Parallelism and perpendicularity between lines and planes in space
An engineer has to run a cable perpendicular to a panel, or check that a rail is parallel to the floor of a factory. In both cases the question is the same: what is the relationship between the direction vector of the line and the normal vector of the plane? And if an object needs to know its "shadow" on a surface, the tool is orthogonal projection, which you will always construct using an auxiliary perpendicular plane.
Line and plane: parallel or perpendicular
Determine the relative position of the line $r$ and the plane $\pi$ by checking the relationship between their characteristic vectors. The line is given by the…
Determine the relative position of the line $r$ and the plane $\pi$ by checking the relationship between their characteristic vectors. The line is given by the continuous equation $r: \dfrac{x-1}{2} = \dfrac{y+3}{-1} = \dfrac{z}{4}$ and the plane by the general equation $\pi: 2x - y + 4z - 5 = 0$.
- a) Write a direction vector $\vec{v}_r$ of the line and a normal vector $\vec{n}$ of the plane.
- b) Is the direction vector of the line parallel to the normal vector of the plane? Justify your answer.
- c) Justify what geometric implication this fact has for the relative position of $r$ and $\pi$.
Given the point $Q(3, -1, 4)$ and the plane $\alpha: 2x + y = 0$, find the equation of the line $r$ passing through $Q$ and perpendicular to the plane $\alpha$. Take…
Given the point $Q(3, -1, 4)$ and the plane $\alpha: 2x + y = 0$, find the equation of the line $r$ passing through $Q$ and perpendicular to the plane $\alpha$. Take special care when identifying the coordinates of the normal vector of the plane, paying attention to the absence of the variable $z$.
Find the value of the parameter $k$ so that the line $r: \dfrac{x-2}{3} = \dfrac{y}{-2} = \dfrac{z+1}{1}$ is parallel to the plane $\pi: kx + 3y + z - 7 = 0$. Remember…
Find the value of the parameter $k$ so that the line $r: \dfrac{x-2}{3} = \dfrac{y}{-2} = \dfrac{z+1}{1}$ is parallel to the plane $\pi: kx + 3y + z - 7 = 0$. Remember that for a line and a plane to be parallel, the scalar product of their characteristic vectors must be equal to zero.
Two planes: parallel or perpendicular
Check whether the plane $\pi_1: 2x - 4y + 2z - 7 = 0$ and the plane $\pi_2: x - 2y + z + 5 = 0$ are parallel. To do this, analyse whether their normal vectors are…
Check whether the plane $\pi_1: 2x - 4y + 2z - 7 = 0$ and the plane $\pi_2: x - 2y + z + 5 = 0$ are parallel. To do this, analyse whether their normal vectors are proportional. If they are parallel, reason whether they are the same plane (coincident) or if they have no points in common.
Determine the relative position of the plane $\pi: mx + y - z = 1$ and the line $r \equiv \frac{x-1}{2} = \frac{y}{m} = \frac{z+3}{3}$ depending on the real parameter…
Determine the relative position of the plane $\pi: mx + y - z = 1$ and the line $r \equiv \frac{x-1}{2} = \frac{y}{m} = \frac{z+3}{3}$ depending on the real parameter $m$. Specify for which values they are perpendicular and for which the line is parallel to the plane.
Orthogonal projection of a point
An architect is designing a ramp modelled as a plane $\pi$ with general equation $x + y + z - 6 = 0$. To install a lighting sensor, they need to know the orthogonal…
An architect is designing a ramp modelled as a plane $\pi$ with general equation $x + y + z - 6 = 0$. To install a lighting sensor, they need to know the orthogonal projection of the point $P(1, 2, 5)$ onto that plane. Calculate the coordinates of the point $Q$ representing this projection (called the foot of the perpendicular).
Calculate the coordinates of the point $Q$, which is the orthogonal projection of the point $P(3, -1, 5)$ onto the line $r$ defined by the following equation in…
Calculate the coordinates of the point $Q$, which is the orthogonal projection of the point $P(3, -1, 5)$ onto the line $r$ defined by the following equation in continuous form:
$$r: \frac{x - 1}{2} = \frac{y + 2}{-1} = \frac{z}{3}$$Find the point $P'$ that represents the orthogonal projection of the point $P(1, 0, -3)$ onto the plane $\pi$ with equation: $$\pi: 2x - 3y + z - 4 = 0$$
Find the point $P'$ that represents the orthogonal projection of the point $P(1, 0, -3)$ onto the plane $\pi$ with equation:
$$\pi: 2x - 3y + z - 4 = 0$$Projection of a line: the projecting plane
The line $r$, which passes through the points $A(1, 0, 0)$ and $B(1, 1, 1)$, is projected onto the plane $\pi: x = 0$. 1. Find the orthogonal projection of the points…
The line $r$, which passes through the points $A(1, 0, 0)$ and $B(1, 1, 1)$, is projected onto the plane $\pi: x = 0$.
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Find the orthogonal projection of the points $A$ and $B$ onto the plane.
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Write the equation of the projected line $r'$ passing through these new points.
Obtain the implicit equation of the projecting plane of the line $r$ onto the plane $\pi$. Remember that the projecting plane is the plane perpendicular to $\pi$ that…
Obtain the implicit equation of the projecting plane of the line $r$ onto the plane $\pi$. Remember that the projecting plane is the plane perpendicular to $\pi$ that contains the line $r$.
$$r: \begin{cases} x - y + 2 = 0 \\ z = 1 \end{cases} \quad \text{and} \quad \pi: 2x + y - z = 5$$3D investigation: when a projection becomes a point
In a 3D geometry tool, create the plane $z=0$, the point $A=(0,0,2)$ and the line $r_a:(x,y,z)=(0,0,2)+t(a,0,1)$ with a slider a that includes 0. Predict what changes when a becomes zero. Construct two points of the line and their orthogonal projections onto the plane; rotate the view to tell an apparent crossing from a real intersection.
Now compare $r:(t,0,0)$ and $s:(0,u,h)$ with a slider h. For $h=0$ they meet at the origin; for $h\ne0$ they are skew: equating coordinates would require $t=0$, $u=0$ and $h=0$ simultaneously. Their distance is $|h|$. Hand in a table of cases, two screenshots and the proof; a single perspective on the screen does not prove that the lines meet.
This diagram distinguishes the two types of projection:
Back to the beginning: To run the cable perpendicular to the panel, you need the direction vector of the line to be proportional to the normal of the plane ($\vec v \parallel \vec n$); to check that the rail is parallel to the floor, their vectors must be perpendicular ($\vec v \cdot \vec n = 0$).