The Student Room Group
Reply 1
Let OP=6i+19j-k+u(i+4j-2k). since OP is perpendicular to L1, it is perpendicular to the direction vector of L1. So OP.(i+4j-2k)=0.

Hope that helps :smile:
Reply 2
vvidetta
Let OP=6i+19j-k+u(i+4j-2k). since OP is perpendicular to L1, it is perpendicular to the direction vector of L1. So OP.(i+4j-2k)=0.

Hope that helps :smile:

hey thanks for reply....still a bit confused soorry lol thanks for your time..
Reply 3
jimber
hey thanks for reply....still a bit confused soorry lol thanks for your time..


What are you confused about? Sorry, that explanation probably wasn't the best. :s-smilie:
Reply 4
jimber
Hey got exam q's for hwk and now im just stuck on these type q's:

(0,-5,11) is on L1
L1 = 6i + 19j - k + u(i + 4j - 2k)
The point P lies on L1 and is such that OP is perpendicular to L1, where O is the origin.
(b) Find the position vector of point P

Any help will be much appreciated :smile: thanks

We know that the point PP lies on L1L_1, and every point (xyz)\begin{pmatrix} x \\ y \\ z \end{pmatrix} in L1L_1 satisfies (xyz)=(6+μ19+4μ12μ)\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 + \mu \\ 19 + 4 \mu \\ -1-2\mu \end{pmatrix}. Since the direction of L1L_1 is (142)\begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix}, and the direction of OP\overrightarrow{OP} is perpendicular to this direction, so if OP=(xyz)=(6+μ19+4μ12μ)\overrightarrow{OP} =\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 + \mu \\ 19 + 4 \mu \\ -1-2\mu \end{pmatrix} then (6+μ19+4μ12μ)(142)=0\begin{pmatrix} 6 + \mu \\ 19 + 4 \mu \\ -1-2\mu \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} = 0.

You can solve this for μ\mu, then you're done.

[Edited a couple of times because I screwed up first time I posted; it should be alright now.]
Reply 5
nuodai
We know that the point PP lies on L1L_1, and every point (xyz)\begin{pmatrix} x \\ y \\ z \end{pmatrix} in L1L_1 satisfies (xyz)=(6+μ19+4μ12μ)\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 + \mu \\ 19 + 4 \mu \\ -1-2\mu \end{pmatrix}. Since the direction of L1L_1 is (142)\begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix}, and the direction of OP\overrightarrow{OP} is perpendicular to this direction, so if OP=(xyz)=(6+μ19+4μ12μ)\overrightarrow{OP} =\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 + \mu \\ 19 + 4 \mu \\ -1-2\mu \end{pmatrix} then (6+μ19+4μ12μ)(142)=0\begin{pmatrix} 6 + \mu \\ 19 + 4 \mu \\ -1-2\mu \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} = 0.

You can solve this for μ\mu, then you're done.

[Edited a couple of times because I screwed up first time I posted; it should be alright now.]

Thanks guys! really helps.// :smile:

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