# Thread: Position vector of point of intersection of line and plane

1. ## Position vector of point of intersection of line and plane

The line r=a+tb and the plane r.n=p. b.n is not equal to zero. Prove that the position vector of the point of intersection of the line and plane is $a+\frac{(p-a.n)b}{b.n}$

I have no idea how to start this.
Thanks

2. Originally Posted by arze
The line r=a+tb and the plane r.n=p. b.n is not equal to zero. Prove that the position vector of the point of intersection of the line and plane is $a+\frac{(p-a.n)b}{b.n}$
Use the equation $\left( {a + tb} \right) \cdot n = p\, \Rightarrow \,a \cdot n + tb \cdot n = p\, \Rightarrow \,t = \frac{{p - a \cdot n}}{{b \cdot n}}$.
That is the $t-$value of the point of intersection.
Now substitute it into the line.