Question 11(a)

Consider the vectors a∼=3i∼+2j∼\underset{\sim}{a} = 3\underset{\sim}{i} + 2\underset{\sim}{j} and b∼=−i∼+4j∼\underset{\sim}{b} = -\underset{\sim}{i} + 4\underset{\sim}{j}

i. Find 2a∼−b∼2\underset{\sim}{a} - \underset{\sim}{b} .

ii. Find a∼⋅b∼\underset{\sim}{a} \cdot \underset{\sim}{b} .

Solution

In this question, we have two vectors which have been written in component form. In part (i), we can individually apply scalar multiplication and perform addition the vector components to obtain the result

2a∼−b∼=2(3i∼+2j∼)−(−i∼+4j∼)2a∼−b∼=6i∼+4j∼+i∼−4j∼2a∼−b∼=(6+1)i∼+(4−4)j∼2a∼−b∼=7i∼\begin{aligned} 2\underset{\sim}{a} - \underset{\sim}{b} &= 2\left(3\underset{\sim}{i} + 2\underset{\sim}{j}\right) - \left(-\underset{\sim}{i} + 4\underset{\sim}{j}\right) \\ 2\underset{\sim}{a} - \underset{\sim}{b} &= 6\underset{\sim}{i} + 4\underset{\sim}{j} +\underset{\sim}{i} - 4\underset{\sim}{j} \\ 2\underset{\sim}{a} - \underset{\sim}{b} &= (6+1)\underset{\sim}{i} + (4-4)\underset{\sim}{j} \\ 2\underset{\sim}{a} - \underset{\sim}{b} &= 7\underset{\sim}{i} \\ \end{aligned}

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