Vector Dot Problem

**parkhid** · October 18th 2010, 10:04 AM

-----
hi .
please help to solve : Vector Dot Problem

if we know that $\vec{A}.\vec{B_1} = \vec{A}.\vec{B_2}$ can we infer from that : $\vec{B_1} = \vec{B_2}$ ?

thanks in advance ...

**Unknown008** · October 18th 2010, 10:14 AM

No, I don't think so...

$\left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right). \left(\begin{array}{c} 2 \\ 1 \\ 2 \end{array} \right) = \left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array} \right) . \left(\begin{array}{c} 4 \\ 0 \\ 2 \end{array} \right)$

But

$\left(\begin{array}{c} 2 \\ 1 \\ 2 \end{array} \right) \neq \left(\begin{array}{c} 4 \\ 0 \\ 2 \end{array} \right)$

EDIT: I'm not sure of that... yet.

EDIT2: Confirmed!

**NOX Andrew** · October 18th 2010, 10:15 AM

No, assuming $\vec{A} \neq \vec{0}$ .

Let $\vec{A} = <a_1,a_2>$ , $\vec{B_1} = <u_1,u_2>$ , and $\vec{B_2} = <v_1,v_2>$ . Then, $\vec{A} \cdot \vec{B_1} = a_1u_1 + a_2u_2$ and $\vec{A} \cdot \vec{B_2} = a_1v_1 + a_2v_2$ .

If $\vec{A} \cdot \vec{B_1} = \vec{A} \cdot \vec{B_2}$ , then $a_1u_1 + a_2u_2 = a_1v_1 + a_2v_2$ . The preceding equation cannot be solved; therefore, the answer is no.

I found a disproof on Wikipedia: Dot product - Wikipedia, the free encyclopedia

**Unknown008** · October 18th 2010, 10:19 AM

Hm... maybe I'm wrong... or is my example an exception?

**NOX Andrew** · October 18th 2010, 10:25 AM

No, you are correct. My reasoning was flawed. Here is a disproof on Wikipedia: Dot product - Wikipedia, the free encyclopedia

Thread: Vector Dot Problem

LinkBack

Thread Tools

Display

Vector Dot Problem

Similar Math Help Forum Discussions

vector problem

[SOLVED] Vector problem

vector problem

vector problem

Vector problem