Question

1.For The Vector a b c and d show that :
(axb).(cxd)+(bxc).(axd)+(cxa).(bxd)=o
2.Find Two Unit vector Perpendicular to both
A=i-2j+3k
B= -2i+4j

Answer 1

Answer on Question #39975 – Math – Other

1. For The Vector $a$ and $b$ show that:

- $(axb).(cxd) + (bxc).(axd) + (cxa).(bxd) = 0$

2. Find Two Unit vector Perpendicular to both

$$A = i \cdot 2j + 3k$$ $$B = -2i + 4j$$

Soluiton:

1. Using property:

- $(axb).(cxd) = (axb) \cdot p = \{let p = cxd\} = a \cdot (bxp) = \{as in a scalar triple product the dot and the cross may be interchanged\} = a \cdot [bx(cxd)] = a \cdot [(b \cdot d)c - (b \cdot c)d] = (a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c) = \begin{vmatrix} a & c & a & d \\ b & c & b & d \end{vmatrix}$ we obtain

- $(axb).(cxd) + (bxc).(axd) + (cxa).(bxd) = (c \cdot b)(a \cdot d) - (c \cdot d)(a \cdot b) + (b \cdot a)(c \cdot d) - (b \cdot d)(c \cdot a) + (a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c) = 0$, that is answer.

2. Let find vector $c = A \times B$. $c$ is Perpendicular to both $A$ and $B$.

$$A \times B = \begin{pmatrix} i & j & k \\ 1 & -2 & 3 \\ -2 & 4 & 0 \end{pmatrix} = i \begin{pmatrix} -2 & 3 \\ 4 & 0 \end{pmatrix} - j \begin{pmatrix} 1 & 3 \\ -2 & 0 \end{pmatrix} + k \begin{pmatrix} 1 & -2 \\ -2 & 4 \end{pmatrix} = i(-12) - j(6) + k(4 - 4) = -12i - 6j.$$

So, vector $c = \{-12; -6\}$ - is perpendicular to the vectors $A$ and $B$.

Since we need Two Unit Perpendicular vectors, so let $c_1 = \{12; 6\}$.

Answer: 2. $\{-12; -6\}$, $\{12; 6\}$