Vectors u and v have components $$ egin{pmatrix} p\\ -2\\ 4 ight) \\ egin{pmatrix} (2p+16)\\ -3\\ 6 ight), p \\in \\mathbb{R} - Scottish Highers Maths - Question 9 - 2022

Vectors are perpendicular if their dot product is zero:

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

We can use the quadratic formula to find the values of p:

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

disc = 16^2 - 4 imes 2 imes 30 = 256 - 240 = 16$$

Calculating the roots:

p = \frac{-16 \pm 4}{4}$$ This gives us: 1. $$p = \frac{-12}{4} = -3$$ 2. $$p = \frac{-20}{4} = -5$$ Thus, the values of p for which u and v are perpendicular are p = -3 and p = -5.

Vectors are parallel if one is a scalar multiple of the other. We set the ratios of corresponding components equal:

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

This gives us:

1. From the first ratio: $3p = 2p + 16 \\ \\implies p = 16$

2. From the second ratio: $6p = -8 \\ \\implies p = -\frac{8}{6} = -\frac{4}{3}$

The values of p for which u and v are parallel are p = 16 and p = -\frac{4}{3}.