• Dec 5th 2006, 08:52 AM
Jenny20
question 1
The nonzero vector U and V are orthogonal if and only if U x V = 0 .

question 2
U x V = 0 if and only if U and V are orthogonal.

Can you explain to me why question 1 is false ?
I know question 2 is true because it is the definition of orthogonal.

Thank you very much.
• Dec 5th 2006, 09:03 AM
ThePerfectHacker
Quote:

Originally Posted by Jenny20
question 1
The nonzero vector U and V are orthogonal if and only if U x V = 0 .

question 2
U x V = 0 if and only if U and V are orthogonal.

Can you explain to me why question 1 is false ?
I know question 2 is true because it is the definition of orthogonal.

Thank you very much.

False, False, False.
Two vectors are othrogonal if and only if their dot product is zero.
• Dec 5th 2006, 10:47 AM
Soroban
Hello, Jenny!

Quote:

1) The nonzero vectors $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ are orthogonal if and only if $\displaystyle \vec{u} \times \vec{v}\:=\:0$

The answer is false . . . Right!

Two vectors are orthogonal if their dot product is zero: .$\displaystyle \vec{u}\cdot\vec{v}\,=\,0$

Quote:

2) $\displaystyle \vec{u} \times \vec{v}\:=\:0$ if and only if $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ are orthogonal.
.
Strange . . . This is identical to question 1.

$\displaystyle \vec{u} \times \vec{v}\,=\,0 \;\;\longleftrightarrow\;\;\vec{u} \parallel \vec{v}$

The cross product is zero if and only if the vectors are parallel.

• Dec 5th 2006, 02:28 PM
Jay Gatsby
Question 2 is not identical to Question 1. The assumption that the vectors are nonzero has been dropped.

It's still false though for essentially the same reasons that Question 1 is false.
• Dec 5th 2006, 02:43 PM
Plato
$\displaystyle \begin{array}{l} I: < 1,0,0 > \quad \& \quad J: < 0,1,0 > \\ I \cdot J = 0\quad \& \quad I \times J = < 0,0,1 > \\ \end{array}$
The above shows that #1 is false.

$\displaystyle \begin{array}{l} A: < 2,0,0 > \quad \& \quad B: < 1,0,0 > \\ A \times B = < 0,0,0 > \quad \& \quad A \cdot B \not= 0 \\ \end{array}.$
The above shows that #2 is false.
• Dec 5th 2006, 07:39 PM
Jenny20
I see.