What is the point of intersection of a line defined by 2 points with the plane $y=0$?

Question

Referred to the origin $O$, the points $P$ and $Q$ have position vectors i + $2$j + $3$k and $3$i + $3$j + $5$k respectively. Find the position vector of the point where the line $PQ$ meets the plane $y=0$ .

I’ve found the direction of line $PQ$:

$\overrightarrow{PQ}$

$= \overrightarrow {OQ} - \overrightarrow {OP}$

$=$ $3$i + $3$j + $5$k - ( i + $2$j + $3$k)

$=$ $2$i + j + $2$k

So, the vector equation of line $PQ$ is r = i + $2$j + $3$k + $\lambda$($2$i + j + $2$k)

To find the position vector of the point of intersection between line $PQ$ and the plane $y=0$, I understand that I need to substitute r = i + $2$j + $3$k + $\lambda$($2$i + j + $2$k) into the equation of plane, r$\cdot$n = $d$, where n is a vector perpendicular to the plane and $d$ is a constant.

Is it right to assume that n= $\left(\begin{matrix}2\\1\\2\\\end{matrix}\right)$ ?

How do I find the value of $d$? How do I apply the information of plane $y=0$?

Any hints would be helpful. Thank you

Answer 1

The line can be parameterised by $p(t) = (1,2,3)+t ( (3,3,5) - (1,2,3)) = (1,2,3)+t(2,1,2)$.

You want to find the $t^*$ such that $[p(t^*)]_2 = 0$.

This gives $0 = 2+t^*$ or $t^* = -2$.

Hence the point is $p(t^*) = (-3,0,-1)$.

Answer 2

All points in line $PQ$ are in the form of $a\vec{P}+b\vec{Q}, a+b=1$.

Now look at the $\hat{j}$ component, since it needs to be zero then $2a+3b=0$. Solving the two equations yields $a=3$ and $b=-2$