\textbf {i} - \textbf {j} + 2\textbf {k}\,\), \(\quad \underline {d} = -4\textbf {i} + 2\textbf {k}\,\) and \(\quad \underline {e} = \textbf {i} - \textbf {j} - 2\textbf {k},\,\) find \(\,\underline {c}\cdot \left ( \underline {d}\times \underline {e}\right )\).

(d)

Given that \(\quad \underline {a} = i\textbf {i} + 2\textbf {j} - 3\textbf {k}\,\) and \(\quad \underline {a} = -2\textbf {i} - \textbf {j} + 3,\textbf {k}\,\) find \(\cot \theta ,\,\) where \(\theta \) is angle between \(\underline {a}\) and \(\underline {b}\)

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