How to find perpendicular vectors in 3D

Question

Find all values of a such that the vector $q = \langle 2, a, –2\rangle$ is perpendicular to the vector $p = \langle –3, a, 5 \rangle$.

Answer 1

Hint: $p \cdot q = 2\cdot(-3) + a\cdot a + (-2)\cdot 5 = a^2 - 16$. $p$ and $q$ are perpendicular if and only if $p \cdot q = 0$. What values of $a$ satisfy $a^2 - 16 = 0$?

Answer 2

The vectors $p$ and $q$ are perpendicular if and only if their dot product $p \cdot q$ is $0$. Taking the dot product gives $$p \cdot q=(2,a,-2)\cdot(-3,a,5)=-6+a^2-10=a^2-16$$ Setting this equal to $0$ gives us $$a^2-16=0 \\a^2=16 \\a=\pm4$$

Answer 3

HINT: $2(-3)+a^2+(-2)5=0$ and the equation $a^2=c$ you should solve.

Answer 4

Have you heard of the 'dot' or 'scalar' product before?

Given two vectors $\vec{a}, \vec{b} \in \mathbb{R}^n$, their dot product is defined as: $$\vec{a} \cdot \vec{b} \ = \ \| \vec{a} \| \|\vec{b} \| \cos \theta \ = \ \pmatrix{a_1 \\ a_2 \\ \dots \\ a_n} \cdot \pmatrix{b_1 \\ b_2 \\ \dots \\ b_n} \ = \sum_{i=1}^{n} a_i b_i $$

Two vectors are perpendicular if and only if their scalar product is equal to $0$ i.e. the angle between them is $\pi/2$ radians.

This should give you enough information to find your answer!