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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 $(\hat{n}\times \vec{d})\times \hat{n}$.

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3 Answers 3

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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 $\vec{PQ}$;

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  • $\begingroup$ 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? $\endgroup$
    – Bunny
    Nov 23, 2021 at 11:11
  • $\begingroup$ @Bunny Actually, the fact is $\hat u_2$ is perpendicular to the plane created by normal of plane$\pi$ and line $r = a + \lambda d$ $\endgroup$
    – Darshan P.
    Nov 23, 2021 at 11:19
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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 $

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  • $\begingroup$ 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? $\endgroup$
    – Bunny
    Nov 23, 2021 at 11:04
  • $\begingroup$ 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 $n$. $\endgroup$ Nov 23, 2021 at 12:00
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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 $\hat{\bf n}$).

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

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