The points $P$ and $Q$ have position vectors, relative to the origin $O$, given by $$ \overrightarrow{OP} = 7\mathbf{i} + 7\mathbf{j} - 5\mathbf{k} \quad\text{and}\quad \overrightarrow{OQ} = -5\mathbf{i} + \mathbf{j} + \mathbf{k}. $$ The mid-point of $PQ$ is the point $A$. The plane $\varPi$ is perpendicular to the line $PQ$ and passes through $A$.
- Find the equation of $\varPi$, giving your answer in the form $ax+by+cz=d$.
- The straight line through $P$ parallel to the $x$-axis meets $\varPi$ at the point $B$. Find the distance $AB$, correct to $3$ significant figures.
I answered the first part, but I don't the second part. Should the $\mathbf{j}$ and $\mathbf{k}$ vectors be zero?
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$\begingroup$Hint: We know $A$, and the line $AB$ is described by this equation: $$l: A+\lambda\begin{pmatrix}1\\0\\0 \end{pmatrix}$$ Once you have this, you will be able to solve for the point $B$, and with your information on $A$, you should be able to work out the distance.
$\endgroup$ 2 $\begingroup$So I believe you have found the equation to be $-2x+y+z=18.$
We want to find coordinates of B, and we are given the information that it is the intersection point between the straight line through P and the plane.
As B lies on the straight line through P parallel to the x-axis, it will have the coordinates of $$B: \begin{pmatrix}7\\7\\-5\end{pmatrix}+\lambda\begin{pmatrix}1\\0\\0 \end{pmatrix}$$
We substitute the $x,y,z$ components of this B coordinates as shown above into the equation you found in part (a), so you would obtain a number for $\lambda$
Then substitute this value of $\lambda$ back into the vector equation I gave for B above, then find the magnitude of the vector AB. Hope it helps
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