Which of he following vectors are parallel or perpendicular to (1, 1, -1)?
(2, 2, -2) d) (1, 0, 1)
(2, -2, 0) e)
(-2, 2, 2) f)
To determine whether a vector is parallel or perpendicular to another vector, we can use the dot product.
The dot product of two vectors is given by the sum of the products of their corresponding components. If the dot product is 0, then the vectors are perpendicular. If the dot product is nonzero, then the vectors are not perpendicular.
Let's calculate the dot product of each vector with (1, 1, -1).
a) (2, 2, -2) dot (1, 1, -1) = 2(1) + 2(1) + (-2)(-1) = 2 + 2 + 2 = 6
b) (1, 0, 1) dot (1, 1, -1) = 1(1) + 0(1) + 1(-1) = 1 - 1 = 0
c) (2, -2, 0) dot (1, 1, -1) = 2(1) + (-2)(1) + 0(-1) = 2 - 2 = 0
d) (-2, 2, 2) dot (1, 1, -1) = (-2)(1) + 2(1) + 2(-1) = -2 + 2 - 2 = -2
From the calculations above, we find that only vector (1, 0, 1) is perpendicular to (1, 1, -1). The dot products of vectors (2, 2, -2) and (-2, 2, 2) with (1, 1, -1) are nonzero, so they are not perpendicular. Therefore, vector (1, 0, 1) is the only vector that is perpendicular to (1, 1, -1).
To determine whether a vector is parallel or perpendicular to (1, 1, -1), we can use the dot product.
The dot product between two vectors u = (a, b, c) and v = (d, e, f) is given by:
u · v = ad + be + cf
If the dot product is zero, then the vectors are perpendicular. If the dot product is non-zero, then the vectors are parallel.
Let's calculate the dot product for each vector:
a) (2, 2, -2) · (1, 1, -1) = (2)(1) + (2)(1) + (-2)(-1) = 2 + 2 + 2 = 6
Since the dot product is non-zero, vector (2, 2, -2) is parallel to (1, 1, -1).
b) (2, -2, 0) · (1, 1, -1) = (2)(1) + (-2)(1) + (0)(-1) = 2 - 2 + 0 = 0
Since the dot product is zero, vector (2, -2, 0) is perpendicular to (1, 1, -1).
c) (-2, 2, 2) · (1, 1, -1) = (-2)(1) + (2)(1) + (2)(-1) = -2 + 2 - 2 = -2
Since the dot product is non-zero, vector (-2, 2, 2) is parallel to (1, 1, -1).
Therefore, the vectors that are parallel to (1, 1, -1) are (2,2,-2) and (-2,2,2), and the vector that is perpendicular to (1, 1, -1) is (2,-2,0).