The numbers in brackets are supposed to be in subscript so V(2) would be V subscript 2 and the number in front is just multiply, all the variables are vectors so they have a arrow on top. Determine if, W(1) = 2V(1) + 3V(2) , W(2) = V(2) + 2V(3) and W(3) = -V(1) - 3V(3) are coplanar i dont understand how to do this. i tried to use the property that if they are coplanar therefore they also must be colinear therefore [Note: C is the only variable that is not a vector it is a unknown integer.] C(1)W(1) + C(2)W(2)....etc = 0 but i cant think of how that would resolve am i on the right track?
If they are coplanar then all three lie in the same plane. If they are collinear then all three lie in the same line. Do every three vectors that lie in the same plane have to lie along the same line? What you probably meant was linearly dependent. To show that they are linearly dependent, you have to show that your equation has a solution for the C's where the C's are not all 0. Just substitute in the definition of the W's and solve. I assume that the question asks whether the W's are coplanar for ALL choice of the V's, so you need to determine whether there is some choice of C's such that each of the V's in the equation cancels to 0.
But i cant solve it there are too many variables. i subbed in the values for W vectors and then i expanded it so now i have like C(1)V(1)+C(1)V(2) etc.. then i added up all the V(1) vectors and ended with sumthing like V(1)[C(1)+C(2)]=0 then that led me to [C(1)+C(2)]=0 cause C would be the number of V and for it to be zero they would have to cancel out. i did the same for all of the V's now i have a set of these i cant solve it.... there are to many variables
From what you've described there aren't too many variables. Assume that the V's are nonzero (i.e. sufficiently general--the problem is not very well defined as you've stated it, so I'm assuming it means for any V's). Then you have V(1)[C(1)+C(2)]=0 implies C(1) + C(2) = 0 Though actually the equation you should have for V(1) is V(1)[2C(1) - C(3)] = 0
Do you mean you don't know how to do 3 equations in 3 unknowns? You want to show that W_{1} = 2V_{1} + 3V_{2} , W_{2} = V_{2} + 2V_{3}, and W_{3} = -V_{1} - 3V_{3} are coplanar. Three vectors are coplanar if and only if they are dependent- that is if there exist some non-trivial ([tex]\alpha[/tex], [tex]\beta[/tex], [tex]\gamma[/tex] not all equal to 0) such that the linear combination [tex]\alpha W_1+ \beta W_2+ \gamma W_3= 0[/tex] [tex]\alpha(2V_1+ 3V_2)+ \beta(V_2+ 2V_3)+ \gamma(-V_1- 3V_3)= 0[/tex] [tex](2\alpha-\gamma)V_1+ (3\alpha+ \beta)V_2+ (3\beta- 3\gamma)V_3= 0[/tex] My first thought was that we would need to know more about V_{1}, V_{2}, and V_{3} but I don't think we do. If V_{1}, V_{2}, and V_{3} are themselves dependent, then they are coplanar and any linear combination of them is coplanar. So the only question is if V_{1}, V_{2}, and V_{3} are independent vectors. If V_{1}, V_{2}, and V_{3} are independent then in order to have [tex](2\alpha-\gamma)V_1+ (3\alpha+ \beta)V_2+ (3\beta- 3\gamma)V_3= 0[/tex] we must have [tex]2\alpha- \gamma= 0[/tex] [tex]3\alpha+ \beta= 0[/tex] [tex]3\beta- 3\gamma= 0[/tex] The three vectors W_{1}, W_{2}, and W_{3} are coplanar if and only if there exist solutions to those equations other than [tex]\alpha= \beta= \gamma= 0[/tex]