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  1. #1
    JMV
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    vector problems

    ive been working on about 20 questions of vectors.. but these 5 i cant seem to get.. any help would be much appreciated!

    1. Deterine the components of a vector with tail at P (1,1,1), magnitude 3, and in the same direction as a= (-3,-4, 12)

    2. The vector v has a magnitude 10 units, positive and equal x and y components, and a z component of 4. Determine v.

    3. the vector u= (1,-2,1) Determine 3 different non-collinear vectors that are perpendicular to u. Then show that the 3 vectors determined in the first part of the question are coplanar.

    4. Prove that (a-b) x (a+b)= 2(axb)

    5. Find the total work done by a 15 N force, F, in the direction of the vector (1,2,2) when it moves particle from O (0,0,0) to P(1,-3,4) and then from P to Q(7,2,5). The distance is measured in metres.
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  2. #2
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    Quote Originally Posted by JMV View Post
    ...
    1. Deterine the components of a vector with tail at P (1,1,1), magnitude 3, and in the same direction as a= (-3,-4, 12)

    2. The vector v has a magnitude 10 units, positive and equal x and y components, and a z component of 4. Determine v.

    3. the vector u= (1,-2,1) Determine 3 different non-collinear vectors that are perpendicular to u. Then show that the 3 vectors determined in the first part of the question are coplanar.
    ...
    to #1:

    |\vec a| = \sqrt{9+16+144}=13

    Therefore a vector which has the same direction as \vec a and has the magnitude 3 has the components: \left(-\frac9{13}~,~-\frac{12}{13}~,~\frac{36}{13}\right)

    to #2:
     \vec v=(x, x, 4)~\implies~|\vec v|=\sqrt{x^2+x^2+16}

    And therefore you have to solve for x:

    2x^2+16=100~\implies~x=\sqrt{84} (since only the positive value should be used). That means:

    \vec v=\left(\sqrt{84}~,~\sqrt{84}~,~4\right)

    to #3:
    Let \vec v \perp \vec u~\implies~\vec v \cdot \vec u = 0. Let \vec v = (r, s, t)

    Then you know:

    (1, -2, 1) \cdot (r,s,t)=r-2s+t

    r-2s+t=0~\implies~t=2s-r and therefore the vector

    \vec v = (r, s, 2s-r)~,~r,s \in \mathbb{R} is perpendicular to \vec u

    Actually this is the equation of a plane passing through the origin with \vec u as it's normal vector.

    Choose numbers for r and s to build vectors and do the required examinations.
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