Using cross product find direction vector of line joining point of intersection of line and plane and foot of perpendicular from line to plane.

Solution 1:

Imagine this line $r=a+\lambda\vec{d}$ is intersecting plane $\pi: \vec{r}.\hat n = k$

• The cross product of $\hat n \times \hat{d} = \hat{u_1}$ this $\hat{u_1}$ will be the unit vector normal to the plane of line containing line $r=a+\lambda\vec{d}$ and plane $\pi: \vec{r}.\hat n = k$

• Now, we take the cross product of $\hat{u_1}$ and $\hat n$: $\hat {u_2} = \hat {u_1} \times \hat n$ will be the required unit vector.

• Note: Here the $\hat {u_2}$ depends on the unit vector $\hat d$ I mean $\hat u_2 = \hat {u_1} \times \hat n $ or $\hat {u_2} = \hat n \times \hat {u_1}$

Find the direction vector of line PQ.

Here, I used unit vectors only as you were interested in the direction of $\vec{PQ}$;

Solution 2:

The segment $PQ$ lies in a plane that is spanned by the perpendicular to the plane $n$ and the direction vector $d$, so the normal to this plane is $n \times d$. But $PQ$ also lies in the plane whose normal is $n$, hence the direction vector of $PQ$ must be along the vector $(n \times d) \times n $

Solution 3:

Let $\alpha$ be the angle between ${\bf n}$ and ${\bf d},$ and $\beta$ be the angle between ${\bf n}$ and the perpendicular to both ${\bf n}$ and ${\bf d}.$

$$( \hat{\bf n} \times {\bf d} ) \times \hat{\bf n}$$

1. has magnitude $$\Big(\Vert\hat{\bf n}\Vert \,\Vert{\bf d}\Vert|\,\sin\alpha|\Big)\,\Vert\hat{\bf n}\Vert\,|\sin\beta|\\ =\Vert\hat{\bf n}\Vert \,\Big(\Vert{\bf d}\Vert|\,\sin\alpha|\Big)\,\Vert\hat{\bf n}\Vert\,|\sin\beta| \\=(1)\,PQ\,(1)\,|\sin90^\circ |\\=PQ;$$

2. • is perpendicular to $\hat{\bf n},$
   • and to the normal of the plane spanned by $\hat{\bf n}$ and ${\bf d},$ i.e., lies in the plane spanned by $\hat{\bf n}$ and ${\bf d};$

   thus is collinear to $\vec{PQ}$ (which is also perpendicular to $\hat{\bf n}$).