Find a unit vector r such that r x (-6i + 3j + 2k) = 0 vector (null vector)
If the vector u = ai+bj+ck, then we have
|i j k|
|a b c| = (2b-3c)i + (-6c-2a)j + (3a+6b)k = 0i+0j+0k
|-6 3 2|
so now solve those three equations for a,b,c
To find a unit vector r such that r x (-6i + 3j + 2k) is equal to the zero vector (null vector), we can use the property of cross product that says if two vectors are parallel, then their cross product is equal to the zero vector.
The given vector (-6i + 3j + 2k) has the coefficients of i, j, and k as -6, 3, and 2 respectively.
Let's assume the unit vector r is ai + bj + ck, where a, b, and c are the coefficients we need to find.
Taking the cross product of r and (-6i + 3j + 2k), we get:
r x (-6i + 3j + 2k) = (bi - cj) * (-6i + 3j + 2k) + (ck - ai) * (-6i + 3j + 2k) + (aj - bk) * (-6i + 3j + 2k)
Expanding the cross product:
r x (-6i + 3j + 2k) = -6bi - 3cj - 2ck + 6cj - 3ak + 2ai - 6aj + 3bk - 2bk
Combining like terms:
r x (-6i + 3j + 2k) = (2a - 3b) i + (-6b - 3a) j + (-2c + 6c) k
For this result to be the zero vector, all the coefficients must be equal to zero:
2a - 3b = 0
-6b - 3a = 0
-2c + 6c = 0
Simplifying these equations:
2a = 3b --> Equation 1
-3a = -6b --> Equation 2
4c = 0
From Equation 2, we can solve for a in terms of b:
a = 2b
Substituting this value of a in Equation 1:
2(2b) = 3b
4b = 3b
b = 0
From Equation 2:
-3a = -6(0)
a = 0
From Equation 3:
4c = 0
c = 0
Therefore, the coefficients of the unit vector r are a = 0, b = 0, and c = 0.
So, the unit vector r is the zero vector: r = 0i + 0j + 0k.