if the position vectors of A,B,C are respectively 2i-j+k,i-3j-5k,3i-4j-4k then cos^2A is

To find cos^2A, we need to find the angle A first.

The position vectors of A, B, and C are given as: A = 2i - j + k B = i - 3j - 5k C = 3i - 4j - 4k

To find the angle between two vectors, we can use the dot product formula: cosθ = (A · B) / (|A| |B|)

First, let's find the dot product of vectors A and B: A · B = (2i - j + k) · (i - 3j - 5k) = 2i · i + 2i · (-3j) + 2i · (-5k) - j · i - j · (-3j) - j · (-5k) + k · i + k · (-3j) + k · (-5k) = 2 + (-6) + (-10) + 0 + 3 + 5 + 0 + 0 + 5 = -1

Next, let's find the magnitudes of vectors A and B: |A| = √(2^2 + (-1)^2 + 1^2) = √6 |B| = √(1^2 + (-3)^2 + (-5)^2) = √35

Now, we can calculate cosθ: cosθ = (-1) / (√6 √35)

Finally, to find cos^2A, we square the value of cosθ: cos^2A = [(-1) / (√6 √35)]^2

Therefore, cos^2A is equal to [(-1) / (√6 √35)]^2.