# Solution verification: given the point $T$ and the line $p$, find the distance between their orthogonal projections onto the plane $\pi$.

Let $$T'$$ and $$p'$$ be the orthogonal projections of the point $$T(-8,2,-3)$$ and the line $$p\ldots\frac{x}4=\frac{y-4}3=\frac{z+1}{-2}$$ onto the plane $$\pi\ldots x-y+3z+8=0$$.

Find the distance between the point $$T'$$ and the line $$p'$$.

My attempt:

In general, a vector perpendicular to the plane $$\varphi= A\cdot x+ B\cdot y+ C\cdot z+D=0$$ is $$(A,B,C)$$, so $$\vec{n}=(1,-1,3)\perp\pi.$$

The equation of a vector $$\vec{v}$$ passing through two points $$P_1(x_1,y_1,z_1)$$ and $$P_2(x_2,y_2,z_3)$$: $$\vec{v}=a\vec{i}+b\vec{j}+c\vec{k}=(x_2-x_1)\vec{i}+(y_2-y_1)\vec{j}+(z_2-z_1)\vec{k}$$ \begin{aligned}\implies x_2&=a+x_1 && y_2= b+y_1&&z_2=c+z_1\\x_2&=-7&&y_2=1&&z_2=0\end{aligned}

So, the orthogonal projection of the point $$T$$ onto the plane $$\pi$$ is $$T'(-7,1,0)$$.

From the equation $$p\ldots\frac{x}4=\frac{y-4}3=\frac{z+1}{-2}$$

we can see the line $$p$$ is determined by the point $$T_0(x_0,y_0,z_0)=(0,4,-1)$$ and by the direction vector $$\vec{q}=a\vec{i}+b\vec{j}+c\vec{k}=4\vec{i}+3\vec{j}-2\vec{k}$$

Since we want to find a plane $$\rho,\ \rho\perp\pi$$, we have to find the normal vector $$\vec{m}\perp\rho,\vec{m}\in\pi$$, so $$\vec{m}=\vec{n}\times\vec{q}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\1&-1&3\\4&3&-2\end{vmatrix}=-7\vec{i}+14\vec{j}+7\vec{k}=7(-\vec{i}+2\vec{j}+1\vec{k})$$

Since $$T_0\in\rho$$, $$-7\cdot0+14\cdot4+7\cdot(-1)+D=0\implies D=-49$$ $$\implies\rho\ldots-7x+14y+7z-49=0$$

Let's find the intersection line of $$\pi$$ and $$\rho$$, which is the orthogonal projection of $$p$$ onto $$\pi$$, by solving the following system: \begin{aligned}x-y+3z+8&=0\\-x+2y+z-7&=0\end{aligned} \begin{aligned}x&=y-3z-8\\x&=2y+z-7\end{aligned} $$\implies y-3z-8=2y+z-7\implies y=-4z-1\implies x=-7z-9$$

Now, we have the parametric equation $$\pi\cap\rho=p'=\{-7z-9,y=-4z-1,z\}$$

Let's express $$p'$$ in terms of a vector: $$p'=(-7z-9)\vec{i}+(-4z-1)\vec{j}+z\vec{k}=\left(-9,-1,0\right)+z\underbrace{\left(-7\vec{i}-4\vec{j}+\vec{k}\right)}_{\text{direction}}$$

So, $$p'$$ is passing through the point $$S(-9,-1,0)$$.

Let $$\vec{a}=-7\vec{i}-4\vec{j}+\vec{k}$$ and

$$\vec{b}=(-7,1,0)-(-9,-1,0)=\left(2,2,0\right)$$ $$d(T',p')=\frac{\|a\times b\|}{\|a\|}$$ $$a\times b=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-7&-4&1\\2&2&0\end{vmatrix}=-2\vec{i}-2\vec{j}-6\vec{k}$$ $$\|a\|=\sqrt{66}$$ $$\|a\times b\|=\sqrt{44}$$ $$\implies \boxed{d(T',p')=\frac{\sqrt{6}}3}$$

$$\pi$$ is yellow and $$\rho$$ is red.

May I ask if this is correct? Is there any more efficient method?

You don't need to find each projection to find the distance between their projections. Just find the normal vector from the point $$T$$ to the line $$p$$, and then find it's distance projected to the plane $$\pi$$, which can be found as follows

Any point on line p is written as $$P_p = (0,4,-1) + \lambda(\frac{4}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{-2}{\sqrt{29}})$$

Hence, the vector from T to $$P_p$$ is expressed as

$$v = (\frac{4\lambda}{\sqrt{29}} + 8, \frac{3\lambda}{\sqrt{29}} + 2, \frac{-2\lambda}{\sqrt{29}} + 2)$$

Now for this to be the shortest, it must be orthonormal to p

Hence

$$(\frac{4\lambda}{\sqrt{29}} + 8).4 + ( \frac{3\lambda}{\sqrt{29}} + 2).3 -(\frac{-2\lambda}{\sqrt{29}} + 2).2 = 0$$

Solve for $$\lambda$$ and put back into $$v$$ to get

$$v = (\frac{96}{29}, -\frac{44}{29}, \frac{126}{29})$$

Now, we need to find the distance of it's projection in $$\pi$$, which can be done by removing the magnitude of it's dot product with the normal of the plane