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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

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

enter image description here $\pi$ is yellow and $\rho$ is red.


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

Thank you in advance!

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1 Answer 1

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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

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