Question

(a) Determine the vector \({{\rm{k}}_i}\) is perpendicular to \({{\rm{v}}_j}\) except at \(i = j\).

(b) Determine the dot product \({{\rm{k}}_i} \cdot {{\rm{v}}_i} = 1\).

(c) Determine the condition \({{\rm{k}}_1} \cdot \left( {{{\rm{k}}_2} \times {{\rm{k}}_3}} \right) = \frac{1}{{{{\rm{v}}_1} \cdot \left( {{{\rm{v}}_2} \times {{\rm{v}}_3}} \right)}}\).

Short answer

(a) The vector \({{\rm{k}}_i}\) is perpendicular to \({{\rm{v}}_j}\) except at \(i = j\)is shown.

(b) The dot product \({{\rm{k}}_i} \cdot {{\rm{v}}_i} = 1\)is shown.

(c) The condition \({{\rm{k}}_1} \cdot \left( {{{\rm{k}}_2} \times {{\rm{k}}_3}} \right) = \frac{1}{{{{\rm{v}}_1} \cdot \left( {{{\rm{v}}_2} \times {{\rm{v}}_3}} \right)}}\) is shown.

Step-by-step solution

Use the three expressions and perform dot multiplication.

(a)

Using the three given expressions for \({{\bf{k}}_1},{{\bf{k}}_2},{{\bf{k}}_3}\)

\({{\bf{k}}_i} = \frac{{{{\bf{v}}_j} \times {{\bf{v}}_k}}}{{{{\bf{v}}_i} \cdot \left( {{{\bf{v}}_j} \times {{\bf{v}}_k}} \right)}}\)

Dot multiplication with \({{\bf{v}}_j}\)

\({{\bf{v}}_j} \cdot {{\bf{k}}_i} = \frac{{{{\bf{v}}_j} \cdot \left( {{{\bf{v}}_j} \times {{\bf{v}}_k}} \right)}}{{{{\bf{v}}_i} \cdot \left( {{{\bf{v}}_j} \times {{\bf{v}}_k}} \right)}}\)

\({{\bf{v}}_j} \cdot {{\bf{k}}_i} = \frac{{\left( {{{\bf{v}}_j} \times {{\bf{v}}_j}} \right) \cdot {{\bf{v}}_k}}}{{{{\bf{v}}_i} \cdot \left( {{{\bf{v}}_j} \times {{\bf{v}}_k}} \right)}}\)

The cross product of parallel vectors is\(0\).

\(\begin{array}{l}{{\bf{v}}_j} \cdot {{\bf{k}}_i} = \frac{{{\bf{0}} \cdot {{\bf{v}}_k}}}{{{{\bf{v}}_i} \cdot \left( {{{\bf{v}}_j} \times {{\bf{v}}_k}} \right)}}\\{{\bf{v}}_j} \cdot {{\bf{k}}_i} = \frac{0}{{{{\bf{v}}_i} \cdot \left( {{{\bf{v}}_j} \times {{\bf{v}}_k}} \right)}}\\{{\bf{v}}_j} \cdot {{\bf{k}}_i} = {\bf{0}}\end{array}\)

Since two vectors have a zero-dot product only when they are perpendicular, we can conclude that \({{\bf{v}}_j}\) and \({{\bf{k}}_i}\) are perpendicular.

Hence proved

Use the three expressions and perform dot multiplication.

(b)

Using the three given expressions for \({{\bf{k}}_1},{{\bf{k}}_2},{{\bf{k}}_3}\)

\({{\bf{k}}_i} = \frac{{{{\bf{v}}_j} \times {{\bf{v}}_k}}}{{{{\bf{v}}_i} \cdot \left( {{{\bf{v}}_j} \times {{\bf{v}}_k}} \right)}}\)

Perform dot multiplication with\({{\bf{v}}_i}\)

\({{\bf{k}}_i} \cdot {{\bf{v}}_i} = \frac{{\left( {{{\bf{v}}_j} \times {{\bf{v}}_k}} \right) \cdot {{\bf{v}}_i}}}{{{{\bf{v}}_i} \cdot \left( {{{\bf{v}}_j} \times {{\bf{v}}_k}} \right)}}\)

\({\bf{u}} \cdot {\bf{v}} = {\bf{v}} \cdot {\bf{u}}\)

\({{\bf{k}}_i} \cdot {{\bf{v}}_i} = \frac{{{{\bf{v}}_i} \cdot \left( {{{\bf{v}}_j} \times {{\bf{v}}_k}} \right)}}{{{{\bf{v}}_i} \cdot \left( {{{\bf{v}}_j} \times {{\bf{v}}_k}} \right)}} = 1\)

Hence proved

Concept of property 6 theorem 11.

(c)

It states that\({\bf{a}} \times ({\bf{b}} \times {\bf{c}}) = ({\bf{a}} \cdot {\bf{c}}){\bf{b}} - ({\bf{a}} \cdot {\bf{b}}){\bf{c}}\).

Use the three expressions and simplify.

\({{\bf{k}}_2} \times {{\bf{k}}_3} = {{\bf{k}}_2} \times \left( {\frac{{{{\bf{v}}_1} \times {{\bf{v}}_2}}}{{{{\bf{v}}_1} \cdot \left( {{{\bf{v}}_2} \times {{\bf{v}}_3}} \right)}}} \right)\)

Note that\({{\bf{v}}_1} \cdot \left( {{{\bf{v}}_2} \times {{\bf{v}}_3}} \right)\)is a scalar

It is because dot product of any two vectors is a scalar

\(\begin{array}{l}{{\bf{k}}_2} \times {{\bf{k}}_3} = \frac{{{{\bf{k}}_2} \times \left( {{{\bf{v}}_1} \times {{\bf{v}}_2}} \right)}}{{{{\bf{v}}_1} \cdot \left( {{{\bf{v}}_2} \times {{\bf{v}}_3}} \right)}}\\{{\bf{k}}_2} \times {{\bf{k}}_3} = \frac{{\left( {{{\bf{k}}_2} \cdot {{\bf{v}}_2}} \right){{\bf{v}}_1} - \left( {{{\bf{k}}_2} \cdot {{\bf{v}}_1}} \right){{\bf{v}}_2}}}{{{{\bf{v}}_1} \cdot \left( {{{\bf{v}}_2} \times {{\bf{v}}_3}} \right)}}\end{array}\)

From part(b), we know that\({{\bf{k}}_i} \cdot {{\bf{v}}_i} = 1\)

From part(a), we know that\({{\bf{k}}_i} \cdot {{\bf{v}}_j} = 0\)if\(i \ne j\)

Therefore,

\(\begin{array}{l}{{\bf{k}}_2} \times {{\bf{k}}_3} = \frac{{(1){{\bf{v}}_1} - (0){{\bf{v}}_2}}}{{{{\bf{v}}_1} \cdot \left( {{{\bf{v}}_2} \times {{\bf{v}}_3}} \right)}}\\{{\bf{k}}_2} \times {{\bf{k}}_3} = \frac{{{{\bf{v}}_1}}}{{{{\bf{v}}_1} \cdot \left( {{{\bf{v}}_2} \times {{\bf{v}}_3}} \right)}}\end{array}\)

Therefore,

\(\begin{array}{l}{{\bf{k}}_1} \cdot \left( {{{\bf{k}}_2} \times {{\bf{k}}_3}} \right) = {{\bf{k}}_1} \cdot \left( {\frac{{{{\bf{v}}_1}}}{{{{\bf{v}}_1} \cdot \left( {{{\bf{v}}_2} \times {{\bf{v}}_3}} \right)}}} \right)\\{{\bf{k}}_1} \cdot \left( {{{\bf{k}}_2} \times {{\bf{k}}_3}} \right) = \frac{{{{\bf{k}}_1} \cdot {{\bf{v}}_1}}}{{{{\bf{v}}_1} \cdot \left( {{{\bf{v}}_2} \times {{\bf{v}}_3}} \right)}}\end{array}\)

From part(b), we know that\({{\bf{k}}_i} \cdot {{\bf{v}}_i} = 1\)

\({{\bf{k}}_1} \cdot \left( {{{\bf{k}}_2} \times {{\bf{k}}_3}} \right) = \frac{1}{{{{\bf{v}}_1} \cdot \left( {{{\bf{v}}_2} \times {{\bf{v}}_3}} \right)}}\)