13. planes in ๐น - arizona state universitysurgent/mat267/examples/planes_r3.pdfย ยท intersecting...
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13. Planes in ๐น๐
Given a point ๐0 = (๐ฅ0, ๐ฆ0, ๐ง0) and a vector ๐ง = โจ๐, ๐, ๐โฉ, a plane that passes through ๐0 and is
normal (orthogonal) to n has the equation
๐(๐ฅ โ ๐ฅ0) + ๐(๐ฆ โ ๐ฆ0) + ๐(๐ง โ ๐ง0) = 0,
which simplifies to ๐๐ฅ + ๐๐ฆ + ๐๐ง = ๐, where ๐ = ๐๐ฅ0 + ๐๐ฆ0 + ๐๐ง0.
Example 13.1: Find the equation of the plane passing through ๐0 = (โ3,9,1) and normal to
๐ง = โจ7,3, โ5โฉ.
Solution: The plane has the equation
7(๐ฅ โ (โ3)) + 3(๐ฆ โ 9) โ 5(๐ง โ 1) = 0.
Simplifying, we have
7(๐ฅ + 3) + 3(๐ฆ โ 9) โ 5(๐ง โ 1) = 0 7๐ฅ + 21 + 3๐ฆ โ 27 โ 5๐ง + 5 = 0
7๐ฅ + 3๐ฆ โ 5๐ง = 1.
Example 13.2: State a vector that is normal to the plane 2๐ฅ โ 4๐ฆ + 3๐ง = 12.
Solution: There are infinitely-many possible vectors. One is โจ2, โ4,3โฉ, which is found by reading
the coefficients of x, y and z. Any non-zero multiple of โจ2, โ4,3โฉ is also a correct answer.
Example 13.3: Find the equation of the line normal to the plane โ2๐ฅ + 7๐ฆ + 4๐ง = โ7 that passes
through the point (1, โ3,5) on the plane.
Solution: The line is parallel to (in the direction of) ๐ง = โจโ2,7,4โฉ, so the line is given by
โจ๐ฅ, ๐ฆ, ๐งโฉ = โจ1, โ3,5โฉ + ๐กโจโ2,7,4โฉ = โจ1 โ 2๐ก, โ3 + 7๐ก, 5 + 4๐กโฉ.
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Example 13.4: Find the point of intersection of the line โจ๐ฅ, ๐ฆ, ๐งโฉ = โจ4 โ ๐ก, 2 + 3๐ก, 3 โ 5๐กโฉ and the
plane 6๐ฅ + 2๐ฆ โ 3๐ง = 10.
Solution: Substitute the equations for x, y and z into the plane, and solve for ๐ก:
6(4 โ ๐ก) + 2(2 + 3๐ก) โ 3(3 โ 5๐ก) = 10 24 โ 6๐ก + 4 + 6๐ก โ 9 + 15๐ก = 10
15๐ก = โ9
๐ก = โ3
5 .
The point of intersection is found by substituting ๐ก = โ3
5 into the equations for the line:
โจ๐ฅ, ๐ฆ, ๐งโฉ = โจ4 โ (โ3
5) , 2 + 3 (โ
3
5) , 3 โ 5 (โ
3
5)โฉ = โจ
23
5,1
5, 6โฉ.
This is a vector. However, if it is referenced from the origin, its head lies at the point (23
5,
1
5, 6).
Example 13.5: Find the equation of the plane passing through the points ๐ด = (1,3,4),
๐ต = (โ3,2,6) and ๐ถ = (1,0, โ6).
Solution: From the three points, form two vectors. For example, vectors AB and AC:
๐๐ = โจโ4, โ1,2โฉ, ๐๐ = โจ0, โ3, โ10โฉ
Next, find a vector ๐ง normal to AB and AC by finding the cross product ๐๐ ร ๐๐:
๐ง = ๐๐ ร ๐๐ = โจ16, โ40,12โฉ
Any non-zero multiple of ๐ง will suffice, so we can divide through by 4, getting ๐ง = โจ4, โ10,3โฉ. Using any one of the three points, we now find the equation of the plane. Weโll use ๐ด first, then
check our work with ๐ต and ๐ถ. Using ๐ด = (1,3,4) and ๐ง = โจ4, โ10,3โฉ, we have
4(๐ฅ โ 1) โ 10(๐ฆ โ 3) + 3(๐ง โ 4) = 0 4๐ฅ โ 4 โ 10๐ฆ + 30 + 3๐ง โ 12 = 0
4๐ฅ โ 10๐ฆ + 3๐ง = โ14.
Verify this is correct by substituting the coordinates for points B and C into the equation:
๐ต = (โ3,2,6): 4(โ3) โ 10(2) + 3(6) = โ12 โ 20 + 18 = โ14, true. ๐ถ = (1,0, โ6): 4(1) โ 10(0) + 3(โ6) = 4 โ 18 = โ14, true.
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Example 13.6: Show that the equation of a plane passing through the axis-intercepts (๐, 0,0),
(0, ๐, 0) and (0,0, ๐) is ๐ฅ
๐+
๐ฆ
๐+
๐ง
๐= 1.
Solution: Form two vectors:
The vector from (๐, 0,0) to (0, ๐, 0) is ๐ฎ = โจ0 โ ๐, ๐ โ 0,0 โ 0โฉ = โจโ๐, ๐, 0โฉ. The vector from (0, ๐, 0) to (0,0, ๐) is ๐ฏ = โจ0 โ 0,0 โ ๐, ๐ โ 0โฉ = โจ0, โ๐, ๐โฉ.
Their cross product is ๐ฎ ร ๐ฏ = โจ๐๐, ๐๐, ๐๐โฉ, and using (๐, 0,0) as a point in the plane, the planeโs
equation is
๐๐(๐ฅ โ ๐) + ๐๐(๐ฆ โ 0) + ๐๐(๐ง โ 0) = 0
๐๐๐ฅ โ ๐๐๐ + ๐๐๐ฆ + ๐๐๐ง = 0
๐๐๐ฅ + ๐๐๐ฆ + ๐๐๐ง = ๐๐๐.
Now, divide through by ๐๐๐:
๐๐๐ฅ
๐๐๐+
๐๐๐ฆ
๐๐๐+
๐๐๐ง
๐๐๐=
๐๐๐
๐๐๐, which simplifies as
๐ฅ
๐+
๐ฆ
๐+
๐ง
๐= 1.
This is a handy formula for finding the equation of a plane given its axis-intercepts.
Example 13.7: Find the angle formed by the planes ๐ฅ + 3๐ฆ โ 2๐ง = 5 and 4๐ฅ โ ๐ฆ + 5๐ง = โ2.
Solution: The respective normal vectors of each plane are ๐ง1 = โจ1,3, โ2โฉ and ๐ง2 = โจ4, โ1,5โฉ.
The angle between these two planes is the same as the angle between the two normal vectors:
๐ = cosโ1 (๐ง1 โ ๐ง2
|๐ง1||๐ง2|) = cosโ1 (
โ9
โ14 โ 42) โ 111.79 degrees
However, planes always intersect at an acute angle (except in the case where they are orthogonal).
The preferred answer is the supplement: 180 โ 111.79 = 68.21 degrees.
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Example 13.8: Find the acute angle formed by vector ๐ฏ = โจ1,4,3โฉ and plane 2๐ฅ + ๐ฆ + 5๐ง = 4.
Solution: The planeโs normal vector is ๐ง = โจ2,1,5โฉ, so the angle between ๐ฏ and ๐ง is
๐ = cosโ1 (๐ฏ โ ๐ง
|๐ฏ||๐ง|) = cosโ1 (
21
โ26 โ 30) โ 41.24ยฐ.
Visually, ๐ฏ and ๐ง are on the same side of the plane, as their angle is acute. Note that 41.24 is the
angle between ๐ฏ and ๐ง, so the angle between ๐ฏ and the plane is the complement, 90 โ 41.24 =
48.76.
Find the angle between v and n first, then find the complement.
Example 13.9: Find the acute angle formed by vector ๐ฏ = โจ2,1, โ8โฉ and plane ๐ฅ + 2๐ฆ + ๐ง = 4.
Solution: The planeโs normal vector is ๐ง = โจ1,2,1โฉ, so the angle between ๐ฏ and ๐ง is
๐ = cosโ1 (๐ฏ โ ๐ง
|๐ฏ||๐ง|) = cosโ1 (
โ4
โ69 โ 6) โ 101.34ยฐ.
In this example, the vector ๐ง is on one side of the plane, and ๐ฏ on the opposite side. Since the angle
from ๐ง to the plane is 90, the remainder, 101.34 โ 90 = 11.34, is the desired angle.
Vectors v and n lie on opposite sides of the plane, so the acute angle v makes
with the plane is found by subtracting 90 degrees.
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Another way to view this is to use ๐ง = โจโ1, โ2, โ1โฉ, which would have given us
๐ = cosโ1 (๐ฏ โ ๐ง
|๐ฏ||๐ง|) = cosโ1 (
4
โ69 โ 6) โ 78.66ยฐ.
Thus, ๐ฏ and ๐ง are on the same side of the plane, and the complement gives us 90 โ 78.66 =
11.34, the desired angle.
Example 13.10: Find the parametric equation of the line formed by the intersection of the planes
๐ฅ โ 3๐ฆ + 2๐ง = โ16 and 2๐ฅ + ๐ฆ โ 7๐ง = 18.
Solution: The planes are not parallel since their normal vectors, ๐ง1 = โจ1, โ3,2โฉ and ๐ง2 =โจ2,1, โ7โฉ, are not scalar multiples of one another. Had the planes been parallel, they would not
intersect (unless they were the same coincident plane). To find the parametric equation of the line,
we need a point on this line of intersection, and a direction vector parallel to the line of intersection.
For a point on the line of intersection, set one variable to 0, and solve for the other two variables.
Let x = 0 in both equations. This results in a system
โ3๐ฆ + 2๐ง = โ16
๐ฆ โ 7๐ง = 18.
The solution is ๐ฆ = 4 and ๐ง = โ2. Thus, (0,4, โ2) is a point on the line of intersection of the two
planes.
For the direction vector, we note that it will be orthogonal to both normal vectors of each plane.
Thus, we can find a vector parallel to the line of intersection by finding the cross product of the
two normal vectors. We get ๐ง1 ร ๐ง2 = โจ19,11,7โฉ. Therefore, the line of intersection of the two
planes is given parametrically by
โจ๐ฅ, ๐ฆ, ๐งโฉ = โจ19๐ก, 4 + 11๐ก, โ2 + 7๐กโฉ.
Example 13.11: Find the equation of the plane passing through ๐0 = (7,3, โ11) and parallel to
the plane 2๐ฅ โ ๐ฆ + 6๐ง = 1.
Solution: The two planes are parallel, so they share the same normal vector ๐ง = โจ2, โ1,6โฉ. Thus,
the planeโs equation is
2(๐ฅ โ 7) โ 1(๐ฆ โ 3) + 6(๐ง โ (โ11)) = 0
2๐ฅ โ 14 โ ๐ฆ + 3 + 6๐ง + 66 = 0 2๐ฅ โ ๐ฆ + 6๐ง = โ55.
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Parametric Representation of a Plane
A plane can also be represented parametrically in vector form. Assume that a plane passes through
๐0 = (๐ฅ0, ๐ฆ0, ๐ง0) and that ๐ฏ1 and ๐ฏ2 are two (non-parallel) vectors that lie within the plane. The
plane can then be written as
โจ๐ฅ, ๐ฆ, ๐งโฉ = ๐ฏ0 + ๐ ๐ฏ1 + ๐ก๐ฏ2,
where ๐ and ๐ก are independent parameter variables that may assume any real-number value, and
that ๐ฏ0 is the vector whose head points to ๐0 when ๐ = 0 and ๐ก = 0.
Example 13.12: Write the plane 3๐ฅ + 2๐ฆ + ๐ง = 5 in parametric form.
Solution: Any of the three variables can be isolated, leaving one dependent variable and two
independent variables. In this case, we isolate z:
๐ง = 5 โ 3๐ฅ โ 2๐ฆ.
Now, let ๐ฅ = ๐ and ๐ฆ = ๐ก. We now have
๐ฅ = ๐
๐ฆ = ๐ก
๐ง = 5 โ 3๐ โ 2๐ก.
Observe what happens when we โzero-fillโ the missing terms:
๐ฅ = 0 + 1๐ + 0๐ก
๐ฆ = 0 + 0๐ + 1๐ก
๐ง = 5 โ 3๐ โ 2๐ก.
The vectors ๐ฏ0, ๐ฏ1 and ๐ฏ2 can be inferred from the columns seen above:
[๐ฅ๐ฆ๐ง
] = [005
] + ๐ [10
โ3] + ๐ก [
01
โ2].
Thus, the plane 3๐ฅ + 2๐ฆ + ๐ง = 5 written in parametric form is
โจ๐ฅ, ๐ฆ, ๐งโฉ = ๐ฏ0 + ๐ ๐ฏ1 + ๐ก๐ฏ2 = โจ0,0,5โฉ + ๐ โจ1,0, โ3โฉ + ๐กโจ0,1, โ2โฉ.
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Example 13.13: Rewrite the plane given by โจ๐ฅ, ๐ฆ, ๐งโฉ = โจ0,0,2โฉ + ๐ โจ1,0,4โฉ + ๐กโจ0,1, โ7โฉ into the
form ๐๐ฅ + ๐๐ฆ + ๐๐ง = ๐.
Solution: Reading across component by component, we have
๐ฅ = 0 + 1๐ + 0๐ก
๐ฆ = 0 + 0๐ + 1๐ก
๐ง = 2 + 4๐ โ 7๐ก.
In this case, we have ๐ฅ = ๐ and ๐ฆ = ๐ก, so the last equation can be written as
๐ง = 2 + 4๐ฅ โ 7๐ฆ.
Thus, the plane โจ๐ฅ, ๐ฆ, ๐งโฉ = โจ0,0,2โฉ + ๐ โจ1,0,4โฉ + ๐กโจ0,1, โ7โฉ can be written as โ4๐ฅ + 7๐ฆ + ๐ง = 2.
Example 13.14: Rewrite the plane given by โจ๐ฅ, ๐ฆ, ๐งโฉ = โจ1,3,4โฉ + ๐ โจ2, โ1,5โฉ + ๐กโจโ3,8,1โฉ into the
form ๐๐ฅ + ๐๐ฆ + ๐๐ง = ๐.
Solution: Reading across component by component, we have
๐ฅ = 1 + 2๐ โ 3๐ก (๐)
๐ฆ = 3 โ ๐ + 8๐ก (๐)
๐ง = 4 + 5๐ + ๐ก. (๐)
We need to write ๐ and ๐ก in terms of ๐ฅ and ๐ฆ. From the first two equations (1) and (2), we have
2๐ โ 3๐ก = ๐ฅ โ 1
โ๐ + 8๐ก = ๐ฆ โ 3.
We can eliminate ๐ by multiplying the bottom row by 2, then adding:
2๐ โ 3๐ก = ๐ฅ โ 1
โ2๐ + 16๐ก = 2๐ฆ โ 6,
which gives 13๐ก = ๐ฅ + 2๐ฆ โ 7. Thus, ๐ก =1
13(๐ฅ + 2๐ฆ โ 7). In a similar way, we can eliminate ๐ก,
in which case we have ๐ =1
13(8๐ฅ + 3๐ฆ โ 17). These are substituted into equation (3):
๐ง = 4 + 5๐ + ๐ก = 4 + 5 (1
13(8๐ฅ + 3๐ฆ โ 17)) + (
1
13(๐ฅ + 2๐ฆ โ 7)).
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This simplifies to
๐ง =41
13๐ฅ +
17
13๐ฆ โ
40
13 .
Multiplying by 13 to clear fractions, we have 13๐ง = 41๐ฅ + 17๐ฆ โ 40. Thus, the plane can be
written as 41๐ฅ + 17๐ฆ โ 13๐ง = 40.
Example 13.15: Find the equation of the plane passing through ๐0 = (โ3,9,1) and normal to
๐ง = โจ7,3, โ5โฉ. (This is a repeat of Example 13.1)
Solution: We need two vectors, ๐ฏ1 and ๐ฏ2, in the plane. These two vectors will be orthogonal to
๐ง. For example, if we choose the x-component of ๐ฏ1 to be 0, then ๐ฏ1 = โจ0,5,3โฉ is orthogonal to n
and hence, in the plane. In a similar way, we could let the y-component of ๐ฏ2 be 0, and so ๐ฏ2 =โจ5,0,7โฉ is orthogonal to n and also in the plane.
Noting that ๐ฏ0 corresponds to point ๐0 = (โ3,9,1), we have
โจ๐ฅ, ๐ฆ, ๐งโฉ = ๐ฏ0 + ๐ ๐ฏ1 + ๐ก๐ฏ2 = โจโ3,9,1โฉ + ๐ โจ0,5,3โฉ + ๐กโจ5,0,7โฉ.
Is this the same as 7๐ฅ + 3๐ฆ โ 5๐ง = 1, the result from Example 13.1? Letโs find out. Reading
component-wise, we have
๐ฅ = โ3 + 0๐ + 5๐ก
๐ฆ = 9 + 5๐ + 0๐ก
๐ง = 1 + 3๐ + 7๐ก.
The first equation gives ๐ก =1
5(๐ฅ + 3), and the second equation gives ๐ =
1
5(๐ฆ โ 9). These are
substituted into the third equation:
๐ง = 1 + 3๐ + 7๐ก = 1 + 3 (1
5(๐ฆ โ 9)) + 7 (
1
5(๐ฅ + 3)).
Clearing parentheses, we have
๐ง =7
5๐ฅ +
3
5๐ฆ โ
1
5 .
Multiplying by 5, we have 5๐ง = 7๐ฅ + 3๐ฆ โ 1. Rearranging terms, we do have 7๐ฅ + 3๐ฆ โ 5๐ง = 1.
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14. Distances in ๐น๐
We use projections to find the shortest distances between a point and a line, between two non-
intersecting lines, between a point and a plane, and between two parallel planes. The shortest
distance is defined to be a line that meets the objects orthogonally.
Example 14.1: Find the shortest distance between the line โจ๐ฅ, ๐ฆ, ๐งโฉ = โจ1,2, โ1โฉ + ๐กโจ2, โ3,4โฉ and
the point ๐ = (4,8,3).
Solution: Choose any point ๐ on the line. For example, when ๐ก = 0, we have ๐ = (1,2, โ1).
Then find the vector from ๐ to ๐, which is ๐ฐ = โจ4 โ 1,8 โ 2,3 โ (โ1)โฉ = โจ3,6,4โฉ. Meanwhile,
the directional vector of the line is ๐ฏ = โจ2, โ3,4โฉ. The projection of w onto v is:
proj๐ฏ ๐ฐ =๐ฏ โ ๐ฐ
๐ฏ โ ๐ฏ๐ฏ =
4
29๐ฏ = โจ
8
29, โ
12
29,16
29โฉ.
The normal vector is found by subtracting proj๐ฏ ๐ฐ from w:
norm๐ฏ ๐ฐ = ๐ฐ โ proj๐ฏ ๐ฐ = โจ3,6,4โฉ โ โจ8
29, โ
12
29,16
29โฉ = โจ
79
29,186
29,100
29โฉ.
The magnitude of this normal vector is the distance from the point ๐ to the line:
|norm๐ฏ ๐ฐ| = โ(79
29)
2
+ (186
29)
2
+ (100
29)
2
โ 7.78 units.
Example 14.2: Find the shortest distance between point ๐ = (1,4,3) and plane ๐ฅ โ 3๐ฆ + 2๐ง = 6.
Solution: Pick any point ๐ in the plane by choosing values for two of the variables and solving
for the third. If ๐ฅ = 0 and ๐ฆ = 0, we get ๐ง = 3, so a point in the plane is ๐ = (0,0,3). The vector
v from ๐ to ๐ is ๐ฏ = โจ1,4,0โฉ. The normal vector to the plane is ๐ง = โจ1, โ3,2โฉ. Project v onto n:
proj๐ง๐ฏ =๐ฏ โ ๐ง
๐ง โ ๐ง๐ง = โ
11
14โจ1, โ3,2โฉ.
The magnitude of this vector is the distance from ๐ = (1,4,3) to the plane ๐ฅ โ 3๐ฆ + 2๐ง = 6:
|proj๐ง๐ฏ| =11
14โ12 + (โ3)2 + 22 =
11
14โ14 โ 2.94 units.
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A Pictorial Guide to Finding the Shortest Distance Between a Point and a Line
1.
Start with a line L, a point Q not on
L, and a direction vector v
parallel to L.
2.
Pick any point P on L.
3.
Form a vector w from P to Q.
4.
Find the projection of w onto v.
5.
Find a vector normal to the
projection, i.e. completing
a right triangle.
6.
The distance from Q to L is the
magnitude of this normal vector.
When a vector w is projected onto a vector v, this produces a vector called proj๐ฏ ๐ฐ. This can be
viewed as one leg of a right triangle, with w being the hypotenuse. The other leg is called norm๐ฏ ๐ฐ.
In vector addition, we have:
๐ฐ = proj๐ฏ ๐ฐ + norm๐ฏ ๐ฐ (Remember, these are vectors.)
Their magnitudes are related by the Pythagorean Theorem:
|๐ฐ|๐ = |proj๐ฏ ๐ฐ|๐ + |norm๐ฏ ๐ฐ|๐ (These are scalars.)
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A Pictorial Guide to Finding the Shortest Distance Between a Point and a Plane
1.
Start with a plane and a point Q
not on the plane.
2.
Find n, the normal vector to the
plane.
3.
Find any point P on the plane, and
then form a vector v from P to Q.
4.
If it helps, you can visualize the
vectors v and n having a common
foot.
5.
Project v onto n.
6.
The distance from Q to the plane is
the magnitude of projn v.
Example 14.3: Find the shortest distance between the parallel planes
๐ฅ โ 3๐ฆ + 2๐ง = 6 and โ2๐ฅ + 6๐ฆ โ 4๐ง = 1.
Solution: Note that the vector ๐ง = โจ1, โ3,2โฉ is orthogonal to both planes. Choose any point on
each plane. From the plane ๐ฅ โ 3๐ฆ + 2๐ง = 6, we use the point ๐ด = (0,0,3), and from the plane
โ2๐ฅ + 6๐ฆ โ 4๐ง = 1, we use ๐ต = (โ1
2, 0, 0). The vector v from ๐ด to ๐ต is ๐ฏ = โจ
1
2, 0, 3โฉ. We then
project v onto n:
proj๐ง๐ฏ =๐ฏ โ ๐ง
๐ง โ ๐ง๐ง =
13
28โจ1, โ3,2โฉ.
The magnitude of this vector is the shortest distance between the two planes:
|proj๐ง๐ฏ| = โ(13
28)
2
+ (โ39
28)
2
+ (26
28)
2
โ 1.737 units.
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Example 14.4: Find the shortest distance between the lines ๐ฟ1: โจ๐ฅ, ๐ฆ, ๐งโฉ = โจ1, โ2,6โฉ + ๐กโจ1, โ5,4โฉ
and ๐ฟ2: โจ๐ฅ, ๐ฆ, ๐งโฉ = โจ0,8,1โฉ + ๐ โจ4, โ7,3โฉ
Solution: These lines are skew. Thus, they will not intersect. Two skew lines in ๐ 3 can always be
placed within two parallel planes, so that this example is nearly identical to the previous example.
From the first line, we have a point ๐ด = (1, โ2,6) and from the second line, we have a point ๐ต =(0,8,1), and the vector v from ๐ด to ๐ต is ๐ฏ = โจโ1,10, โ5โฉ.
The normal vector n to the two planes that each contain one of the lines is found by finding the
cross product of the two direction vectors of each line. From the first line, we have ๐ฏ1 = โจ1, โ5,4โฉ and from the second line, we have ๐ฏ2 = โจ4, โ7,3โฉ. Their cross product is
๐ง = ๐ฏ1 ร ๐ฏ2 = โจ13, 13, 13โฉ.
However, we can use any non-zero multiple of n for this example, so ๐ง = โจ1,1,1โฉ will suffice,
since the projection step will extend n to the desired length. Now project v onto n:
proj๐ง๐ฏ =๐ฏ โ ๐ง
๐ง โ ๐ง๐ง =
4
3โจ1,1,1โฉ.
The distance is the magnitude of this projection vector:
|proj๐ง๐ฏ| =4
3โ12 + 12 + 12 โ 2.309 units.
ยฉ 2009-2016 Scott Surgent ([email protected])