The angle between A25 mi 45 mj and the positive x axis is 29

The angle between A=(25 m)i +(45 m)j and the positive x axis is: 29degree 61degree 151degree 209degree 241degree The vector V_3 in the diagram is equal to: Let 5 = (1 m)i +(2 m)j + (2 m)k and t = (3 m)i +(4 m)k. The angle between thes vectors is given by: cos^-1(14/15) cos^-1 (1/225) cos^-1(104/225) cos^-1(11/15) cannot be found since S and T do not lie in the same plane The value of i =(j times k) is:

Solution

22. A = 25i + 45j

x = i

A.x = |A||x| cos@

(25i + 45j ) . ( i )   = (sqrt(25^2 + 45^2)) (1 ) cos@

25 = 51.48 cos@

cos@ = 0.486

@ = 60.95 deg

Ans(B)

23. V2 - V1 = V3

Abs (C)

24. S = 1i + 2j + 2k

r = 3i + 4k

S. R = |S||R| cos@

(1i + 2j + 2k ) ( 3i + 4k) = (sqrt(1^2 + 2^2 + 2^2) sqrt(3^2 + 4^2)) cos@

(1 x 3) + (2 x 0 ) + (2 x 4) = (3 x 5 ) cos@

cos@ = 11/15

@ = cos^-1(11/15)

Abns(D)

25. i (j x k )

j x k = i

i.i = 1

Ans(B)