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 vector of A is given as 2i - j + k.

To find the magnitude of vector A, we can use the formula |A| = sqrt(Ax^2 + Ay^2 + Az^2), where Ax, Ay, and Az are the components of vector A.

In this case, Ax = 2, Ay = -1, and Az = 1.

So, |A| = sqrt((2)^2 + (-1)^2 + (1)^2) = sqrt(4 + 1 + 1) = sqrt(6).

Next, we need to find the dot product of vectors A and B.

The position vector of B is given as i - 3j - 5k.

The dot product of two vectors A and B is given by the formula A · B = Ax * Bx + Ay * By + Az * Bz, where Bx, By, and Bz are the components of vector B.

In this case, Bx = 1, By = -3, and Bz = -5.

So, A · B = (2 * 1) + (-1 * -3) + (1 * -5) = 2 + 3 - 5 = 0.

Now, we can find the magnitude of vector B using the same formula as before.

|B| = sqrt((1)^2 + (-3)^2 + (-5)^2) = sqrt(1 + 9 + 25) = sqrt(35).

Finally, we can find the cosine of angle A using the formula cos(A) = (A · B) / (|A| * |B|).

cos(A) = 0 / (sqrt(6) * sqrt(35)) = 0.

Therefore, cos^2A = (cos(A))^2 = (0)^2 = 0.