(−120 k) x (Dxi + Dyj) + (120j −160k) x T i + (80k − 200i) x (−720j) = 0
I need help with simple 3d cross product problem. please explain in detail. thanks
thumb rules for cross product
i^ x j^ = k^
j^ x k^ = i^
k^x i^ = j^
-(a x b ) = b x a
(−120 k) x (Dxi + Dy j)
= -120Dx j^ + 120Dy i^
(120j −160k) x T i
= -120T k^ - 160T j^
(80k − 200i) x (−720j)
= 57600 i^ + 144000k^
since RHS is zero so each component (that is coefficients of i^ , j^ , k^ has to be zero )
equating coefficient of i^ = 0
57600+120 Dy = 0
Dy = -57600/120 =-480
equating coefficient of j^ = 0
-160T - 120 Dx = 0
equating coefficient of k^ = 0
-120T + 144000 = 0
T = 1200
-160T - 120 Dx = 0
-160*1200 -120 Dx = 0
Dx = 1600
Get Answers For Free
Most questions answered within 1 hours.