PinkFlamingo Messages 19 Reaction score 0 Thread starter May 17, 2004 #1 Could someone refresh my memory how to find the angle between 2 vectors in 3d, say: a= 4i + j + 2k b= -i +2j + k Thanks!
Could someone refresh my memory how to find the angle between 2 vectors in 3d, say: a= 4i + j + 2k b= -i +2j + k Thanks!
enigma Staff Emeritus Science Advisor Gold Member Messages 1,739 Reaction score 20 May 17, 2004 #2 dot product a.b = |a||b|cos (ang)
PinkFlamingo Messages 19 Reaction score 0 May 17, 2004 #3 So how would I find the angle for the two I posted?
cookiemonster Messages 977 Reaction score 0 May 17, 2004 #4 He just told you. Solve for theta and plug and chug. [tex]\theta = \arccos{\left( \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|\right)}[/tex] [tex]\vec{a}\cdot\vec{b} = a_1b_1 + a_2b_2 + a_3b_3[/tex] and [tex]|\vec{a}| = \sqrt{{a_1}^2 + {a_2}^2 + {a_3}^2}[/tex] if [itex]\vec{a} = <a_1,a_2,a_3>[/itex] and likewise for b. cookiemonster Last edited: May 17, 2004
He just told you. Solve for theta and plug and chug. [tex]\theta = \arccos{\left( \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|\right)}[/tex] [tex]\vec{a}\cdot\vec{b} = a_1b_1 + a_2b_2 + a_3b_3[/tex] and [tex]|\vec{a}| = \sqrt{{a_1}^2 + {a_2}^2 + {a_3}^2}[/tex] if [itex]\vec{a} = <a_1,a_2,a_3>[/itex] and likewise for b. cookiemonster