Find a unit vector perpendicular to a=(4,-3,1) and b=(2,3,-1)
What I did:
a x b = ((-3)(-1)-(3)(1), (1)(2)-(-1)(4), (4)(3) - (2)(-3))
=(0,6,18)
Answer is (0, 1/square root 10, 3/square root 10)
your initial vector is correct
magnitude of (0,6,18) = √(0+36+324) = √360
= 6√10
so I see that your unit vector is also correct
Perhaps you are looking at some answer key and the answer there looks different.
They might have rationalized the denominator
1/√10
= 1/√10 *√10/√10
= √10/10
3/√10 = 3√10/10
alternate form of your unit vector:
(0 , √10/10 , 3√10/10 )
To find a unit vector perpendicular to vectors a and b, you can use the cross product. The cross product of two vectors in three-dimensional space gives you a vector that is perpendicular to both of the original vectors.
First, let's calculate the cross product of a and b using the formula:
a x b = ((a2b3 - a3b2), (a3b1 - a1b3), (a1b2 - a2b1))
Plugging in the values for a = (4, -3, 1) and b = (2, 3, -1):
a x b = ((-3)(-1) - (3)(1), (1)(2) - (-1)(4), (4)(3) - (2)(-3))
= (0, 6, 18)
Now, to convert this vector into a unit vector, you need to divide each component by the magnitude of the vector.
The magnitude of vector v = (v1, v2, v3) is given by:
|v| = sqrt(v1^2 + v2^2 + v3^2)
In our case, the magnitude of vector (0, 6, 18) is:
|v| = sqrt(0^2 + 6^2 + 18^2)
= sqrt(0 + 36 + 324)
= sqrt(360)
= 6√10
Finally, divide each component of the vector (0, 6, 18) by the magnitude 6√10 to get the unit vector:
(0, 6/6√10, 18/6√10)
Simplifying the fractions:
(0, 1/√10, 3/√10)
Therefore, a unit vector perpendicular to vectors a and b is (0, 1/√10, 3/√10).