• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q52E

Expert-verified
Found in: Page 566

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Prove the formula $$(a \times b) \cdot (c \times d) = \left| {\begin{array}{*{20}{c}}{a \cdot c}&{b \cdot c}\\{a \cdot d}&{b \cdot d}\end{array}} \right|$$.

The formula $$(a \times b) \cdot (c \times d) = \left| {\begin{array}{*{20}{l}}{a \cdot c}&{b \cdot c}\\{a \cdot d}&{b \cdot d}\end{array}} \right|$$ is proved.

See the step by step solution

## Step 1: Concept of vectors.

A vector is an object that has both a magnitude as well as a direction.

## Step 2: Write out $${\bf{a}} \times {\bf{b}}$$and $${\bf{c}} \times {\bf{d}}$$ in component form.

$${\bf{a}} \times {\bf{b}} = \left\langle {{a_2}{b_3} - {a_3}{b_2},{a_3}{b_1} - {a_1}{b_3},{a_1}{b_2} - {a_2}{b_1}} \right\rangle$$

$${\bf{c}} \times {\bf{d}} = \left\langle {{c_2}{d_3} - {c_3}{d_2},{c_3}{d_1} - {c_1}{d_3},{c_1}{d_2} - {c_2}{d_1}} \right\rangle$$

$$({\bf{a}} \times {\bf{b}}) \cdot ({\bf{c}} \times {\bf{d}})$$

$$= \left( {{a_2}{b_3} - {a_3}{b_2}} \right)\left( {{c_2}{d_3} - {c_3}{d_2}} \right) + \left( {{a_3}{b_1} - {a_1}{b_3}} \right)\left( {{c_3}{d_1} - {c_1}{d_3}} \right)$$

$$\quad + \left( {{a_1}{b_2} - {a_2}{b_1}} \right)\left( {{c_1}{d_2} - {c_2}{d_1}} \right)$$

$$\begin{array}{l} = {a_2}{b_3}{c_2}{d_3} - {a_2}{b_3}{c_3}{d_2} - {a_3}{b_2}{c_2}{d_3} + {a_3}{b_2}{c_3}{d_2}\\ + {a_3}{b_1}{c_3}{d_1} - {a_3}{b_1}{c_1}{d_3} - {a_1}{b_3}{c_3}{d_1} + {a_1}{b_3}{c_1}{d_3}\\ + {a_1}{b_2}{c_1}{d_2} - {a_1}{b_2}{c_2}{d_1} - {a_2}{b_1}{c_1}{d_2} + {a_2}{b_1}{c_2}{d_1}\end{array}$$

$$\begin{array}{l} = {a_1}{b_2}{c_1}{d_2} + {a_1}{b_3}{c_1}{d_3} + {a_2}{b_1}{c_2}{d_1} + {a_2}{b_3}{c_2}{d_3}\\ + {a_3}{b_1}{c_3}{d_1} + {a_3}{b_2}{c_3}{d_2} - {a_2}{b_1}{c_1}{d_2} - {a_3}{b_1}{c_1}{d_3}\\ - {a_1}{b_2}{c_2}{d_1} - {a_3}{b_2}{c_2}{d_3} - {a_1}{b_3}{c_3}{d_1} - {a_2}{b_3}{c_3}{d_2}\end{array}$$

## Step 3: Write the determinant of the right side of the equation in terms of components.

$$\left| {\begin{array}{*{20}{c}}{{\bf{a}} \cdot {\bf{c}}}&{{\bf{b}} \cdot {\bf{c}}}\\{{\bf{a}} \cdot {\bf{d}}}&{{\bf{b}} \cdot {\bf{d}}}\end{array}} \right| = ({\bf{a}} \cdot {\bf{c}})({\bf{b}} \cdot {\bf{d}}) - ({\bf{b}} \cdot {\bf{c}})({\bf{a}} \cdot {\bf{d}})$$

$$\begin{array}{l} = \left( {{a_1}{c_1} + {a_2}{c_2} + {a_3}{c_3}} \right)\left( {{b_1}{d_1} + {b_2}{d_2} + {b_3}{d_3}} \right)\\ - \left( {{b_1}{c_1} + {b_2}{c_2} + {b_3}{c_3}} \right)\left( {{a_1}{d_1} + {a_2}{d_2} + {a_3}{d_3}} \right)\end{array}$$

$$\begin{array}{l} = {a_1}{c_1}{b_1}{d_1} + {a_1}{c_1}{b_2}{d_2} + {a_1}{c_1}{b_3}{d_3}\\ + {a_2}{c_2}{b_1}{d_1} + {a_2}{c_2}{b_2}{d_2} + {a_2}{c_2}{b_3}{d_3}\\ + {a_3}{c_3}{b_1}{d_1} + {a_3}{c_3}{b_2}{d_2} + {a_3}{c_3}{b_3}{d_3}\\ - {b_1}{c_1}{a_1}{d_1} - {b_1}{c_1}{a_2}{d_2} - {b_1}{c_1}{a_3}{d_3}\\ - {b_2}{c_2}{a_1}{d_1} - {b_2}{c_2}{a_2}{d_2} - {b_2}{c_2}{a_3}{d_3}\\ - {b_3}{c_3}{a_1}{d_1} - {b_3}{c_3}{a_2}{d_2} - {b_3}{c_3}{a_3}{d_3}\end{array}$$

Sort into “abcd”

$$\begin{array}{l} = {a_1}{b_1}{c_1}{d_1} + {a_1}{b_2}{c_1}{d_2} + {a_1}{b_3}{c_1}{d_3}\\ + {a_2}{b_1}{c_2}{d_1} + {a_2}{b_2}{c_2}{d_2} + {a_2}{b_3}{c_2}{d_3}\\ + {a_3}{b_1}{c_3}{d_1} + {a_3}{b_2}{c_3}{d_2} + {a_3}{b_3}{c_3}{d_3}\\ - {a_1}{b_1}{d_1}{c_1} - {a_2}{b_1}{c_1}{d_2} - {a_3}{b_1}{c_1}{d_3}\\ - {a_1}{b_2}{c_2}{d_1} - {a_2}{b_2}{c_2}{d_2} - {a_3}{b_2}{c_2}{d_3}\\ - {a_1}{b_3}{c_3}{d_1} - {a_2}{b_3}{c_3}{d_2} - {a_3}{b_3}{c_3}{d_3}\end{array}$$

only the "1111", "2222", "3333" terms cancel out.

$$\begin{array}{l} = {a_1}{b_2}{c_1}{d_2} + {a_1}{b_3}{c_1}{d_3} + {a_2}{b_1}{c_2}{d_1}\\ + {a_2}{b_3}{c_2}{d_3} + {a_3}{b_1}{c_3}{d_1} + {a_3}{b_2}{c_3}{d_2}\\ - {a_2}{b_1}{c_1}{d_2} - {a_3}{b_1}{c_1}{d_3} - {a_1}{b_2}{c_2}{d_1}\\ - {a_3}{b_2}{c_2}{d_3} - {a_1}{b_3}{c_3}{d_1} - {a_2}{b_3}{c_3}{d_2}\end{array}$$

## Step 4: Write the determinant of the right side of the equation in terms of components.

$$({\bf{a}} \times {\bf{b}}) \cdot ({\bf{c}} \times {\bf{d}}) = \left| {\begin{array}{*{20}{c}}{{\bf{a}} \cdot {\bf{c}}}&{{\bf{b}} \cdot {\bf{c}}}\\{{\bf{a}} \cdot {\bf{d}}}&{{\bf{b}} \cdot {\bf{d}}}\end{array}} \right|$$

So, it’s true.

Hence, the elements of the determinant are just scalars

$$\left| {\begin{array}{*{20}{l}}{{\bf{a}} \cdot {\bf{c}}}&{{\bf{b}} \cdot {\bf{c}}}\\{{\bf{a}} \cdot {\bf{d}}}&{{\bf{b}} \cdot {\bf{d}}}\end{array}} \right| = ({\bf{a}} \cdot {\bf{c}})({\bf{b}} \cdot {\bf{d}}) - ({\bf{b}} \cdot {\bf{c}})({\bf{a}} \cdot {\bf{d}})$$