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

A line with equation $$r=a+\lambda\vec{d}$$ meets plane $$\pi$$ with equation $$r.\hat{n}=k$$ at point P. Point Q lies in $$\pi$$ and is the foot of the perpendicular from A to $$\pi$$. Find the direction vector of line PQ.

By solving $$(a+\lambda\vec{d}).\hat{n}=k$$, I was able to find the position vector of P. Then by finding the intersection of line AQ and plane I was able to find the position vector of Q and hence the direction vector PQ.
However, the answer can be found simply by finding $$(\hat{n}\times \vec{d})\times \hat{n}$$ where $$\times$$ is cross-product. I don't understand why.
Here's what I know : The cross-product of 2 vectors gives a 3rd vector perpendicular to the 2 vectors. Line PQ lies on plane so direction vector PQ $$\perp \hat{n}$$. Also, AQ is parallel to $$\hat{n}$$.

The first part $$w=(\hat{n}\times \vec{d})$$ gives a vector perpendicular to line and parallel to plane. Won't $$w\times n$$ give a vector perpendicular to the plane again? I can't understand the geometric interpretation of $$(\hat{n}\times \vec{d})\times \hat{n}$$.

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 $$\vec{PQ}$$;

• I understood the first bullet point but not the second. How did you know that the cross product of u1 and u2 gives the direction vector of PQ? Nov 23, 2021 at 11:11
• @Bunny Actually, the fact is $\hat u_2$ is perpendicular to the plane created by normal of plane$\pi$ and line $r = a + \lambda d$ Nov 23, 2021 at 11:19

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 $$(n \times d) \times n$$

• Did the fact that Q is the foot of the perpendicular imply that the direction vector d spans the plane containing segment PQ? Or is the information given about foot of perpendicular irrelevant? Nov 23, 2021 at 11:04
• That plane is the plane containing the triangle $APQ$ so it is spanned by $AP$ which is along vector $d$, and $AQ$ which is along $n$. Nov 23, 2021 at 12:00

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

• 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 $$\hat{\bf n}$$).

• Can you elaborate how you obtained magnitude in number 1. Which term gave PQ? Nov 23, 2021 at 11:20
• No I did not downvote any answer. After drawing the triangle, I understood what you meant. Thank you. Nov 23, 2021 at 12:12