Line perpendicular to a plane through a point in 3D

Key takeaway: This 3D geometry problem relies on the fact that a plane’s normal vector gives the direction of any line perpendicular to the plane. Once the line is written, the intersection point and distance follow by substitution and vector geometry.

Use the plane’s normal vector

The plane is

$3x+2y+z=6.$

A normal vector to this plane is

$\mathbf{n}=\begin{pmatrix}3\\2\\1\end{pmatrix}.$

A line perpendicular to the plane must have this same direction vector. Since it passes through $A(7,5,3)$, its equation is

$\mathbf{r}=\begin{pmatrix}7\\5\\3\end{pmatrix}+\lambda\begin{pmatrix}3\\2\\1\end{pmatrix}.$

That answers part (i).

Find the point of intersection

For a point on the line,

$x=7+3\lambda,\quad y=5+2\lambda,\quad z=3+\lambda.$

Substitute into the plane equation:

$3(7+3\lambda)+2(5+2\lambda)+(3+\lambda)=6.$

Simplify:

$21+9\lambda+10+4\lambda+3+\lambda=6$ $34+14\lambda=6$ $14\lambda=-28$ $\lambda=-2.$

So the intersection point is

$P=(7-6,\,5-4,\,3-2)=(1,1,1).$

Find the distance $AP$

The distance from $A(7,5,3)$ to $P(1,1,1)$ is

$AP=\sqrt{(7-1)^2+(5-1)^2+(3-1)^2}$ $=\sqrt{36+16+4}=\sqrt{56}=2\sqrt{14}.$

So

$\boxed{\mathbf{r}=\begin{pmatrix}7\\5\\3\end{pmatrix}+\lambda\begin{pmatrix}3\\2\\1\end{pmatrix},\quad P=(1,1,1),\quad AP=2\sqrt{14}}.$

Key geometric idea

A line perpendicular to a plane is always parallel to the plane’s normal vector. That is why the normal vector gives the line direction immediately. The rest of the solution is just substitution into the plane equation and then a distance formula.

When working in 3D, it helps to verify the intersection point lies on the plane. For $P=(1,1,1)$, we get $3(1)+2(1)+1=6$, which confirms the result.

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Pitfalls the pros know 👇 A common error is using the coefficients of the plane equation as a point instead of a normal vector. The ordered triple $(3,2,1)$ is not on the plane automatically; it is the direction perpendicular to the plane. Another frequent mistake is forgetting that the line must pass through $A(7,5,3)$, so the vector form must start at that point. For the distance, some students try to use the plane equation directly without finding the foot of the perpendicular, but here the intersection point is exactly that foot, so the distance is the length of segment $AP$.

What if the problem changes? If the plane were written as $3x+2y+z=12$ instead of $6$, the line through $A$ would stay the same because the normal vector is unchanged, but the intersection point would move. You would substitute the same parametric line into the new plane equation and solve for a different parameter value. If the point were changed to $A(7,5,3)$ and the plane were parallel to the original one, the distance calculation would instead represent the shortest distance from the point to a parallel plane, not the length of a segment between two points on the same perpendicular line.

Tags: normal vector, parametric line equation, point-to-plane distance