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With respect to the origin $O$, the points $A$ and $B$ have position vectors given by $\overrightarrow{OA} = i + 2j + 2k$ and $\overrightarrow{OB} = 3i + 4j$. The point $P$ lies on the line $AB$ and $OP$ is perpendicular to $AB$.

$i)$ Find a vector equation of the line AB.

$ii)$ Find the position vector of $P$.

$iii)$ Find the equation of the plane which contains AB and which is perpendicular to the plane OAB, giving your answer in the form $ax + by + cz = d$.

I did part $i)$ and got $r = i + 2j + 2k +\lambda(2i + 2j - 2k)$

How can I do part $ii)$, I think once I do part $ii)$ I will be able to do $iii)$, any help please?

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2 Answers

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Let $\mathbf{OP} = <a,b,c> $.

We know that $AB$ and $\mathbf{OP}$ must be perpendicular, so the dot product of $\mathbf{OP}$ and $<2,2,-2>$ is $0$ ($<2,2,-2>$ plays the role of $\mathbf{v}$ in the diagram below)

$$<2,2,-2> \bullet <a,b,c> = 2a+2b-2c = 0$$ $$ \Rightarrow a+b-c=0$$ We also have that for some value of $\lambda$ (let's just call it $t$) $$<1,2,2> + <2,2,-2>t $$ $$=<1+2t, 2+2t, 2-2t> = <a,b,c>$$ because $P$ lies in the $AB $

Combining the two equations, we get $$1+2t +2 +2t - 2 +2t = 0 \Rightarrow 6t = -1$$ $$\Rightarrow t = -1/6$$

So $\mathbf{OP} = <2/3, 5/3, 7/3> = \frac13 <2,5,7>$

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hints...You have the correct answer for part (i)

For part (ii) since $P$ lies on the line $AB$ then $$\overrightarrow{OP}=i+2j+2k+\lambda(2i+2j-2k)$$ for some $\lambda$

Therefore find $\lambda$ so that $\overrightarrow{OP}\cdot(2i+2j-2k)=0$

For part (iii) the normal to the plane is $$n=\overrightarrow{OA}\times\overrightarrow{OB}$$

Then use the equation of the plane $r\cdot n=a\cdot n$

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