Vectors u and v have components $$ egin{pmatrix} p \\ -2 \\ 4 ight) \\ egin{pmatrix} (2p+16) \\ -3 \\ 6\end{pmatrix}, p \in \mathbb{R} - Scottish Highers Maths - Question 9 - 2019

For vectors u and v to be perpendicular, their dot product must equal zero:

$2p^2 + 16p + 30 = 0.$

To solve this quadratic equation, we can apply the quadratic formula:

$p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 2, b = 16, c = 30.$

Calculating the discriminant: $b^2 - 4ac = 16^2 - 4(2)(30) = 256 - 240 = 16$

Substituting back, we get: $p = \frac{-16 \pm 4}{4}.$ This results in:

1. $p = -3$
2. $p = -5$.

Thus, the values of p for which u and v are perpendicular are $p = -3$ and $p = -5$.

For vectors u and v to be parallel, their scalar multiples must be equal, which can be expressed as:

$\frac{p}{2p + 16} = \frac{-2}{-3} = \frac{4}{6}.$

Cross-multiplying gives us: $3p = 2(2p + 16).$

Expanding and rearranging yields: $3p = 4p + 32 \Rightarrow -p = 32 \Rightarrow p = -32.$

Therefore, the value of p for which u and v are parallel is $p = -32$.