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(-1,-1,3) x (-1,0,2)

Guest Mar 22, 2015

Best Answer 

 #1
avatar+91479 
+10

(-1,-1,3) x (-1,0,2)

 

$$\\\begin{pmatrix}
-1&-1&3 \\
\end{pmatrix}
&\times \begin{pmatrix}
-1& \\
0&\\
2&
\end{pmatrix} \\\\
=(-1*-1)+(-1*0)+(3*2))\\=1+0+6\\=7$$

Melody  Mar 22, 2015
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4+0 Answers

 #1
avatar+91479 
+10
Best Answer

(-1,-1,3) x (-1,0,2)

 

$$\\\begin{pmatrix}
-1&-1&3 \\
\end{pmatrix}
&\times \begin{pmatrix}
-1& \\
0&\\
2&
\end{pmatrix} \\\\
=(-1*-1)+(-1*0)+(3*2))\\=1+0+6\\=7$$

Melody  Mar 22, 2015
 #2
avatar+81063 
+5

 

 

Melody has given the dot product - a scalar - of these two vectors...

However....if the cross-product was wanted, we have

(-1,-1,3) x (-1,0,2) = (u1, u2, u3) x (v1, v2, v3)  = (s1, s2, s3)

 

\begin{align} s_1 &= u_2v_3-u_3v_2\\ s_2 &= u_3v_1-u_1v_3\\ s_3 &= u_1v_2-u_2v_1 \end{align}

 

s1 = [-1*2] - [3*0] = -2 - 0  = -2

s2 = [3*-1] - [-1*2]  =  -3 - (-2)  = -3 + 2 = -1

s3 = [-1*0] - [-1* -1 ] = 0 - 1  = -1

The cross-product produces a vector in three-space that is perpendicular to the given vectors, namely..... (-2, -1, -1)

 

   

CPhill  Mar 22, 2015
 #3
avatar+26406 
+5

Here's a picture of the vectors for the cross-product:

 cross product

.

Alan  Mar 22, 2015
 #4
avatar+81063 
0

Thanks, Alan.....

 

  

CPhill  Mar 22, 2015

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