3D lines equation corresponding to the plane x+y+z=3

Question

Perpendicular are drawn from the point on the line $\frac{{x + 2}}{2} = \frac{{y + 1}}{{ - 1}} = \frac{z}{3}$ to the plane $x+y+z=3$. Then the foot of the perpendicular lies on the line

(A) $\frac{x}{5} = \frac{{y - 1}}{8} = \frac{{z - 2}}{{-13}}$

(B) $\frac{x}{2} = \frac{{y - 1}}{3} = \frac{{z - 2}}{{-5}}$

(C) $\frac{x}{4} = \frac{{y - 1}}{3} = \frac{{z - 2}}{{-7}}$

(D) $\frac{x}{2} = \frac{{y - 1}}{-7} = \frac{{z - 2}}{{5}}$

The official answer id (D)

The points on the line $\frac{{x + 2}}{2} = \frac{{y + 1}}{{ - 1}} = \frac{z}{3}$ is represented by $(-2+2t,-1-t,3t)$

Hence the line equation is of the type $\frac{{x -(2t-2)}}{1} = \frac{{y -(-t-1)}}{{ 1}} = \frac{z-(3t)}{1}$. How to I proceed from here

Answer 1

You need to find a value of $t$: $$-2+2t-1-t+3t=3,$$ which gives $$t=\frac{3}{2}$$ and we obtain the following equation: $$\frac{x-1}{1}=\frac{y+\frac{5}{2}}{1}=\frac{z-\frac{9}{2}}{1},$$ but $(1,1,1)$ is not a vector, which is parallel to the needed line.

The right way it's the following.

We got already $\left(1,-\frac{5}{2}, \frac{9}{2}\right)$ as the point on the needed line.

We'll find another point.

The line $(-2+t,-1+t,t)$ is perpendicular to the plane and intersects with the given line in $(-2,-1,0)$.

Now, the equation $$-2+t-1+t+t=3$$ gives us $t=2$ and the second point: $(0,1,2)$ and the needed equation: $$(x,y,z)=(0,1,2)+t(2,-7,5).$$

Answer 2

Here is a "all-vectorial" solution.

Let us call $L$ the line.

$V=\begin{pmatrix}2\\-1\\3\end{pmatrix}$ is a directing vector of line L and $U=\dfrac{1}{\sqrt{3}}\begin{pmatrix}1\\1\\1\end{pmatrix}$ is a unit vector of the normal to the plane.

Consider the orthogonal decomposition $V=V_1+V_2$ where $V_1$ is the projection onto the line directed by $U$ and $V_2$ is the projection onto the plane.

As $V_1=\underbrace{(V.U)}_{\text{dot product}}U=\dfrac{4}{3}\begin{pmatrix}1\\1\\1\end{pmatrix}$, we get:

$V_2=V-V_1=\begin{pmatrix} \ \ 2/3\\-7/3\\ \ \ 5/3\end{pmatrix}$ which is proportional to $\begin{pmatrix} \ \ 2\\-7\\ \ \ 5\end{pmatrix}$ corresponding indeed to the fourth solution.

Answer 3

The foot of perpendicular lies on the projection of line $(-2-1,0)+\lambda(2,-1,3)$ onto the plane $x+y+z=3$.

The projection points along $\vec n\times \vec n \times \vec v$ where $\vec v$ is direction of given line ($\vec v=(2,-1,3)$).

The direction of projection is obtained as ($2,-7,5$).