Suggested languages for you:

Americas

Europe

Q22E

Expert-verified
Found in: Page 565

### Essential Calculus: Early Transcendentals

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

# Show the $$(a \times b) \cdot b = 0$$ for all vectors $${\bf{a}}$$ and $${\bf{b}}$$ in $${V_3}$$.

The condition $$({\rm{a}} \times {\rm{b}}) \cdot {\rm{b}} = 0$$ is shown for all vectors $${\bf{a}}$$ and $${\bf{b}}$$ in $${V_3}$$.

See the step by step solution

## Step 1: Formula used

Consider, $${\rm{a}} = \left\langle {{a_1},{a_2},{a_3}} \right\rangle$$ and $${\rm{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle$$.

## Step 2: Find the cross product $$({\bf{a}} \times {\bf{b}}) \cdot {\bf{b}}$$

If $${\bf{a}} = \left\langle {{a_1},{a_2},{a_3}} \right\rangle ,{\bf{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle$$ and $${\bf{c}} = \left\langle {{c_1},{c_2},{c_3}} \right\rangle$$, then

$$({\bf{a}} \times {\bf{b}}) \cdot {\bf{c}} = \left| {\begin{array}{*{20}{l}}{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|$$

Therefore, if $${\bf{a}} = \left\langle {{a_1},{a_2},{a_3}} \right\rangle ,{\bf{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle$$, we can write

$$({\bf{a}} \times {\bf{b}}) \cdot {\bf{b}} = \left| {\begin{array}{*{20}{l}}{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right|$$

Subtract Row 2 from Row 3

$$({\bf{a}} \times {\bf{b}}) \cdot {\bf{b}} = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\\0&0&0\end{array}} \right|$$

If we expand the determinant along the bottom row, we will get

$$\begin{array}{l}({\bf{a}} \times {\bf{b}}) \cdot {\bf{b}} = {( - 1)^{3 + 1}} \cdot 0 \cdot \left| {\begin{array}{*{20}{l}}{{a_2}}&{{a_3}}\\{{b_2}}&{{b_3}}\end{array}} \right| + {( - 1)^{3 + 2}} \cdot 0 \cdot \left| {\begin{array}{*{20}{c}}{{a_1}}&{{a_3}}\\{{b_1}}&{{b_3}}\end{array}} \right| + {( - 1)^{3 + 3}} \cdot 0 \cdot \left| {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}\\{{b_1}}&{{b_2}}\end{array}} \right|\\({\bf{a}} \times {\bf{b}}) \cdot {\bf{b}} = 0 + 0 + 0\\({\bf{a}} \times {\bf{b}}) \cdot {\bf{b}} = 0\end{array}$$

Hence, the condition $$({\rm{a}} \times {\rm{b}}) \cdot {\rm{b}} = 0$$ is shown for all vectors $${\bf{a}}$$ and $${\bf{b}}$$ in $${V_3}$$.