Question 12(a) | BetaMath

Question 12(a)

The vectors $\displaystyle \begin{pmatrix} a^2 \\ 2 \end{pmatrix}$ and $\displaystyle \begin{pmatrix} a + 5 \\ a - 4 \end{pmatrix}$ are perpendicular.

Find the possible values of $a$ .

Solution

We construct a cubic in the unknown $a$ by computing a dot product, which must be equal to $0$ for these vectors to be perpendicular. We then solve this cubic for the possible values of $a$ which satisfy this equation

\begin{aligned} 0 &= \displaystyle \begin{pmatrix} a^2 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} a + 5 \\ a - 4 \end{pmatrix} \\ 0 &= a^2(a+5) + 2(a - 4) \\ 0 &= a^3 + 5a^2 + 2a - 8 \\ \end{aligned}

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