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C4 Position Vector

Relative to an origin O, the point A and B have position vectors 3i + 2j +3k and i + 3j + 4k respectively.

(i) Find a vector equation of the line passing through A and B

r=(323)+t(−211) \mathbf{r} = \begin{pmatrix} 3 \\ 2 \\ 3 \end{pmatrix} + t\begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}

(ii) Find the position vector of the point P on AB such that OP is perpendicular to AB

I'm fairly sure it has something to do with dot product equalling zero. But I'm not sure how to set it up. Could anyone explain how to do this. Thanks in advance.
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Avatar for Ch1pp0
Ch1pp0
12
a∙b=∣a∣∣b∣∗cos(θ)[br]θ=90[br]cos(θ)=0[br][br]a \bullet b= \mid a \mid \mid b \mid * cos (\theta)[br]\theta = 90[br]cos(\theta) =0[br][br]

Call x, y, z the coordinates of P:
(−211)∙(xyz)=0[br]−2x+y+z=0 \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} \bullet \begin{pmatrix} x \\ y \\ z \end{pmatrix} = 0 [br]-2x + y + z = 0

(323)+t(−211)=(xyz)[br][br]3−2t=x[br]2+t=y[br]3+t=z[br]\begin{pmatrix} 3 \\ 2 \\ 3 \end{pmatrix} + t\begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}[br][br]3-2t = x[br]2+t = y[br]3+t = z[br]
Plumb those values of x, y and z into the -2x + y + z = 0 equation.
Find t from that, use t in your formula from earlier:
Original post by Freerider101
(i)
(edited 12 years ago)

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