Knowing the angles between planes and lines is necessary when studying the properties of volumetric figures, such as pyramids or prisms. It is not difficult to calculate this angle. It is enough to know the equations of the line and the plane and the corresponding formula. Consider the issue of finding the marked angle in the article.
Mathematical description of a line and a plane
Both geometric objects mentioned in the title of the item are described by equations of different types. Here we will not give all of them, but only describe the vector expression for the line and the general equation for the plane, since it is these types of equalities that are convenient to use when calculating the angle between the line and the plane.
The equation in a vector form for a line in three-dimensional space has the following notation:
(x; y; z) = (x 0 ; y 0 ; z 0 ) + α * (a; b; c)
The first term in brackets is a certain known point that belongs to a line. The second term is the coordinates of a vector directed along a straight line. Parameter α can take any numerical values, due to which the coordinates of all points of this line are calculated.
The plane by the general equation is given in the following form:
A * x + B * y + C * z + D = 0
Uppercase Latin letters are certain fixed numbers. The advantage of this form of writing in relation to other types of expressions for a plane is that it is convenient to find the coordinates of a vector perpendicular to this two-dimensional object from it. These coordinates are equal to:
n¯ (A; B; C)
The recorded vector is called normal or guide for the plane.
Note that knowledge of this vector allows us to write a family of parallel planes that differ from each other by the free term D. To uniquely determine a plane, in addition to n¯, you need to know one more point that belongs to it.
Line and plane in space
The calculation of the angle between the line and the plane can be carried out, if we understand what options exist in principle in the mutual arrangement of these objects. There are only three of these options:
- the straight line of the plane is parallel, but does not lie in it;
- all points of the line are also points of the plane;
- line intersect with the plane.
The first two options correspond to an angle of 0 o between the considered geometric objects. In the case of intersection, the angle is different from zero, but it is always less than or equal to the right angle. If at the intersection of the plane and the line at one point, the angle is 90 o , then they are considered mutually perpendicular.
In the figure above, the line and the plane are parallel to each other, and in the diagram below they intersect.
The formula for the angle between a line and a plane
We obtain the formula for the quantity in question in general form. To do this, we rewrite the presented equations of the line and the plane:
(x; y; z) = (x 0 ; y 0 ; z 0 ) + λ * (a; b; c);
A * x + B * y + C * z + D = 0
For simplicity, suppose that a straight plane intersects. According to the definition, the angle between them is the angle between this line and its projection onto the plane. To obtain a projection, it is enough to lower the perpendicular from an arbitrary point on the plane to the plane, and then draw a line through the resulting point on the plane and the intersection point. The corresponding figure is shown below, where the symbol α denotes the desired angle.
The figure also shows the direction vector of the straight line v¯ and the normal n¯, the angle β between them is marked. The figure shows that these angles are connected with each other by the expression:
β + α = 90 o
How to find β, any student who is familiar with the properties of a scalar product can answer. To do this, it is enough to calculate it and divide by the product of the modules of the corresponding vectors, that is:
cos (β) = | (n¯ * v¯) | / (| n¯ | * | v¯ |)
Please note that the numerator contains the product module. This allows you to find only right angles of intersection.
From trigonometric formulas the following equality is known:
cos (β) = cos (90 o - α) = sin (α)
Then the desired angle can be calculated by the formula:
α = arcsin (| (n¯ * v¯) | / (| n¯ | * | v¯ |)
If we substitute the coordinates of the vectors for the line and plane written above, we get the final formula:
α = arcsin (| (A * a + B * b + C * c) | / (√ (A 2 + B 2 + C 2 ) * √ (a 2 + b 2 + c 2 ))
We show how to use it in solving problems.
Plane and line and the value of the angle of their intersection
It is necessary to find the angle between the line and the plane given by the expressions:
(x; y; z) = (1; 1; 0) + λ * (2; -1; 3);
x + y - 2z + 1 = 0
It is convenient to use the above formula for α if we preliminarily calculate the moduli of vectors and their scalar product. Let's do it:
n¯ (1; 1; -2);
v¯ (2; -1; 3);
(n¯ * v¯) = ((1; 1; -2) * (2; -1; 3)) = -5;
| n¯ | = √ (1 + 1 + 4) = √6;
| v¯ | = √ (4 + 1 + 9) = √14
Now the found values can be substituted into the formula for α:
α = arcsin (| -5 | / (√6 * √14)) = 33.06 o
Thus, we have shown that the plane and the line really intersect, and the angle between them is approximately equal to 33 o .
The intersection of a straight coordinate plane
Now we will solve such a problem. A straight line is given, which is defined as follows:
(x; y; z) = (1; 0; 0) + λ * (2; 0; -1)
It is necessary to find the angles of its intersection with three coordinate planes.
First you need to mathematically write down the expressions for the indicated planes. They have the form:
x = 0 (yz plane);
y = 0 (xz plane);
z = 0 (xy plane)
For each of them, we write the coordinates of the normal vector:
n¯ (1; 0; 0) for x = 0;
n¯ (0; 1; 0) for y = 0;
n¯ (0; 0; 1) for z = 0
It can be seen that the lengths of all normal vectors are equal to unity. We find scalar products for each of them with a directing vector of the line:
for x = 0: ((2; 0; -1) * (1; 0; 0)) = 2;
for y = 0: ((2; 0; -1) * (0; 1; 0)) = 0;
for z = 0: ((2; 0; -1) * (0; 0; 1)) = -1
The guide module for the direct vector is:
| (2; 0; -1) | = √5
We substitute the calculated values into the formula, we obtain the intersection angles:
with x = 0: α = arcsin (| 2 | / √5) ≈ 63.4 o ;
with y = 0: α = arcsin (| 0 | / √5) = 0 o ;
with z = 0: α = arcsin (| -1 | / √5) ≈ 26.6 o
Thus, the given line intersects only the yz and xy planes, and it is parallel to the xz plane.