Find Perpendicular Direction Vector for (1, 5, -1)

[masterofthewave124]

is there a quick way to find a perpendicular direction vector to the direction vector (1, 5, -1)?

for example in 2D, i know you just switch the coordinates and the sign of one of them.

 

[StatusX]

There is a whole plane of vectors perpendicular to a given vector in 3D. Do you know how to find the equation of this plane? Another way would be to take the cross product of this vector with any other vector (as long as it isn't a scalar multiple of this one).

 

[masterofthewave124]

yeah there's infinite vectors perpendicular to this one so i would have to employ one of the techniques you mentioned? no quick way?

 

[HallsofIvy]

Do you understand what "a whole plane of vectors" means? What StatusX is saying is that there is NO single answer to this no matter what method you use!

Of course, since two vectors are perpendicular if and only if their dot product is 0, if you really need just "a perpendicular vector" as you say, you can "switch the coordinates and the sign of one of them" as long as you take the third coordinate 0:
$(1, 5, -1)\cdot(1, 0 1)= 0$ so (1, 0, 1) is perpendicular to (1, 5, -1).

It's also true that $(1, 5, -1)\cdot(5, -1, 0)= 0$ so (5, -1, 0) is also perpendicular to (1, 5, -1).

Also, $(1, 5, -1)\cdot(0, 1, 5)= 0$ so (0, 1, 5) is perpendicular to (1, 5, -1). Do you see the point?

 

[masterofthewave124]

yep thanks that's what i was looking for!

the entire question was: find an equation that passes through the point (2,6,5) and is perpendicular to the line (x, y, z) = (5, 2, 4) + t(1, 5, -1).

so it didn't really matter which perpendicular vector i had, as long as i had a direction vector for my line equation.

 

[masterofthewave124]

ok another question,

given the 3 equations:
3x−3y−2z=14
5x+y−6z=10
x−2y+4z=9

a) show that the three planes intersect at a single point
b) find the coordinates of the intersection point

for a), should i go on to prove that the normals of each vector are not parallel nor coplanar.

so n1 = (3,-3,-2), n2 = (5,1,-6), n3 = (1,-2,4) and then go on to show that n1 and n2 are not scalar multiples; similarly n1 and n3 and n2 and n3. to show that they are not coplanar, i just have to prove that n1 cannot be written as a combination of the other two. is there a simpler method that could combine a) and b) because for b) i would have to go through a long set of matrixes ending up with the same conclusion, that they interesect a single point...

 

[StatusX]

I'm not sure exactly what they're after for the first part. Clearly by finding the intersection point you are showing one (and only one) such point exists, and doing anything else would be redundant. But if they insist you first use a quicker method to verify the point exists, the only one I can think of is finding the determinant of the corresponding matrix. If this is non-zero, the matrix is invertible, and so there is a solution.

 

[HallsofIvy]

yep thanks that's what i was looking for!

the entire question was: find an equation that passes through the point (2,6,5) and is perpendicular to the line (x, y, z) = (5, 2, 4) + t(1, 5, -1).

so it didn't really matter which perpendicular vector i had, as long as i had a direction vector for my line equation.

I hope that wasn't the actual question! An equation doesn't pass through any points- it just sits there on the paper! Presumably the question was "Find an equation of the line that passes through the point (2, 6, 5) and is perpendicular to the line (x,y,z)= (5,2,4)+ t(1,5,-1).
Unfortunately, just finding a perpendicular to the line doesn't help you- it may not contain (2,6,5).
The plane that contains the point (2,6,5) and is perpendicular to that line has equation 1(x-2)+ 5(y- 6)- (z-5)= 0 since a normal vector to the plane is parallel to the line. Replace x by 5+ t, y by 2+ 5t, and z by 4- t in that and solve for t to find the point on the given line in that plane. Now you know two points your perpendicular line must go through.

 

[HallsofIvy]

ok another question,

given the 3 equations:
3x−3y−2z=14
5x+y−6z=10
x−2y+4z=9

a) show that the three planes intersect at a single point
b) find the coordinates of the intersection point

for a), should i go on to prove that the normals of each vector are not parallel nor coplanar.

so n1 = (3,-3,-2), n2 = (5,1,-6), n3 = (1,-2,4) and then go on to show that n1 and n2 are not scalar multiples; similarly n1 and n3 and n2 and n3. to show that they are not coplanar, i just have to prove that n1 cannot be written as a combination of the other two. is there a simpler method that could combine a) and b) because for b) i would have to go through a long set of matrixes ending up with the same conclusion, that they interesect a single point...

I don't think you need to worry about vectors at all. You have 3 equations of 3 planes- 3 equations in 3 unknowns. Just solve the three equations simultaneously. By doing b) you have proved a).

 

[masterofthewave124]

I hope that wasn't the actual question! An equation doesn't pass through any points- it just sits there on the paper! Presumably the question was "Find an equation of the line that passes through the point (2, 6, 5) and is perpendicular to the line (x,y,z)= (5,2,4)+ t(1,5,-1).
Unfortunately, just finding a perpendicular to the line doesn't help you- it may not contain (2,6,5).
The plane that contains the point (2,6,5) and is perpendicular to that line has equation 1(x-2)+ 5(y- 6)- (z-5)= 0 since a normal vector to the plane is parallel to the line. Replace x by 5+ t, y by 2+ 5t, and z by 4- t in that and solve for t to find the point on the given line in that plane. Now you know two points your perpendicular line must go through.

that was kind of hard to follow. would it make a difference if i told you that the question asked for a vector equation? by definition, all you need to form a VE is a point and a direction vector.

so (x,y,z) = (2,6,5) + t(5,-1,0) wouldn't be a correct VE? (there are no stipulations that say t can't be 0).

 

[masterofthewave124]

wrt my other question, yes i think combining is the most efficient way to do it.1. In the yz-plane find:
a) a vector equation of the line 3y+2z=6
b) a vector equation of the line y=(3/4)z−2

i'm having some difficulty with this question, mostly because aren't both those equations already in the yz plane?also,

2. Determine the equation of the plane in the form Ax + By + Cz + D = 0, that passes though the point P(6, −1, 1), has z-intercept of −4, and is parallel to the line (x−1)/2 = (y+1)/2 = z/-1

to find the cartesian plane vector, i have (A,B,C) = (2,2,-1) and i have a point (6, −1, 1) i can use to sub in and find a value for D. what's the significance of the z-intercept?

 

[arunbg]

1)The reason they specified yz plane is because in 3D the given equations are those of a plane, and not a line. This is simply 2D .You can do the question by simply finding two points on this line ( substitution would be easiest) and writing down the vector equation as in 3D

2)A z intercept of -4 tells us that the plane passes through the point
(0,0,-4). Can you take it from here ?

Arun

 

[HallsofIvy]

that was kind of hard to follow. would it make a difference if i told you that the question asked for a vector equation? by definition, all you need to form a VE is a point and a direction vector.

so (x,y,z) = (2,6,5) + t(5,-1,0) wouldn't be a correct VE? (there are no stipulations that say t can't be 0).

"Vector equation" or "scalar equation" makes no difference.

One more time:

There is a whole plane of vectors perpendicular to a given vector in 3D.

Any reason for choosing (5, -1, 0) out of the infinite number of vectors that are perpendicular to the given line?

Where do the lines given by (x,y,z) = (2,6,5) + t(5,-1,0) and (x, y, z) = (5, 2, 4) + t(1, 5, -1) intersect? So as not to confuse the two parameters, rewrite the first as (x,y,z) = (2,6,5) + s(5,-1,0). At the point of intersection we have x= 2+ 5s= 5+ t, y= 6- s= 4+ 5t, and z= 5= 4-t.
From the last, t= 4- 5= -1. From the first, then, 2+ 5s= 5-1= 4 so 5s= 2 and s= 2/5. But then y= 6-(2/5)= 28/5 which is not 4+ 5(-1)= -1.

The two lines do not intersect. Yes, there is a line in the direction of the vector (5, -1, 0) and so perpendicular to the given line but it does not pass through (2, 6, 5). The line (2,6,5)+ t(5,-1,0) is parallel to that line and skew to the given line.

The plane that contains the point (2,6,5) and is perpendicular to that line has equation 1(x-2)+ 5(y- 6)- (z-5)= 0 since a normal vector to the plane is parallel to the line. Replace x by 5+ t, y by 2+ 5t, and z by 4- t in that and solve for t to find the point on the given line in that plane.

Okay, making those replacements, (5+t- 2)+ 5(2+5t- 6)- (4-t-5)= (t+ 3)+ (25t-20)+ (t+ 1)= 27t- 15= 0 so t= 15/27= 5/9. That means that x= 5+ 5/9= 50/9, y= 2+ 5(5/9)= 43/9, z= 4- 5/9= 31/9 is the point where the given line crosses the perpendicular plane containing (2, 6, 5)- the point where the perpendicular from (2, 6, 5) intersects the given line.
Find the vector equation of the line containing (2, 6, 5) and (50/9, 43/9, 31/9).

 

[HallsofIvy]

wrt my other question, yes i think combining is the most efficient way to do it.


1. In the yz-plane find:
a) a vector equation of the line 3y+2z=6
b) a vector equation of the line y=(3/4)z−2

i'm having some difficulty with this question, mostly because aren't both those equations already in the yz plane?

arunbg has already answered this but I will also point out that finding the "vector equation of a line" simply means writing the y and z (and x in 3D) coordinates in terms of a single parameter t. In b, since y is already written as a function of z, you can simply take z itself as a parameter:
x= 0, y= (3/4)t- 2, z= t or (x,y,z)= (0,-2,0)+ t(0,3/4,1).

also,

2. Determine the equation of the plane in the form Ax + By + Cz + D = 0, that passes though the point P(6, −1, 1), has z-intercept of −4, and is parallel to the line (x−1)/2 = (y+1)/2 = z/-1

to find the cartesian plane vector, i have (A,B,C) = (2,2,-1) and i have a point (6, −1, 1) i can use to sub in and find a value for D. what's the significance of the z-intercept?

No, you do not have (A,B,C)= (2,2,-1)! That vector is parallel to the given line and so parallel to the plane. You want (A,B,C) perpendicular to the plane. You know that (2,2,-1) is parallel to the plane and that, since the plane contains both (6, -1, -1) and (0, 0, -4) (the z-intercept), the vector (6-0, -1-0, -1-(-4))= (6, -1, 3) is in the plane (and so parallel to the plane). A vector perpendicular to the plane must be perpendicular to both of those. How do you find a vector perpendicular to both of two given vectors?

 

[masterofthewave124]

thank you guys, i have finished these questions successfully.

Given the plane equations p1: 3x + 2y – 7z = 3 and p2: 4x – 5y + z = 2
a) Find 3 times equation 1 plus (-2) times equation 2.
b) What is the geometric significance of the equation found in part (a)?

i found a) to be x + 16y - 23z = 5

for b), is the geometric significance that all three of these planes will intersect at the same point?