Chapter 13 - Vector Geometry - 13.2 Vectors in Three Dimensions - Exercises - Page 659: 9

$$\sqrt{26}.$$

Work Step by Step

First, we have $$\overrightarrow{ OR}=R-O=(1,4,3)-(0,0,0)=\langle 1,4,3 \rangle.$$ Now, $$\|\overrightarrow{ OR}\|=\sqrt{1+16+9}=\sqrt{26}.$$

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