problem involving vectors, planes and points

**milkyway** · August 26th 2009, 01:53 AM

question:
1. according to the 'flat earth club', Earth is a plane described by the equation: x + y + z = 18.
also according to the flat earth club, earth will be destroyed on 31.8.09 by an explosion that will spontaneously occur at the so-called ' armageddon point' A = (1,1,1) in space. it so happens that melbourne (considered as a point) will be the first place on earth that is destroyed by this explosion.
a) calculate the coordinates of melbourne
b) what is the distance between A(the armageddon point) and melbourne. what is the distance between A and the earth?

2. a) according to the flat mars society. mars is also a plane, given by the equation: 4x +3y - z = -3.
find a parametric equation of the line in which earth (as in question 1) intersects mars.
b) how far away is canberra, given by the point (-5,10,13), from this line?

3. Two perfectly round pieces of rock are hurtling through space. The first, Superman’s
holiday asteroid, has radius 0.3, and is traveling on the line given by the parametric
equation
x(t) = 2 + t, y(t) = −1 − t, z(t) = t.
The second rock, made of kryptonite and set in motion by one of Superman’s enemies,
has radius 0.1 and is traveling on a line given by the parametric equation
x(s) = 3 − s, y(s) = 1, z(s) = 1 + s.
Calculate the distance between the two lines (8 marks) and use this distance to prove
that Superman has nothing to worry about. (2 marks)

**adkinsjr** · August 26th 2009, 06:09 PM

Originally Posted by milkyway

question:
1. according to the 'flat earth club', Earth is a plane described by the equation: x + y + z = 18.
also according to the flat earth club, earth will be destroyed on 31.8.09 by an explosion that will spontaneously occur at the so-called ' armageddon point' A = (1,1,1) in space. it so happens that melbourne (considered as a point) will be the first place on earth that is destroyed by this explosion.
a) calculate the coordinates of melbourne
b) what is the distance between A(the armageddon point) and melbourne. what is the distance between A and the earth?

2. a) according to the flat mars society. mars is also a plane, given by the equation: 4x +3y - z = -3.
find a parametric equation of the line in which earth (as in question 1) intersects mars.
b) how far away is canberra, given by the point (-5,10,13), from this line?

3. Two perfectly round pieces of rock are hurtling through space. The first, Superman’s
holiday asteroid, has radius 0.3, and is traveling on the line given by the parametric
equation
x(t) = 2 + t, y(t) = −1 − t, z(t) = t.
The second rock, made of kryptonite and set in motion by one of Superman’s enemies,
has radius 0.1 and is traveling on a line given by the parametric equation
x(s) = 3 − s, y(s) = 1, z(s) = 1 + s.
Calculate the distance between the two lines (8 marks) and use this distance to prove
that Superman has nothing to worry about. (2 marks)

lol, this is a unique one. I'll work on some of it. Since the equation of the plane is $x+y+z=18$ it is easy to find points. We know that the point $P(6,6,6)$

is on the plane. The armageddon point' A = (1,1,1) in space is on the terminal end of the vector $n=<1,1,1>$ normal to the plane. I think the melbourne point should be the point through which the normal vector will pass through the plane, since this would correspond to the shortest distance between melbourne and "A," Ths distance turns out to be:

$D=\frac{\mid21\mid}{\sqrt{3}}$

I got this from an elementry formula, I will not derive it here. I'll just write it down:

For the distance D between a point $P_o(x_1,y_1,z_1)$ and a plane $ax_1+by_1+cz_1+d =0$

$D=\frac{\mid ax_1+by_1+cz_1+d \mid}{\sqrt{a^2+b^2+c^2}}$ , where, in our case, ${a,b,c}=1$ are the coefficients of the terms of the plane $x+y+z -18=0$

The distance between $A$ and melbourne was calculated by the formula I wrote above. The distance between $A$ and Earth is the same as the distance between $A$ and melbourne since $A(1,1,1)$ can be expressed as a normal vector $n=<1,1,1>$

To find the melbourne point we can draw from the origin of our coordinate system, a vector $r_o$ with the terminal end meeting the plain at the melbourne point $P_o$ . and therefore, also meeting the tail of $n=<1,1,1>$ . We can draw another vector $r_1(6,6,6)$ from the origin to the point $P(6,6,6)$ . From $P_o$ we draw a vector $r_{\perp}$ . to $P(6,6,6)$ . Since $r_{\perp}=r_1 -r_o$ is on the surface of the plane, it is perpendicular to $n=<1,1,1>$ . We may write:

$n\cdot r_{\perp}=n\cdot (r_1-r_o)=0$ Now all we have to do is solve the vector equation for $r_o$ . I will leave that to you.

**aidan** · August 27th 2009, 04:28 AM

this is a unique one

Hello milkyway
The teacher who created the scenarios is brilliant.
It just makes it more appealing.
Very interesting.

**milkyway** · August 30th 2009, 12:13 AM

thanks! i figured it all out

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