A typical geometric problem is finding the angle between lines. On a plane, if the equations of lines are known, they can be drawn and the angle measured with a protractor. However, this method is laborious and not always possible. To find out the named angle, it is not necessary to draw straight lines, it can be calculated. This article will answer how this is done.
A straight line and its vector equation
Any straight line can be represented as a vector that starts at -∞ and ends at +∞. In this case, the vector passes through some point in space. Thus, all vectors that can be drawn between any two points on a straight line will be parallel to each other. This definition allows you to set the equation of a straight line in vector form:
(x; y; z)=(x0; y0; z0) + α(a; b; c)
Here, the vector with coordinates (a; b; c) is the guide for this line passing through the point (x0; y0; z0). The α parameter allows you to transfer the specified point to any other for this line. This equation is intuitive and easy to work with both in 3D space and on a plane. For a plane, it will not contain the z coordinates and the third direction vector component.
The convenience of performing calculations and studying the relative position of straight lines due to the use of a vector equation is due to the fact that its directing vector is known. Its coordinates are used to calculate the angle between lines and the distance between them.
General equation for a straight line on a plane
Let's write explicitly the vector equation of the straight line for the two-dimensional case. It looks like:
x=x0+ αa;
y=y0+ αb
Now we calculate the parameter α for each equality and equate the right parts of the obtained equalities:
α=(x - x0)/a;
α=(y - y0)/b;
(x - x0)/a=(y - y0)/b
Opening the brackets and transferring all terms to one side of equality, we get:
1/ax +(-1/b)y+y0/b- x0/a=0=>
Ax + By + C=0, where A=1/a, B=-1/b, C=y0/b- x 0/a
The resulting expression is called the general equation for a straight line given in two-dimensional space (in three-dimensional this equation corresponds to a plane parallel to the z-axis, not a straight line).
If we explicitly write y through x in this expression, then we get the following form, knowneach student:
y=kx + p, where k=-A/B, p=-C/B
This linear equation uniquely defines a straight line on the plane. It is very easy to draw it according to the well-known equation, for this you should put x=0 and y=0 in turn, mark the corresponding points in the coordinate system and draw a straight line connecting the obtained points.
Formula of the angle between lines
On a plane, two lines can either intersect or be parallel to each other. In space, to these options is added the possibility of the existence of skew lines. Whatever version of the relative position of these one-dimensional geometric objects is implemented, the angle between them can always be determined by the following formula:
φ=arccos(|(v1¯v2¯)|/(|v1 ¯||v2¯|))
Where v1¯ and v2¯ are the guide vectors for line 1 and 2 respectively. The numerator is the modulus of the dot product to exclude obtuse angles and take into account only sharp ones.
The vectors v1¯ and v2¯ can be given by two or three coordinates, while the formula for the angle φ remains unchanged.
Parallelism and perpendicularity of lines
If the angle between 2 lines calculated by the formula above is 0o, then they are said to be parallel. To determine whether the lines are parallel or not, you can not calculate the angleφ, it suffices to show that one direction vector can be represented through a similar vector of another line, that is:
v1¯=qv2¯
Here q is some real number.
If the equations of lines are given as:
y=k1x + p1,
y=k2x + p2,
then they will be parallel only when the coefficients of x are equal, that is:
k1=k2
This fact can be proved if we consider how the coefficient k is expressed in terms of the coordinates of the directing vector of the straight line.
If the angle of intersection between lines is 90o, then they are called perpendicular. To determine the perpendicularity of lines, it is also not necessary to calculate the angle φ, for this it is enough to calculate only the scalar product of the vectors v1¯ and v2¯. It must be zero.
In the case of intersecting straight lines in space, the formula for the angle φ can also be used. In this case, the result should be correctly interpreted. The calculated φ shows the angle between the direction vectors of lines that do not intersect and are not parallel.
Task 1. Perpendicular lines
It is known that the equations of lines have the form:
(x; y)=(1; 2) + α(1; 2);
(x; y)=(-4; 7) + β(-4; 2)
It is necessary to determine whether these lines areperpendicular.
As mentioned above, to answer the question, it is enough to calculate the scalar product of the vectors of the guides, which correspond to the coordinates (1; 2) and (-4; 2). We have:
(1; 2)(-4; 2)=1(-4) + 22=0
Since we got 0, this means that the considered lines intersect at a right angle, that is, they are perpendicular.
Task 2. Line intersection angle
It is known that two equations for straight lines have the following form:
y=2x - 1;
y=-x + 3
It is necessary to find the angle between the lines.
Since the coefficients of x have different values, these lines are not parallel. To find the angle that is formed when they intersect, we translate each of the equations into a vector form.
For the first line we get:
(x; y)=(x; 2x - 1)
On the right side of the equation, we got a vector whose coordinates depend on x. Let's represent it as a sum of two vectors, and the coordinates of the first will contain the variable x, and the coordinates of the second will consist exclusively of numbers:
(x; y)=(x; 2x) + (0; - 1)=x(1; 2) + (0; - 1)
Since x takes arbitrary values, it can be replaced by the parameter α. The vector equation for the first line becomes:
(x; y)=(0; - 1) + α(1; 2)
We do the same actions with the second equation of the line, we get:
(x; y)=(x; -x + 3)=(x; -x) + (0; 3)=x(1; -1) + (0; 3)=>
(x; y)=(0; 3) + β(1; -1)
We rewrote the original equations in vector form. Now you can use the formula for the angle of intersection, substituting in it the coordinates of the directing vectors of the lines:
(1; 2)(1; -1)=-1;
|(1; 2)|=√5;
|(1; -1)|=√2;
φ=arccos(|-1|/(√5√2))=71, 565o
Thus, the lines under consideration intersect at an angle of 71.565o, or 1.249 radians.
This problem could have been solved differently. To do this, it was necessary to take two arbitrary points of each straight line, compose direct vectors from them, and then use the formula for φ.
