Find a non-zero vector orthogonal to (1, 2, -1)
To find a vector orthogonal to a given vector, we can take the cross product of the given vector with any arbitrary vector. Let's choose the vector (1, 0, 0) as our arbitrary vector.
The cross product of vectors (a, b, c) and (x, y, z) is given by:
(a, b, c) × (x, y, z) = (bz - cy, cx - az, ay - bx)
Using this formula, we can calculate the cross product of (1, 2, -1) and (1, 0, 0):
(1, 2, -1) × (1, 0, 0) = (2(0) - (-1)(0), (-1)(1) - 1(0), 1(0) - 2(1))
= (0, -1, -2)
Therefore, the vector (0, -1, -2) is orthogonal to (1, 2, -1).
To find a vector that is orthogonal (or perpendicular) to a given vector, we can use the fact that the dot product of two orthogonal vectors is zero.
Given the vector v = (1, 2, -1), we need to find a vector u = (x, y, z) that is orthogonal to v.
To find such a vector, we can use the dot product property and set the dot product of v and u to zero:
v · u = 0
(1 * x) + (2 * y) + (-1 * z) = 0
x + 2y - z = 0
Since we need a non-zero vector u, we can choose arbitrary values for x, y, and z as long as they satisfy the equation above. Let's set x = 1 and y = 0:
1 + 2(0) - z = 0
1 - z = 0
z = 1
Therefore, a vector orthogonal to (1, 2, -1) is (1, 0, 1).