If coordinates of A , B and C are ( –1, 0, –3), (2, 3, –1) and (–1, 4, 2) respectively. Find a unit vector perpendicular to both 𝐴𝐵⃗⃗⃗⃗⃗ and 𝐴𝐶⃗⃗⃗⃗⃗ .

Sure, here are the steps to find a unit vector perpendicular to both AB and AC:

Step 1: Find the vectors AB and AC.

AB = B - A = (2 - (-1), 3 - 0, -1 - (-3)) = (3, 3, 2) AC = C - A = (-1 - (-1), 4 - 0, 2 - (-3)) = (0, 4, 5)

Step 2: Find the cross product of AB and AC. This will give us a vector that is perpendicular to both AB and AC.

AB x AC = (AB_yAC_z - AB_zAC_y, AB_zAC_x - AB_xAC_z, AB_xAC_y - AB_yAC_x) = (35 - 24, 20 - 35, 34 - 30) = (7, -15, 12)

Step 3: Find the magnitude of the cross product.

|AB x AC| = sqrt((7)^2 + (-15)^2 + (12)^2) = sqrt(49 + 225 + 144) = sqrt(418)

Step 4: Divide the cross product by its magnitude to get the unit vector.

Unit Vector = (7/sqrt(418), -15/sqrt(418), 12/sqrt(418))

So, the unit vector perpendicular to both AB and AC is approximately (0.342, -0.732, 0.585).