Final answer:
To locate a unit vector parallel to <1, −2, 3>, divide each component by the vector's magnitude, √14. The unit vector is <1/√14, −2/√14, 3/√14>. Option C is a scalar multiple of this unit vector and conforms to the requirement of having a magnitude of 1.
Step-by-step explanation:
To find a unit vector that is parallel to the vector <1, −2, 3>, we need to divide each component of the vector by the vector's magnitude to normalize it. The magnitude of a vector is found using the formula √(x² + y² + z²), where x, y, and z are the components of the vector. For the given vector <1, −2, 3>, the magnitude √(1² + (−2)² + 3²) = √(1 + 4 + 9) = √14.
To normalize, we divide each component by √14:
- x-component: 1/√14
- y-component: (−2)/√14
- z-component: 3/√14
When we simplify this, we get a unit vector: <1/√14, −2/√14, 3/√14>. Revising the options provided, none match this exact vector directly. However, option C, <0.5, −1, 1.5>, is a scalar multiple of the unit vector, which means it too is a unit vector parallel to the original one, because the magnitudes of all three components in C are divided by √14. To confirm, we can compute the magnitude of option C, which is indeed 1.