## anonymous one year ago ques

1. 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

2. 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

3. anonymous

|dw:1438410330020:dw|

4. imqwerty

take 'a' common from 2nd row and 'b' common from 3rd row nd then try to solve :)

5. 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

6. anonymous

damn im really bad at drawing

7. imqwerty

lol

8. anonymous

|dw:1438410602789:dw|

9. Michele_Laino

hint: if I develop the determinant along the first row, I get this: $\Large \vec a\left\{ {\left( {\vec a \cdot \vec b} \right)\left( {\vec b \cdot \vec c} \right) - \left( {\vec a \cdot \vec c} \right)\left( {\vec b \cdot \vec b} \right)} \right\} - \vec b\left\{ {...} \right\} + \vec c\left\{ {...} \right\}$

10. imqwerty

11. 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.

12. anonymous

(AEI+BFG+CDH)-(CEG+AFH+BDI)|dw:1438410761644:dw|

13. anonymous

can I arrange $(\vec a.\vec c)(\vec b . \vec b)=(\vec a . \vec b)(\vec b . \vec c)$ ??

14. 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

15. 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 $\left[\begin{matrix}\vec a .(\vec b \times \vec c) & \vec b .(\vec c \times \vec a) & \vec c.(\vec a \times \vec b) \\ \vec a .(\vec a .(\vec b \times \vec c)) & \vec a .(\vec b .(\vec c \times \vec a)) & \vec a .(\vec c . (\vec a \times \vec b)) \\ \vec b.(\vec a .(\vec b \times \vec c)) & \vec b .(\vec b .(\vec c \times \vec a)) & \vec b.(\vec c.(\vec a \times \vec b))\end{matrix}\right]$

16. anonymous

becomes 0 ??