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Which of the following is a unit vector that is parallel to the vector ⟨1,−2,3⟩?

A. (1/3,-2/3,3/3)
B. (1,-2,3)
C. (0.5,-1,1.5)
D. (2,-4,6)

User Daljeet
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1 Answer

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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.

User Krlm
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