# Chapter 1 - Vectors - 1.1 The Geometry and Algebra of Vectors - Exercises 1.1 - Page 17: 16

$$-a+b+6c$$

#### Work Step by Step

We're given: $$-3(a-c)+2(a+2b)+3(c-b)$$ Using the distributivity of scalar-vector multiplication over vector addition (Theorem 1.1.e), we have: $$(-3)a+(-3)(-1)c+(2)a+(2)(2)b+(3)c+(3)(-1)b$$ Through scalar multiplication (Theorem 1.1.g), we get: $$(-3)a+(3)c+(2)a+(4)b+(3)c+(-3)b$$ Then, rearrange using the commutativity of vector addition (Theorem 1.1.a) $$(-3)a+(2)a+(4)b+(-3)b+(3)c+(3)c$$ Using the distributivity of scalar-vector multiplication over scalar addition, we get: $$(-3+2)a+(4-3)b+(3+3)c$$ Lastly, we simplify by scalar addition to get: $$-a+b+6c$$

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