anonymous
  • anonymous
ques
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
How can I solve the following determinant??Do I use dot product or cross product ?? \[\left[\begin{matrix}\vec a & \vec b & \vec c \\ \vec a . \vec a & \vec a . \vec b & \vec a . \vec c \\ \vec b . \vec a & \vec b . \vec b & \vec b . \vec c\end{matrix}\right]\] I have made a matrix but I mean to find the determinant, there's no tool for determinant in the equation box I did something like taking a and b out from the 2nd and 3rd row and it becomes 0
Michele_Laino
  • Michele_Laino
hint: I think that you have to develop the determinant along the first row of your matrix, what you should get is a vector
anonymous
  • anonymous
|dw:1438410330020:dw|

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imqwerty
  • imqwerty
take 'a' common from 2nd row and 'b' common from 3rd row nd then try to solve :)
anonymous
  • anonymous
I was thinking of expanding but if I expand would I use dot product or cross poduct??I have no idea....it is also given that\[\vec a\] \[\vec b\] \[\vec c\] are coplanar
anonymous
  • anonymous
damn im really bad at drawing
imqwerty
  • imqwerty
lol
anonymous
  • anonymous
|dw:1438410602789:dw|
Michele_Laino
  • Michele_Laino
hint: if I develop the determinant along the first row, I get this: \[\Large \vec a\left\{ {\left( {\vec a \cdot \vec b} \right)\left( {\vec b \cdot \vec c} \right) - \left( {\vec a \cdot \vec c} \right)\left( {\vec b \cdot \vec b} \right)} \right\} - \vec b\left\{ {...} \right\} + \vec c\left\{ {...} \right\}\]
imqwerty
  • imqwerty
is the answer 0
Astrophysics
  • Astrophysics
I don't really see what the problem is, is it just knowing how to do a determinant or something else? When you do take a cross product you are essentially taking the determinant.
anonymous
  • anonymous
(AEI+BFG+CDH)-(CEG+AFH+BDI)|dw:1438410761644:dw|
anonymous
  • anonymous
can I arrange \[(\vec a.\vec c)(\vec b . \vec b)=(\vec a . \vec b)(\vec b . \vec c)\] ??
anonymous
  • anonymous
How can I solve \[(\vec a . \vec b)(\vec b . \vec c)-(\vec a . \vec c)(\vec b . \vec b)\] If I take the components it will take too long
anonymous
  • anonymous
I think I solved it, since a b and c are coplanar, dotting 1st column with b cross c, second column with c cross a and third column with a cross b we get \[\left[\begin{matrix}\vec a .(\vec b \times \vec c) & \vec b .(\vec c \times \vec a) & \vec c.(\vec a \times \vec b) \\ \vec a .(\vec a .(\vec b \times \vec c)) & \vec a .(\vec b .(\vec c \times \vec a)) & \vec a .(\vec c . (\vec a \times \vec b)) \\ \vec b.(\vec a .(\vec b \times \vec c)) & \vec b .(\vec b .(\vec c \times \vec a)) & \vec b.(\vec c.(\vec a \times \vec b))\end{matrix}\right]\]
anonymous
  • anonymous
becomes 0 ??

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