Given the coordinates of the vector AB (4, -5,7) and the coordinates of the point A (2,1, -8) find the coordinates

Given the coordinates of the vector AB (4, -5,7) and the coordinates of the point A (2,1, -8) find the coordinates of the point B and the projection of AB onto the Oz axis.

Based on the definition of the vector, we find the coordinates of the point B – add the coordinates of the point A to the coordinates of the vector AB, we get:
B = AB (4; -5; 7) + A (2; 1; -8) = (4; -5; 7) + (2; 1; -8) = (4 + 2; -5 + 1; 7 + (-8)) = (6; -4; -1).
Let us find the projection of the AB vector onto the Oz axis using the formula:
PrA’B’AB = AB * A’B ‘/ | A’B’ |.
The direction vector for the Oz axis is equal to the vector A’B ‘, where the coordinates of points A’ (0; 0; -8) and B ‘(0; 0; -1). Since the projection onto the Oz axis, then, respectively, the coordinates x and y are 0. Therefore, the vector A’B ‘has coordinates: (0; 0; -1 – (-8)) = (0; 0; 7).
Let’s find the dot product of vectors:
AB x A’B ‘= ABx * A’B’x + ABy * A’B’y + ABz * A’B’z = 4 * 0 + (-5) * 0 + 7 * 7 = 0 – 0 + 49 = 49.
The coordinates of the vector A’B ‘(0; 0; 7), then the modulus of the vector A’B’ is:
| A’B ‘| = √ (A’B’x ^ 2 + A’By ‘^ 2 + A’B’z ^ 2) = √ (0 ^ 2 + 0 ^ 2 + 7 ^ 2) = √ (0 + 0 + 49) = √49 = 7.
Eventually:
PrA’B’AB = AB * A’B ‘/ | A’B’ | = 49/7 = 7.
Answer: coordinates of point B (6; -4; -1); the projection of the vector AB onto the Oz axis is 7.