LovesTheTeam.com

Instructions

1

Let two lines in the space be given by the canonical equations: (x-x1) / q1 = (y-y1) / w1 = (z-z1) / e1; (x-x2) / q2 = (y-y2) / w2 = ( z-z2) / e2.

2

The numbers q, w and e, represented in the denominators, are the coordinates of the directing vectors to these lines. A non-zero vector is referred to as a guide, which lies on this straight or is parallel to it.

3

The cosine of the angle between the lines has the formula: cosλ = ± (q1 · q2 + w1 · w2 + e1 · e2) / √ [(q1) ² + (w1) ² + (e1) ²] · [(q2) ² + (w2 ) ² + (e2) ²].

4

Direct, given by canonical equations,are mutually perpendicular if and only if their directing vectors are orthogonal. That is, the angle between the straight lines (which is the angle between the directing vectors) is 90 °. The cosine of the angle in this case is zero. Since the cosine is expressed by a fraction, its equality to zero is equivalent to the zero denominator. In coordinates this will be written as q1 · q2 + w1 · w2 + e1 · e2 = 0.

5

For straight lines in the plane, the chain of reasoning looks similar, but the perpendicularity condition will be slightly more simplified: q1 · q2 + w1 · w2 = 0, since the third coordinate is absent.

6

Now let the lines be given by the general equations: J1 · x + K1 · y + L1 · z = 0; J2 · x + K2 · y + L2 · z = 0.

7

Here, the coefficients J, K, L are the coordinates of the normal vectors. The normal is the unit vector perpendicular to straight.

8

The cosine of the angle between the lines is now written in the following form: cosλ = (J1 · J2 + K1 · K2 + L1 · L2) / √ [(J1) ² + (K1) ² + (L1) ²] · [(J2) ² + (K2) ² + (L2) ²].

9

The lines are mutually perpendicular in the case when the normal vectors are orthogonal. In vector form, respectively, this condition looks like this: J1 · J2 + K1 · K2 + L1 · L2 = 0.

10

The straight lines in the plane given by the general equations are perpendicular when J1 · J2 + K1 · K2 = 0.

Tip 2: How to find the equation of a straight line

It is often known that y depends linearly on x, and a graph of this dependence is given. In this case it is possible to learn the equation straight. First you need to select on straight two points.

Instructions

1

In the figure, we chose points A and B. It is convenient to choose the points of intersection with the axes. Two points are sufficient to pinpoint the line.

2

Find the coordinates of the selected points. To do this, lower the perpendiculars from the points on the coordinate axes and write down the numbers from the scale. So for point B from our example, the coordinate x is -2, and the coordinate y is 0. Similarly, for point A, the coordinates are (2; 3).

3

It is known that the equation straight has the form y = kx + b. We substitute in the equation in the general form, the coordinates of the selected points, then for point A we obtain the equation: 3 = 2k + b. For point B we get another the equation: 0 = -2k + b. Obviously, we have a system of two equations with two unknowns: k and b.

4

Next, we solve the system in any convenient way. In our case, we can add the equations of the system, since the unknown k occurs in both equations with coefficients that are the same in absolute value but opposite in sign. Then we get 3 + 0 = 2k - 2k + b + b, or, which is the same: 3 = 2b. Thus, b = 3/2. We substitute the found value of b into any of the equations to find k. Then 0 = -2k + 3/2, k = 3/4.

5

We substitute the found k and b in the equation general form and obtain the desired the equation straight: y = 3x / 4 + 3/2.