Chapter 1: Q1.5P (page 8)

Prove the BAC-CAB rule by writing out both sides in component form.

Short Answer

Expert verified

The BAC-CAB rule, $A \times B \times C = B (A . C) - C (A . B)$ , is proven.

Step by step solution

Explain the concept and assume the vectors

To proveBAC-CAB rule, $A \times B \times C = B (A . C) - C (A . B)$ , firstly evaluate its left side,and right side saperately, then the results of both sides mst be compared.

Let the vectors are defined as $A = A_{x} i + A_{y} j + A_{z} k, B = B_{x} i + B_{y} j + B_{z} k$ , and $C = C_{x} i + C_{y} j + C_{z} k$

find the value of A×B×C

$B \times C = |\begin{matrix} i & j & k \\ B_{x} & B_{y} & B_{z} \\ C_{x} & C_{y} & C_{z} \end{matrix}| = (B_{y} C_{z} - B_{z} C_{y}) i - (C_{z} - B_{z} - C_{x}) j + (B_{x} C_{y} - B_{y} C_{x}) k$

Now use the above result to evaluate A x B x C as,

$\begin{array}{rcl} A \times B \times C & = & |\begin{matrix} i & j & k \\ A_{x} & A_{y} & A_{z} \\ B_{y} C_{z} - B_{z} C_{y} & B_{x} C_{z} - B_{z} C_{x} & B_{x} C_{y} - B_{y} C_{x} \end{matrix}| \\ = & (A_{y} B_{x} C_{y} - A_{y} B_{y} C_{x} - A_{z} B_{z} C_{x} + A_{z} B_{x} C_{z}) i \\ - (- A_{x} B_{x} C_{y} - A_{x} B_{y} C_{x} - A_{z} B_{y} C_{z} + A_{z} B_{z} C_{y}) j \\ + (A_{x} B_{z} C_{x} - A_{x} B_{x} C_{z} - A_{y} B_{y} C_{z} + A_{y} B_{z} C_{y}) k \end{array}$

Find the value of B (A - C) - C (A - B)

Substitute $A_{x} i + A_{y} j + A_{z} k$ for A, $B_{x} i + B_{y} j + B_{z} k$ for B, and $C_{x} i + C_{y} j + C_{z} k$ for C into $B (A \cdot C) - C (A \cdot B)$

$B (A \cdot C) - C (A \cdot B) = (B_{x} A_{y} C_{y} - C_{x} A_{y} B_{y} - C_{x} A_{z} B_{z} + B_{x} A_{z} C_{z}) i + (- C_{y} A_{x} B_{x} + B_{y} A_{x} C_{x} + B_{y} A_{z} C_{z} - C_{y} A_{z} B_{z}) j + (B_{z} A_{x} C_{x} - C_{z} A_{y} B_{y} - C_{z} A_{y} B_{y} + B_{z} A_{y} C_{y}) k$ …… (2)

Draw the conclusion

From equation (1) and (2) it can be concluded that the result of $B (A \cdot C) - C (A \cdot B)$ and $A \times B \times C$ is same. Thus $A \times B \times C = B (A \cdot C) - C (A \cdot B)$