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    Hi.
    I am a bit confused on this vector question 5(iii).
    I got an answer of -i+j+k
    But it was meant to be -2i+2j+2k.
    I also only did this through trial and error and the mark scheme suggested that i use the dot product between the (position vector and direction vector) and the direction vector.
    https://983c9f06eb1f75af6e83364e092b...20C4%20OCR.pdf
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    (Original post by SamuelN98)
    Hi.
    I am a bit confused on this vector question 5(iii).
    I got an answer of -i+j+k
    But it was meant to be -2i+2j+2k.
    I also only did this through trial and error and the mark scheme suggested that i use the dot product between the (position vector and direction vector) and the direction vector.
    https://983c9f06eb1f75af6e83364e092b...20C4%20OCR.pdf
    We know from the given information that the position vector of A lies on the line l_{1}, so it must be true that the vector \mathbf{OA}= \begin{pmatrix} 4+3s\\6+2s \\ 4+s \end{pmatrix} for some arbitrary value of s, s \subs\in \mathbb{R}. (Remember that as you input different values of s, you obtain different position vectors on the line.)

    Now we know that this position vector must be perpendicular to the line, so the scalar product of the position vector \mathbf{OA}, which we have obtained previously, and the direction vector of the line must be 0.

    \displaystyle \Rightarrow \begin{pmatrix} 4+3s\\6+2s \\ 4+s \end{pmatrix} \cdot \begin{pmatrix} 3\\2 \\1 \end{pmatrix} = 0

    You will now obtain a value for the scalar s which will give you the position vector of A once inputted into the equation of your line.
 
 
 
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