Find two unit vectors each of which is perpendicular to vectors (1, 1, 0) and (1,0,1)
What I did:
a x b = (1, -1, -1)
|axb|=square toot 3
correct answer is +/- (1/√3 - 1/√3 - 1/√3)
or
=+/- 1/√3 (1, -1, -1)
To find two unit vectors that are perpendicular to both vectors (1, 1, 0) and (1, 0, 1), we can use the cross product of these two vectors.
The cross product of two vectors, A and B, is a vector that is perpendicular to both A and B. In this case, we can take the cross product of vectors (1, 1, 0) and (1, 0, 1) to find a vector that is perpendicular to both.
Let's calculate the cross product:
(1, 1, 0) x (1, 0, 1) = ((1*1) - (0*0), (0*1) - (1*1), (1*0) - (1*1))
= (1 - 0, 0 - 1, 0 - 1)
= (1, -1, -1)
Now, to find a unit vector in the direction of (1, -1, -1), we need to divide the vector by its magnitude. The magnitude of a vector (x, y, z) is given by the formula: sqrt(x^2 + y^2 + z^2).
Magnitude of (1, -1, -1) = sqrt(1^2 + (-1)^2 + (-1)^2)
= sqrt(1 + 1 + 1)
= sqrt(3)
To get a unit vector, divide the vector (1, -1, -1) by its magnitude:
(1/√3, -1/√3, -1/√3)
Therefore, one unit vector perpendicular to both vectors (1, 1, 0) and (1, 0, 1) is (1/√3, -1/√3, -1/√3).
To get another unit vector perpendicular to both vectors, you can simply change the signs of any two components. So, the other unit vector perpendicular to (1, 1, 0) and (1, 0, 1) can be (-1/√3, 1/√3, 1/√3).
Hence, the two unit vectors perpendicular to vectors (1, 1, 0) and (1, 0, 1) are (+/- 1/√3, -1/√3, -1/√3) and (+/- 1/√3, 1/√3, 1/√3).