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Use two methods to determine a unit vector perpendicular to both (2,1,-3) and (-1, 7, 4)
Using Cross Product:
Let v be a perpendicular vector to the given vectors
v = a*b
= (2,1,-3)(-1,7,4)
=((1)(4)-(7)(-3), (-3)(-1)-(4)(2), (2)(7)-(-1)(1))
=(4+21, 3-8, 14+1)
=(25, -5, 15)
= (5, -1, 3)
= |(5,-1, 3)|
= sqr(25+1+9
=sqr(35)
Let u be a unit vector in the direction of (5, -1, 3)
u = v/|v| = 1/sqr(35)(5, -1, 3) = (5/sqr(35), -1/sqr(35), 3/sqr(35))
Using dot product:
Let p = (x, y, z) represent the vector that is perpendicular to the vectors.
a = (2, 1, -3) and b = (-1, 7, 4)
Then p*a = 0 and p*b = 0
(x, y, z)(2, 1, -3) = 0 and (x, y, z)(-1, 7, 4) = 0
2x + y - 3z = 0
-x + 7y + 4Z = 0
Isolate x in equation (1) to get x = -y/2 + 3z/2
Substitute this into equation (2)
-(-y/2 + 3z/2) + 7y +4z = 0
y/2 - 3z/2 +7y +4z = 0
z = 0 --> substitute into equation (1) gives 2x + y = 0 and so x = -y/2
If you let x = 1, then y = -2
Therefore p = (x, y, z) = (1, -2, 0), which gives a different vector from the one found using cross product.
Determine the magnitude of this vector and then find a unit vector as required.
|(1, -2, 0)| = sqr(1+4) = sqr(5)
Let u be a unit vector in the direction of (1, -2, 0)
u = v/|v| = 1/sqr(5)(1, -2, 0) = (1/sqr(5), -2/sqr(5), 0)
The two answers should match up but they don't and I am not sure what I am doing wrong. I have been over it a number of times but I just can not see where the error lies. Any help would be greatly appreciated
Using Cross Product:
Let v be a perpendicular vector to the given vectors
v = a*b
= (2,1,-3)(-1,7,4)
=((1)(4)-(7)(-3), (-3)(-1)-(4)(2), (2)(7)-(-1)(1))
=(4+21, 3-8, 14+1)
=(25, -5, 15)
= (5, -1, 3)
= |(5,-1, 3)|
= sqr(25+1+9
=sqr(35)
Let u be a unit vector in the direction of (5, -1, 3)
u = v/|v| = 1/sqr(35)(5, -1, 3) = (5/sqr(35), -1/sqr(35), 3/sqr(35))
Using dot product:
Let p = (x, y, z) represent the vector that is perpendicular to the vectors.
a = (2, 1, -3) and b = (-1, 7, 4)
Then p*a = 0 and p*b = 0
(x, y, z)(2, 1, -3) = 0 and (x, y, z)(-1, 7, 4) = 0
2x + y - 3z = 0
-x + 7y + 4Z = 0
Isolate x in equation (1) to get x = -y/2 + 3z/2
Substitute this into equation (2)
-(-y/2 + 3z/2) + 7y +4z = 0
y/2 - 3z/2 +7y +4z = 0
z = 0 --> substitute into equation (1) gives 2x + y = 0 and so x = -y/2
If you let x = 1, then y = -2
Therefore p = (x, y, z) = (1, -2, 0), which gives a different vector from the one found using cross product.
Determine the magnitude of this vector and then find a unit vector as required.
|(1, -2, 0)| = sqr(1+4) = sqr(5)
Let u be a unit vector in the direction of (1, -2, 0)
u = v/|v| = 1/sqr(5)(1, -2, 0) = (1/sqr(5), -2/sqr(5), 0)
The two answers should match up but they don't and I am not sure what I am doing wrong. I have been over it a number of times but I just can not see where the error lies. Any help would be greatly appreciated