The position vector of the points A,B and C are respectively: a=2i ̂-3j ̂+k ̂,b=4i ̂+2j ̂+5k ̂ and c=3i ̂-2j ̂+2k ̂. Find the position vector: (i) P_AB (ii) P_BC (iii) P_CA?
If you mean P_AB is the vector from A to B, then that is just
b-a = 2i+5j+4k
do the others in like wise
To find the position vectors P_AB, P_BC, and P_CA, we need to calculate the vector differences of the given points.
The position vector P_AB gives the direction and distance from point A to point B. To find it:
P_AB = B - A
Substituting the values of points A and B:
P_AB = (4i ̂ + 2j ̂ + 5k ̂) - (2i ̂ - 3j ̂ + k ̂)
Simplifying the vector subtraction:
P_AB = (4i ̂ - 2i ̂) + (2j ̂ - (-3j ̂)) + (5k ̂ - k ̂)
P_AB = 2i ̂ + 5j ̂ + 4k ̂
Therefore, the position vector P_AB is 2i ̂ + 5j ̂ + 4k ̂
Similarly, for P_BC:
P_BC = C - B
P_BC = (3i ̂ - 2j ̂ + 2k ̂) - (4i ̂ + 2j ̂ + 5k ̂)
P_BC = (3i ̂ - 4i ̂) + (-2j ̂ - 2j ̂) + (2k ̂ - 5k ̂)
P_BC = -i ̂ - 4j ̂ - 3k ̂
Therefore, the position vector P_BC is -i ̂ - 4j ̂ - 3k ̂
Lastly, for P_CA:
P_CA = A - C
P_CA = (2i ̂ - 3j ̂ + k ̂) - (3i ̂ - 2j ̂ + 2k ̂)
P_CA = (2i ̂ - 3i ̂) + (-3j ̂ - (-2j ̂)) + (k ̂ - 2k ̂)
P_CA = -i ̂ + j ̂ - k ̂
Therefore, the position vector P_CA is -i ̂ + j ̂ - k ̂