Root vector

[nana]

maybe some one can help me in solving this root vector count.

Determined between two large vectors of equal area = F. When comparing between the magnitude and the magnitude of the two vectors is equal to the root 3, then the angle that forms the two vectors is?

Given a vector a = 3i-4j + pk and b = 2i + 2j-3k if the vector projection length a on b is 4 root 17 p value?

[Guest]

I don't know what is meant by the "area" of a vector so I don't know what it means to say that two vectors have the "equal area, F". Did you possibly mean that the two vectors have equal length and that the area between the two vectors is F?

And by "When comparing between the magnitude and the magnitude of the two vectors is equal to the root 3" do you mean comparing the magnitude of that area with the common magnitude of the two vectors?

If the two vectors have equal length then connecting their "tips" we get an isosceles triangle. Taking there common length to be "s", the angle between them to be [tex]\theta[/tex], and the length of the base to be "t", by the cosine law, [tex]t^2= s^2+ s^2- 2(s)(s)cos(\theta)= 2s^2(1- cos(\theta))[/tex]. Further, drawing the perpendicular from the vertex of the triangle to the opposite side divides it into two right triangles with hypotenuse "s" and one leg t/2; By the Pythagorean theorem, the height of the isosceles triangle, the other leg of that right triangle, is [tex]h= \sqrt{s^2+ t^2/4}[/tex] so the isosceles triangle has area [tex](1/2)(t)(\sqrt{s^2+ t^2/4})= F[/tex]. Solve the two equations, [tex]t^2= s^2+ s^2- 2(s)(s)cos(\theta)= 2s^2(1- cos(\theta))[/tex] and [tex](1/2)(t)(\sqrt{s^2+ t^2/4})= F[/tex], for s in terms of F and [tex]\theta[/tex].


For the second problem, the "vector projection of a on b" is given by the dot product of a with b, divided by the length of b. Here, a= 3i- 4j+ pk and b= 2i+ 2j- 3k so [tex]a\cdot b= 3(2)- 4(2)+ p(-3)= -2- 3p. The length of b is [tex]\sqrt{2(2)+ 2(2)+ (-3)(-3)}= \sqrt{4+ 4+ 9}= \sqrt{17}[/tex]. So the "vector projection of a on b" is [tex]\frac{-2- 3p}{\sqrt{17}}= 4\sqrt{17}[/tex]. Solve that equation for p.