you don't have to help me on the 2 questions, just one im happy enough, the thing is i do not understand vectors much, so theres no point of getting answers without understanding, but if you can do at least one question, with your knowledge i would be grateful :) thanks!!
1.
Prove that the line x = 1 + 2t, y = -1 + 3t, z = 2 + 4t is parallel to the plane x - 2y + z = 6. What is the distance between the line and the plane?
2.
Let A, B and C be three non-parallel vectors in R3. Express A x (B x C)
in the form α*B + β*C with scalars α and β :
(i) Explain why it is clear that A x (B x C) must be a linear combination
of B and C;
(ii) Choose an orthonormal basis ijk adapted to the problem and prove it by
standard properties of the vector product.
(Note: there is a way of choosing the orthonormal basis in a way which makes
all the calculations very easy
Vector Question?
2007-11-26 3:27 pm
回答 (3)
2007-11-26 5:46 pm
✔ 最佳答案
the vector of the line 〈 2 , 3, 4 〉the normal vector of the plane 〈 1 , -2 , 1 〉
it will be easy to show that their dot product is 0. thus the two vectors are perpendicular...
to get the distance... choose the point (1,-1,2) on the line...
the distance formula is
d = |(1) - 2(-1) + 2 - 6|/ √ (1² + (-2)² + 1²)
§
2007-11-27 11:11 am
1.
Prove that the line x = 1 + 2t, y = -1 + 3t, z = 2 + 4t is parallel to the plane x - 2y + z = 6. What is the distance between the line and the plane?
The directional vector v, of the line is:
v = <2, 3, 4>
The normal vector n, to the plane is:
n = <1, -2, 1>
If the line is parallel to the plane its vector will lie in the plane. This means it will be perpendicular to the normal vector of the plane. So the dot product of n and v should be zero.
v • n = <2, 3, 4> • <1, -2, 1> = 2 - 6 + 4 = 0
So the two vectors are perpendicular. Therefore the line is parallel to the plane. Since the line and plane are parallel, every point in the line is the same distance from the plane. So pick any point on the line and use the distance formula from a point to a plane. Let t = 0. Then a point on the line is:
P = P(1, -1, 2)
The equation of the plane is:
x - 2y + z = 6
x - 2y + z - 6 = 0
The distance of the point from the plane
distance = | 1*1 - 2*(-1) + 1*2 - 6 | / â[1² + (-2)² + 1²]
distance = | 1 + 2 + 2 - 6 | / â[1 + 4 + 1]
distance = 1 / â6
Prove that the line x = 1 + 2t, y = -1 + 3t, z = 2 + 4t is parallel to the plane x - 2y + z = 6. What is the distance between the line and the plane?
The directional vector v, of the line is:
v = <2, 3, 4>
The normal vector n, to the plane is:
n = <1, -2, 1>
If the line is parallel to the plane its vector will lie in the plane. This means it will be perpendicular to the normal vector of the plane. So the dot product of n and v should be zero.
v • n = <2, 3, 4> • <1, -2, 1> = 2 - 6 + 4 = 0
So the two vectors are perpendicular. Therefore the line is parallel to the plane. Since the line and plane are parallel, every point in the line is the same distance from the plane. So pick any point on the line and use the distance formula from a point to a plane. Let t = 0. Then a point on the line is:
P = P(1, -1, 2)
The equation of the plane is:
x - 2y + z = 6
x - 2y + z - 6 = 0
The distance of the point from the plane
distance = | 1*1 - 2*(-1) + 1*2 - 6 | / â[1² + (-2)² + 1²]
distance = | 1 + 2 + 2 - 6 | / â[1 + 4 + 1]
distance = 1 / â6
2007-11-26 3:57 pm
1. I forgot how to find the distance.. But i could do the first half
from x = 1 + 2t, y = -1 + 3t, z = 2 + 4t , you know that the vector thats runs along that line is v1 = 2i+3j+4k
also, from x - 2y + z = 6 , you know the vector that runs NORMAL to the plane is v2= i-2j+k.
By DOT PRODUCT v1 and v2, you get 2-6+4 = hence the two vectors are perpendicular to each other, i.e. the line is parallel to the plane.
from x = 1 + 2t, y = -1 + 3t, z = 2 + 4t , you know that the vector thats runs along that line is v1 = 2i+3j+4k
also, from x - 2y + z = 6 , you know the vector that runs NORMAL to the plane is v2= i-2j+k.
By DOT PRODUCT v1 and v2, you get 2-6+4 = hence the two vectors are perpendicular to each other, i.e. the line is parallel to the plane.
參考: limited knowledge of mself
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