Question

(−120 k) x
(D_{x}i +
D_{y}j) + (120j
−160k) x T i + (80k − 200i) x (−720j) = 0

I need help with simple 3d cross product problem. please explain in detail. thanks

Answer #1

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

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