Intersections and distances involving planesAQA A-Level Further Maths: Revision notes

To find the intersection of a line $\mathbf r=\mathbf a+t\mathbf b$ with a plane $ax+by+cz=d$:

1. Write the line in parametric form, $x=a_1+tb_1$, $y=a_2+tb_2$, $z=a_3+tb_3$. If the line is given as $\frac{x-a_1}{b_1}=\frac{y-a_2}{b_2}=\frac{z-a_3}{b_3}$, set each fraction equal to $t$.
2. Substitute into the plane equation to get an equation in $t$ alone.
3. Solve for $t$, then substitute back to get the point of intersection.

Example: $l$: $(1+2t,\ -2+t,\ 3-t)$ and $\Pi$: $3x-y+2z=17$. Then $11+3t=17$, $t=2$, and the point is $(5,0,1)$. Check by substituting the point into the plane equation.

After substitution the equation in $t$ can behave in three ways:

• One solution for $t$: the line meets the plane at one point.
• A contradiction such as $0t=5$: the line is parallel to the plane and does not lie in it, so there is no intersection.
• An identity such as $0t=0$: the line lies in the plane.

The line is parallel to the plane when $\mathbf b\cdot\mathbf n=0$, where $\mathbf n$ is the plane's normal. Then test one point of the line in the plane equation: if it satisfies it, the line lies in the plane; if not, the line is parallel but outside it. Example: $\mathbf b=(1,-1,-2)$ and $\mathbf n=(3,-1,2)$ give $\mathbf b\cdot\mathbf n=0$, and the point $(0,1,1)$ gives $1\ne17$, so no intersection.

For a point $(x_1,y_1,z_1)$ and plane $ax+by+cz=d$, the shortest distance is $\text{distance}=\frac{|ax_1+by_1+cz_1-d|}{\sqrt{a^2+b^2+c^2}}.$ In vector form, for a plane $\mathbf r\cdot\mathbf n=d$, it is $\dfrac{|\mathbf p\cdot\mathbf n-d|}{|\mathbf n|}$ for the point with position vector $\mathbf p$. Example: the distance from $(1,2,0)$ to $2x-y+2z=9$ is $\frac{|2-2+0-9|}{3}=3$. The modulus matters: the sign only tells you which side of the plane the point is on.

To find the point of the plane closest to $P$ (the foot of the perpendicular), draw the line through $P$ in the direction of the normal, $\mathbf r=\mathbf p+t\mathbf n$, and find where it meets the plane using the method above. The distance is $|t||\mathbf n|$, which agrees with the distance formula.

For $(1,2,0)$ and $2x-y+2z=9$: the line is $(1+2t,\ 2-t,\ 2t)$, so $9t=9$, $t=1$, and the foot is $(3,1,2)$. The reflection of $P$ in the plane is a further equal distance along the normal: $\mathbf p+2t\mathbf n=(5,0,4)$.

A drone starts at $A(-2,4,10)$ and flies along $(-2+\lambda,\ 4-2\lambda,\ 10-3\lambda)$ towards the plane $2x+3y+z=4$. Substituting, $18-7\lambda=4$, so $\lambda=2$ and it reaches the roof at $(0,0,4)$ after $2\sqrt{14}=7.48$ m. The shortest distance from $A$ to the roof is $\frac{14}{\sqrt{14}}=\sqrt{14}=3.74$ m, at the foot of the perpendicular $(-4,1,9)$, which is shorter than the path actually flown.