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    so basically I need help in C and D

    for C
    they told the line 2 is parallel to L1 so direction will be same but what will be the points
    r=(?)+u(direction)

    for D
    I have to use modulus but how to find B coordinates ?It is one mark but still I got stuck
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    (Original post by Qer)

    so basically I need help in C and D

    for C
    they told the line 2 is parallel to L1 so direction will be same but what will be the points
    r=(?)+u(direction)

    for D
    I have to use modulus but how to find B coordinates ?It is one mark but still I got stuck
    Yes the direction is the same. You only need one point to define the line since you know the direction. You know it passes through B, so find its coordinates since you know its related to the point A, namely \overrightarrow{OB} = 3\overrightarrow{OA}
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    (Original post by RDKGames)
    Yes the direction is the same. You only need one point to define the line since you know the direction. You know it passes through B, so find its coordinates since you know its related to the point A, namely \overrightarrow{OB} = 3\overrightarrow{OA}
    thanks
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    so if A was for instance ( 2, -3, 8 ) then B would be ( 6, -9, 24 )
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    can you guys also help me in part e
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    I draw a diagram something like that
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    (Original post by Qer)
    can you guys also help me in part e

    I draw a diagram something like that
    It says that \overrightarrow{OX} and l_2 are perpendicular. This means that the angle between the vector \overrightarrow{OX} and the direction vector of l_2 is 90, which hence implies that the dot product between the two is 0.

    You know the dir. vector already.
    To get the vector \overrightarrow{OX}, just begin by saying that X has general coordinates \overrightarrow{OX}=(3+5\lambda, -4\lambda, -3 -3\lambda) for some \lambda \in \mathbb{R} since it lies on l_2.
    So, just use the dot product condition mentioned earlier to get the value of \lambda that defines X, then just find the length of that vector.

    A DIFFERENT APPROACH would be to find cosine of the acute angle \alpha between the line l_2 and vector \overrightarrow{OB}. Then use the fact that |\overrightarrow{OA}| = |\overrightarrow{OB}| \sin \alpha (seen clearly from your diagram, in the right-angled triangle OXB)
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    (Original post by RDKGames)
    It says that \overrightarrow{OX} and l_2 are perpendicular. This means that the angle between the vector \overrightarrow{OX} and the direction vector of l_2 is 90, which hence implies that the dot product between the two is 0.

    You know the dir. vector already.
    To get the vector \overrightarrow{OX}, just begin by saying that X has general coordinates \overrightarrow{OX}=(3+5\lambda, -4\lambda, -3 -3\lambda) for some \lambda \in \mathbb{R} since it lies on l_2.
    So, just use the dot product condition mentioned earlier to get the value of \lambda that defines X, then just find the length of that vector.

    A DIFFERENT APPROACH would be to find cosine of the acute angle \alpha between the line l_2 and vector \overrightarrow{OB}. Then use the fact that |\overrightarrow{OA}| = |\overrightarrow{OB}| \sin \alpha (seen clearly from your diagram, in the right-angled triangle OXB)

    thank you
 
 
 
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