For a point $P(x,y,z)$ in 3D, the magnitude of the position vector $\overrightarrow{OP}$ is 
A$x+y+z$
B$\sqrt{x^2+y^2-z^2}$
C$\sqrt{x^2+y^2+z^2}$
D$xy+yz+zx$
Answer & Solution
Correct answer: C. $\sqrt{x^2+y^2+z^2}$
The 3D position vector from the origin to $P(x,y,z)$ has magnitude equal to the distance from the origin, which is $\sqrt{x^2+y^2+z^2}$.
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