Here. are some discussions about the Lines and Planes so Read the Context given below:

Unlike a plane, a line in three dimensions does have an obvious direction, namely, the direction of any vector parallel to it. A line can be defined and uniquely identified by providing one point on the line and a vector parallel to the line (in one of two possible directions) that is, the line consists of exactly those points we can reach by starting at the point and going for some distance in the direction of the vector. Let’s see how we can translate this into more mathematical language.

Lines and planes are probably the easiest curves and places in a three-dimensional area and they will seem important as we seek to understand curves and more complex areas.

The number of lines in two dimensions is ax + by = c; it is reasonable to expect that a line in three dimensions is given an ax + by + cz = d; it makes sense, but it’s not right – it turns out that this is the plane number.

The plane does not have a clear “direction” as does the line It is possible that the plane is aligned with the direction in a very useful way, however: there are two direct directions to the aircraft. Any vector with one of these two indicators is said to be normal in a plane.

Suppose two points (v1, v2, v3) and (w1, w2, w3) are in the plane; then the vector ⟨w1 – v1, w2 – v2, w3 – v3⟩ is like a plane; especially, when this vector is placed with its tail in (v1, v2, v3) its head is in (w1, w2, w3) and lies on the plane. As a result, any vector relative to the plane corresponds to ⟨w1 – v1, w2 – v2, w3 – v3⟩. It is easy to see that an airplane contains those points (w1, w2, w3) where w1 – v1, w2 – v2, w3 – v3⟩ corresponds to the average in aircraft, as shown in Figure 12.5.1.That is, suppose we know that ⟨a, b, c⟩ is common in a plane containing a point (v1, v2, v3).Then (x, y, z) is in the plane when and when Ophelia, b, c⟩ corresponds to ⟨x – v1, y – v2, z – v3⟩. Next, we know that this is especially true when ⟨a, b, c⟩⋅⟨x – v1, y – v2, z – v3⟩ = 0. So, (x, y, z) you are on a plane if and only if ⟨A, b, c⟩⋅⟨x – v1, y – v2, z – v3⟩a (x – v1) + b (y – v2) + c (z – v3) ax + ngo + cz – av1 – bv2 −cv3ax + by + cz = 0 = 0 = 0 = av1 + bv2 + cv3. To work backwards, note that if (x, y, z) is a point that satisfies the ax + by + cz = d then ax + ngu + czax + ngu + cz – da (x – d / a) + b (y – 0) + c (z – 0) ⟨a, b, c⟩⋅⟨x – d / a, y, z ⟩ = D = 0 = 0 = 0.

That is, ⟨a, b, c⟩ corresponds to the verb with the tail in (d / a, 0,0) and the head in (x, y, z). This means that points (x, y, z) satisfy the equation ax + in + cz = d forming a plane in ngokuyaa, b, c⟩. (This does not apply if = 0, but in that case we can use b or c in the paragraph of. That is, it can be (x – 0) + b (y – d / b) + c (z −0) = 0 or a (x – 0) + b (y – 0) + c (z – d / c) = 0.)

Figure 12.5.1. Vehicle defined by vectors according to standard.

Therefore, when given the vector ⟨a, b, c⟩ we know that all planes corresponding to this vector have ax + form + by + cz = d, and any part of this form is a plane pointing to ⟨a, b, c ⟩.

In single variable calculus, we find that a unique function is intrinsically linear. In other words, if we zoom in on the map of the variance function at some point, the map looks like a tangent to the function at that point. Line functions play an important role in single variable calculus, which is useful in accessing variance functions and roots of functions (see Newton’s method) and in arriving at first-order solutions of variance equations (see Euler’s method). In Miscellaneous Calculus, we’re going to study curves in space; Curves also differ in the form of a local line. Furthermore, when we analyze the functions of the two variables, we can see that if the surface defined by the function looks like an enlarged plane (the touch plane), then such a function is analogous to a local.

Consequently, we need to understand both lines and planes in space because it also defines linear functions. (Note that if a function is a polyhedron, all its terms are less than or equal to the numerator. For example, this refers to a single variable linear function and a two-variable linear function, but not a linear one because both In, the sum of the failures of its factors.) We begin our work from scratch with familiar ideas.

The intersection of two planes

Lines and Planes

If two planes intersect, the junction is always dashed the vector equation for the line of intersection is given

r = r_0 + tvr = r0+ TV

where r_0r 0

A point on the line and vv is the result of the transverse vector product of the normal vectors of the two planes.

The parameter equations for the line of intersection are given

x = ax = a, y = by = b, and z = cz = c

The vector equations aa, bb and cc are r = a\bold i + b\bold j + c\bold kr = ai + bj + ck.

Finding the parameter equations representing the line of intersection of two planes

In Parameters, an example of the problem of how to find the fort where two planes meet

For example

Find the parameter equation for the line of intersection of the planes.

2x + y-z = 32x + y-z = 3

x-y + z = 3x-y + z = 3

We have to find the vector equation of the line of intersection to achieve this, we first need to find the cross-product vv of the normal vectors of a given plane.

normal vectors for planes

2x + y -z = 32x + y -z = 3 For the plane, the normal vector \ langle2,1, -1 \ rangelea⟨2,1, −1⟩

x -y + z = 3x -y + z = 3 For the plane, the normal vector b \langle1, -1,1 \rangleb⟨1, −1,1⟩

The cross product of the normal vectors is we need a point at the intersection to achieve this, we will use the given plane equations as the system of linear equations. If we set z = 0z = 0 in both the equations, we get

2x + y-z = 32x + y-z = 3

2x + y-0 = 32x + y-0 = 3

2x + y = 32x + y = 3


x-y + z = 3x-y + z = 3

x-y + 0 = 3x-y + 0 = 3

X-Y = 3X-Y = 3

Intersection parameter line of two planes. Look at the line of intersection, find a point in the first line, and find the cross product of the normal vectors Now we will put the equations together.

(2x + x) + (y – y) = 3 + 3 (2x + x) + (y – y) = 3 + 3

3x + 0 = 63x + 0 = 6

x = 2x = 2

If we put x = 2x = 2 again in x-y = 3x-y = 3, we get

2-y = 32-y = 3

-y = 1 -y = 1

y = -1y = -1

Combine these values ​​and the point on the line of intersection

(2, -1,0) (2, 1,0)

r_0=2\bold i-\bold j+0\bold kr

0= 2i – j + 0k

r_0 = \ lang 2, -1,0 \ rangler

0= 2, −1,0⟩

Now we have vv and r_0r. can mix

0 In the vector equation.

r = r_0 + tvr = r

0+ TV

r = (2\bold i-\bold j+0\bold k) + t (0\bold i-3\bold j-3\bold k) r = (2i-j + 0k) + t(0i-3j -3k)

r=2\bold i-\bold j+0\bold k+0\bold it-3\bold jt-3\bold ktr = 2i-j+0k+0it-3jt-3kt

r=2\bold i-\bold j-3\bold jt-3\bold ktr=2i-j-3jt-3kt

r = (2) \ bolus i + (-1-3t) \ bolus j + (-3t) \bold kr = (2) i + (-1−3t) j + (-3t) k

Using the vector equation for the hand cutting line, the parameter equations for both lines can be found. In our vector equation r = a \ turn i + b \ bold j + c \ bold kr = ai + bj + ck matches r = (2) \ turn i + (-1-3t) \bold j + (– 3t) ) \bold kr = (2) i + (- 1−3t) j + (- 3t) k, we can say that

a = 2a = 2

b = -1-3tb = -1−3t

c = -3tc = -3t

Therefore, the parameter equation for the line of intersection

x = 2x = 2

y = -1-3ty = -1−3t

z = -3tz = -3t


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