Let the vector u=[2,-1,2] and vector v= sq7/2sq5,sq7/sq5,3/2] be two vectors in three dimensional space.... sq means square root...
(i) Find the angle between vectors u and v
(i) Find the unit vectors that are orthogonal to both vector u and v...meaning perpendicular to those two vectors ...
I'd really appreciate step by step explanation on how to do these questions so I know how to approach them on tests...this question is in a book I bought but no solution on it since its one of those odd numbered ones. Thank you!
*****WHat are the Two unit vectors in the answer that i would provide??? Thats where im stuck now with my answer******
(i) There are various ways to find the angles, one of which is using the scalar product.
We know that the scalar product of u and v can be computed in two ways:
First way: u.v = xx' + yy' + zz' where (x,y,z) are the coordinates of u and (x',y',z') are the coordinates of v.
Second way: u.v = |u|.|v|.cos ? where |u| and |v| are the magnitudes of u and v and ? is the angle between the two vectors.
So we conclude that xx' + yy' + zz' = |u|.|v|.cos ?
Thus, cos ? = (xx' + yy' + zz') / (|u|.|v|)
We have xx' + yy' + zz' = 2*?7 / (2 ?5) - ?7 / ?5 + 2 * 3/2 = 3
Let's find |u| and |v|
|u| = ?(x^2 + y^2 + z^2) = ?(4 + 1 + 4) = 3
|v| = ?(x' ^2 + y' ^2 + z' ^2) = ?(7/20 + 7/5 + 9/4) = ?(80/20) = 2
So back to cos ? = 3 / (3*2) = 1/2
So what is the angle that has a cosine of 1/2? ?/3 or -?/3. The bottom line is that it's 60 degrees.
(ii)
We can use the cross product because we know that a vector perpendicular to both u and v will be perpendicular to the plane containing u and v. But we don't have to; we can also use scalar product in the following way
Let w = (x, y, z) represent an arbitrary vector orthogonal to both u and v.
We know that the the scalar product of u and w and that of v and w are both zero since u and w, and v and w form a 90 degree angle.
So u.w = 0 and v.w = 0
So 2x - y + 2z = 0
And ?7 / (2 ?5) x + (?7 / ?5) y + (3/2) z = 0
So w must satisfy both of these equation. So to find what is w we need to solve them. But these are 2 equations with 3 unknowns, so what we'll do is we'll express x and y in terms of z
Multiply the second equation by - 4 ?5 / ?7 to get:
-2x - 4y - 6 ?5/?7 z = 0
Now add the first equation to it:
- 5y + (2 - 6?5 / ?7) z = 0
=> y = (2/5 - 6 / ?35) z
Replace this expression of y in the first equation:
2x - (2/5 - 6 / ?35) z + 2z = 0
=> x = (1/5 - 3 / ?35 - 2) z
x = (-9/5 - 3 / ?35) z
(check my calculations)
So w is of the form:
w = [ (-9/5 - 3 / ?35) z, (2/5 - 6 / ?35) z, z]
So it looks like :
w = z* [ (-9/5 - 3 / ?35), (2/5 - 6 / ?35), 1]
Where z is a parameter. That is, for any value of z, w will still be orthogonal (maybe not zero?, doesn't matter).
Know let's find for which value of z is this vector w a unit vector:
We want: magnitude of w = 1
So:
|z| ?[(-9/5 - 3 / ?35)^2 + (2/5 - 6 / ?35)^2 + 1] = 1
Get some nasty value of |z| => two values of z => two vector w
(i) Find the angle between vectors u and v
(i) Find the unit vectors that are orthogonal to both vector u and v...meaning perpendicular to those two vectors ...
I'd really appreciate step by step explanation on how to do these questions so I know how to approach them on tests...this question is in a book I bought but no solution on it since its one of those odd numbered ones. Thank you!
*****WHat are the Two unit vectors in the answer that i would provide??? Thats where im stuck now with my answer******
(i) There are various ways to find the angles, one of which is using the scalar product.
We know that the scalar product of u and v can be computed in two ways:
First way: u.v = xx' + yy' + zz' where (x,y,z) are the coordinates of u and (x',y',z') are the coordinates of v.
Second way: u.v = |u|.|v|.cos ? where |u| and |v| are the magnitudes of u and v and ? is the angle between the two vectors.
So we conclude that xx' + yy' + zz' = |u|.|v|.cos ?
Thus, cos ? = (xx' + yy' + zz') / (|u|.|v|)
We have xx' + yy' + zz' = 2*?7 / (2 ?5) - ?7 / ?5 + 2 * 3/2 = 3
Let's find |u| and |v|
|u| = ?(x^2 + y^2 + z^2) = ?(4 + 1 + 4) = 3
|v| = ?(x' ^2 + y' ^2 + z' ^2) = ?(7/20 + 7/5 + 9/4) = ?(80/20) = 2
So back to cos ? = 3 / (3*2) = 1/2
So what is the angle that has a cosine of 1/2? ?/3 or -?/3. The bottom line is that it's 60 degrees.
(ii)
We can use the cross product because we know that a vector perpendicular to both u and v will be perpendicular to the plane containing u and v. But we don't have to; we can also use scalar product in the following way
Let w = (x, y, z) represent an arbitrary vector orthogonal to both u and v.
We know that the the scalar product of u and w and that of v and w are both zero since u and w, and v and w form a 90 degree angle.
So u.w = 0 and v.w = 0
So 2x - y + 2z = 0
And ?7 / (2 ?5) x + (?7 / ?5) y + (3/2) z = 0
So w must satisfy both of these equation. So to find what is w we need to solve them. But these are 2 equations with 3 unknowns, so what we'll do is we'll express x and y in terms of z
Multiply the second equation by - 4 ?5 / ?7 to get:
-2x - 4y - 6 ?5/?7 z = 0
Now add the first equation to it:
- 5y + (2 - 6?5 / ?7) z = 0
=> y = (2/5 - 6 / ?35) z
Replace this expression of y in the first equation:
2x - (2/5 - 6 / ?35) z + 2z = 0
=> x = (1/5 - 3 / ?35 - 2) z
x = (-9/5 - 3 / ?35) z
(check my calculations)
So w is of the form:
w = [ (-9/5 - 3 / ?35) z, (2/5 - 6 / ?35) z, z]
So it looks like :
w = z* [ (-9/5 - 3 / ?35), (2/5 - 6 / ?35), 1]
Where z is a parameter. That is, for any value of z, w will still be orthogonal (maybe not zero?, doesn't matter).
Know let's find for which value of z is this vector w a unit vector:
We want: magnitude of w = 1
So:
|z| ?[(-9/5 - 3 / ?35)^2 + (2/5 - 6 / ?35)^2 + 1] = 1
Get some nasty value of |z| => two values of z => two vector w