Vectors u and v have components $$ egin{pmatrix} p \\ -2 \\ 4 \\ ight) ext{ and } egin{pmatrix} 2p+16 \\ -3 \\ 6 \\ ight), p eg ext{R.} $$ (a) (i) Find an expression for u.v - Scottish Highers Maths - Question 9 - 2022

Vectors u and v are perpendicular if their dot product is zero. Therefore, we set the expression from part (i) to zero:

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

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

$$p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Where:

• $a = 2$
• $b = 16$
• $c = 30$

Plugging in the values:

$$p = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 2 \cdot 30}}{2 \cdot 2}$$

Calculating the discriminant:

$$16^2 - 4 \cdot 2 \cdot 30 = 256 - 240 = 16$$

Since the discriminant is positive, we have two distinct real solutions:

$$p = \frac{-16 \pm 4}{4}$$

Calculating further, we find:

• $p_1 = \frac{-12}{4} = -3$
• $p_2 = \frac{-20}{4} = -5$

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

Vectors u and v are parallel if one is a scalar multiple of the other. This can be expressed as:

$$u = k v ext{ for some scalar } k$$

This leads to the following equalities based on corresponding components:

1. rac{p}{2p+16} = rac{-2}{-3} = rac{4}{6}

From the first equality, we cross-multiply:

$$3p = 2p + 16 ightarrow p = 16$$

From the second equality (after simplification), we get:

$$2p + 16 = 2 ightarrow 2p = 2 - 16 ightarrow 2p = -14 ightarrow p = -7$$

Thus, the value of p for which u and v are parallel can be either $p = 16$ or $p = -7$.