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# Edexcel A2 maths C4 vectors Watch

1. A=(3,-4,2)
B=(1,0,-2)
L passes through A and has equation
r=(3,-4,2)+x(-3,5,2)
AB=(-2,4,-4)
C lies on L such that angle ABC is a right angle. Find the coordinates of the point C.

The answer is (-3,6,6) but I have no idea how to get to it :/

Any help would be muchly appreciated
2. (Original post by hkooba)
A=(3,-4,2)
B=(1,0,-2)
L passes through A and has equation
r=(3,-4,2)+x(-3,5,2)
AB=(-2,4,-4)
C lies on L such that angle ABC is a right angle. Find the coordinates of the point C.

The answer is (-3,6,6) but I have no idea how to get to it :/

Any help would be muchly appreciated
http://www.thestudentroom.co.uk/forumdisplay.php?f=38

If you post in maths, you are more likely to get a response. I have asked for this to be moved there for you.
3. A general point on the line L has coordinates:

$\begin{pmatrix} 3-3x\\ -4+5x\\ 2+2x \end{pmatrix}$

Call this point C, with position vector $\vec{OC}$

The angle between vectors $\vec{AB}$ and $\vec{BC}$ is $90^{\circ}$

So first we have to find $\vec{BC}$.

$\vec{BC} = \vec{OC} - \vec{OB}$

$\vec{BC} = \begin{pmatrix} 3-3x\\ -4+5x\\ 2+2x \end{pmatrix}-\begin{pmatrix} 1\\ 0\\ -2 \end{pmatrix}$

$\vec{BC} = \begin{pmatrix} 2-3x\\ -4+5x\\ 4+2x \end{pmatrix}$

Now, using the formula for the angle between two vectors:

$\cos \theta = \frac{\vec{AB}\cdot \vec{BC}}{\left | \vec{AB} \right |\left | \vec{BC} \right |}$

Given that the angle between the vectors is 90, cos90 = 0.

$\vec{AB}\cdot \vec{BC} = \begin{pmatrix} -2\\ 4\\ -4 \end{pmatrix}\cdot \begin{pmatrix} 2-3x\\ -4+5x\\ 4+2x \end{pmatrix}=0$

$18x-36=0$

$x=2$

So the point C on line L is where $x=2$.

$C\begin{pmatrix} 3-3(2)\\ -4+5(2)\\ 2+2(2) \end{pmatrix}$

Therefore,

$C\begin{pmatrix} -3\\ 6\\ 6 \end{pmatrix}$
4. (Original post by The_Big_E)
A general point on the line L has coordinates:

$\begin{pmatrix} 3-3x\\ -4+5x\\ 2+2x \end{pmatrix}$

Call this point C, with position vector $\vec{OC}$

The angle between vectors $\vec{AB}$ and $\vec{BC}$ is $90^{\circ}$

So first we have to find $\vec{BC}$.

$\vec{BC} = \vec{OC} - \vec{OB}$

$\vec{BC} = \begin{pmatrix} 3-3x\\ -4+5x\\ 2+2x \end{pmatrix}-\begin{pmatrix} 1\\ 0\\ -2 \end{pmatrix}$

$\vec{BC} = \begin{pmatrix} 2-3x\\ -4+5x\\ 4+2x \end{pmatrix}$

Now, using the formula for the angle between two vectors:

$\cos \theta = \frac{\vec{AB}\cdot \vec{BC}}{\left | \vec{AB} \right |\left | \vec{BC} \right |}$

Given that the angle between the vectors is 90, cos90 = 0.

$\vec{AB}\cdot \vec{BC} = \begin{pmatrix} -2\\ 4\\ -4 \end{pmatrix}\cdot \begin{pmatrix} 2-3x\\ -4+5x\\ 4+2x \end{pmatrix}=0$

$18x-36=0$

$x=2$

So the point C on line L is where $x=2$.

$C\begin{pmatrix} 3-3(2)\\ -4+5(2)\\ 2+2(2) \end{pmatrix}$

Therefore,

$C\begin{pmatrix} -3\\ 6\\ 6 \end{pmatrix}$
Thank you, I get it now

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