Calculus II - Vector Arithmetic

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Section 11.2 : Vector Arithmetic

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5. Determine if \(\vec a = \left\langle {3, - 5,1} \right\rangle \) and \(\vec b = \left\langle {6, - 2,2} \right\rangle \) are parallel vectors.

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Recall that two vectors are parallel if they are scalar multiples of each other. In other words, these two vectors will be scalar multiples if we can find a number \(k\) such that,

\[\vec a = k\,\vec b\] Show Step 2

Let’s just take a look at the first component from each vector. It is obvious that \(6 = 2\left( 3 \right)\). So, to convert the first components we’d need to multiply \(\vec a\) by 2. However, if we did that we’d get,

\[2\vec a = \left\langle {6, - 10,2} \right\rangle \ne \vec b\]

This is clearly not \(\vec b\). The first component is correct and the third component is correct but the second isn’t correct. Therefore, there is no single number, \(k\), that we can use to convert \(\vec a\) into \(\vec b\) through scalar multiplication.

This in turn means that \(\vec a\) and \(\vec b\) cannot possibly be parallel.