Find the unit vector perpendicular to each of the vectors 2i+j+k and i-j+2k?
does "cross product" or "vector product" ring a bell?
the cross product of two vectors is perpendicular to both
+i +j +k
+2 +1 +1
+1 -1 +2
(2+1)i +(1-4)j + (-2-1)k
= 3i -3j -3 k
To find the unit vector perpendicular to two vectors, you can use the cross product (also known as the vector product) of the given vectors.
The cross product of two vectors, A and B, denoted as A × B, is a vector that is perpendicular (or orthogonal) to both A and B.
Let's calculate the cross product of the given vectors:
Vector A = 2i + j + k
Vector B = i - j + 2k
To find the cross product A × B, we can use the following formula:
A × B = (AyBz - AzBy)i + (AzBx - AxBz)j + (AxBy - AyBx)k
Now, let's substitute the values:
Ax = 2, Ay = 1, Az = 1 (from vector A)
Bx = 1, By = -1, Bz = 2 (from vector B)
A × B = ((1)(2) - (1)(-1))i + ((1)(1) - (2)(2))j + ((2)(-1) - (1)(1))k
= (2 + 1)i + (1 - 4)j + (-2 - 1)k
= 3i - 3j - 3k
So, the cross product A × B is equal to 3i - 3j - 3k.
To obtain the unit vector (also known as a normalized vector) perpendicular to A and B, we divide the cross product by its magnitude. The magnitude of a vector is the square root of the sum of the squares of its components.
The magnitude of 3i - 3j - 3k is √(3^2 + (-3)^2 + (-3)^2) = √27 = 3√3.
Therefore, the unit vector perpendicular to the given vectors is:
(3i - 3j - 3k) / (3√3)
Simplifying, we get:
(1/√3)i - (1/√3)j - (1/√3)k
Hence, the unit vector perpendicular to vectors 2i + j + k and i - j + 2k is (1/√3)i - (1/√3)j - (1/√3)k.