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    Given vector V=2i+3j-6k,find
    a) the magnitude of V
    b)the unit vector in direction of V
    c) the vector in the direction opposite to V and with magnitude 10.5
    d)the anglews which V make with the coordinate axis
    Cannot do part c) and d)

    2. Octavius the spider spins two successive threads along vectors a=i-3j+2k and b=4i+j-3k.Find
    a)the angle through which Octavius turns at the junction of the two threads
    b) the vector he must travel to form a triangular web
    c)the angle this third side makes with the x-axis
    Help in all parts
    4.Vector V makes angle 50 degrees and 70 degrees with the positive x and y directions. Find two possible valvues of V if V magnitude=6.
    Answers: 1c=-3i+4.5j+9k 1d=73.4, 64.4, 149
    2a=105 2b=-5i+2j+k 2c=156
    4)46.7 or 133.3
    THANK YOU
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    (Original post by Asazycat)
    Given vector V=2i+3j-6k,find
    a) the magnitude of V
    b)the unit vector in direction of V
    c) the vector in the direction opposite to V and with magnitude 10.5
    d)the anglews which V make with the coordinate axis
    Cannot do part c) and d)

    2. Octavius the spider spins two successive threads along vectors a=i-3j+2k and b=4i+j-3k.Find
    a)the angle through which Octavius turns at the junction of the two threads
    b) the vector he must travel to form a triangular web
    c)the angle this third side makes with the x-axis
    Help in all parts
    4.Vector V makes angle 50 degrees and 70 degrees with the positive x and y directions. Find two possible valvues of V if V magnitude=6.
    Answers: 1c=-3i+4.5j+9k 1d=73.4, 64.4, 149
    2a=105 2b=-5i+2j+k 2c=156
    4)46.7 or 133.3
    THANK YOU
    1c) Take the unit vector you find in part b. If you multiply it by -1, then the vector will go in the opposite direction. If you multiply by some scalar number, that number will be the magnitude of the vector.

    1d) Take |{\bf V}|. Then quite simply the angles the vector makes between the {\bf i}, {\bf j}, and {\bf k} axis is \cos(\theta)=\frac{\text{the amount by which the vector goes along that axis}}{|{\bf V}|} then obviously you can get your angle \theta

    2a) So find the angle between the two vectors. \cos(\theta)=\frac{{\bf a}\cdot {\bf b}}{|a||b|} ring any bells?

    2b) Say the triangle is ABC with A being the position vector {\bf a} and likewise for B. Then if he travels along A, then along B, what vector must be go along to get back to A?

    2c) Same principle as 1c

    4) I dont understand the question... please elaborate?
 
 
 
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