Mathematics

Mathematics

Problem

Find the position vector of the point P(3,−2,5)P(3,-2,5). a) 5i−2j+3k5\mathbf i - 2\mathbf j + 3\mathbf k b) 3i−2j+5k3\mathbf i - 2\mathbf j + 5\mathbf k c) −3i+2j−5k-3\mathbf i + 2\mathbf j - 5\mathbf k d) 338i−238j+538k\dfrac{3}{\sqrt{38}}\mathbf i - \dfrac{2}{\sqrt{38}}\mathbf j + \dfrac{5}{\sqrt{38}}\mathbf k
Intuitive explanation

The position vector of a point is simply the vector drawn from the origin to that point—it captures both the direction and distance from (0,0,0)(0,0,0) to your target. For any point with coordinates (x,y,z)(x, y, z), the position vector is just ⟨x,y,z⟩\langle x, y, z \rangle or xi+yj+zkx\mathbf{i} + y\mathbf{j} + z\mathbf{k}. This is one of the most direct translations in 3D geometry: read the coordinates straight off and write them as a vector.

Intuitive explanation

The position vector of a point is simply the vector drawn from the origin to that point—it captures both the direction and distance from (0,0,0)(0,0,0) to your target. For any point with coordinates (x,y,z)(x, y, z), the position vector is just ⟨x,y,z⟩\langle x, y, z \rangle or xi+yj+zkx\mathbf{i} + y\mathbf{j} + z\mathbf{k}. This is one of the most direct translations in 3D geometry: read the coordinates straight off and write them as a vector.

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