a = 3i + 2j b=-9i + t. Determine t
Angle between vector a and b is 60
Given the vectors u = (-3,0,1) v= (-1,-2,2) determine
unit vector in the direction of v
v x u
Find the value of the direction angles of both vectors
The formula for the angle between two vectors a & b is
$\displaystyle \cos \theta = \frac{a \cdot b}{|a||b|}$.
$\displaystyle \cos 60 = \frac{3(-9) + 2t}{\sqrt{3^2 + 2^2}\sqrt{(-9)^2 + t^2}}$
$\displaystyle \frac{1}{2} = \frac{-27 + 2t}{\sqrt{13}\sqrt{81 + t^2}}$
$\displaystyle \frac{1}{2} = \frac{-27 + 2t}{\sqrt{1053 + 13t^2}}$
$\displaystyle \sqrt{1053 + 13t^2} = 2(-27 + 2t)$
$\displaystyle \sqrt{13t^2 + 1053} = 4t - 54$
$\displaystyle 13t^2 + 1053 = 16t^2 - 432t + 2916$
$\displaystyle 0 = 3t^2 - 432t + 1863$
$\displaystyle 0 = 3(t^2 - 144t + 621)$
$\displaystyle 0 = t^2 - 144t + 621$
$\displaystyle t = \frac{144 \pm \sqrt{(-144)^2 - 4(1)(621)}}{2(1)}$
$\displaystyle t = \frac{144 \pm \sqrt{18252}}{2}$
$\displaystyle t = \frac{144 \pm 78\sqrt{3}}{2}$
$\displaystyle t = 72 \pm 39\sqrt{3}$
So t = 4.450 or 139.550, but it looks like t = 4.450 is extraneous (the angle turns out to be 120 degrees).
I think that's right.
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Be careful with notation. For vectors, use "<>", not parentheses.
v = <-1, -2, 2>
unit vector in the direction of v = $\displaystyle \frac{v}{|v|}$
$\displaystyle \frac{<-1, -2, 2>}{\sqrt{(-1)^2 + (-2)^2 + 2^2}}$
$\displaystyle \frac{<-1, -2, 2>}{\sqrt{9}}$
$\displaystyle \frac{<-1, -2, 2>}{3}$
$\displaystyle \left\langle -\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right\rangle$
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