Question

How do you find two unit vectors that are orthogonal to $\left( -3,4 \right)$?

Answer 1

We first form the vector with the given coordinate point of $\left( -3,4 \right)$. We find the orthogonal or perpendicular vectors to $\left( -3,4 \right)$ in that same plane. We also know that any vector perpendicular to the X-Y plane is orthogonal to $-3\widehat{i}+4\widehat{j}$. We take all the unit vectors along those orthogonal vectors.

Complete step by step answer:
We need to find two unit vectors that are orthogonal to $\left( -3,4 \right)$.
First, we try to form the vector with the given coordinate point of $\left( -3,4 \right)$.
The vector is indicated by $-3\widehat{i}+4\widehat{j}$. This vector lies on the X-Y plane.
Any arbitrary vector on the X-Y plane will be defined as $a\widehat{i}+b\widehat{j}$.
Let’s assume the vector $a\widehat{i}+b\widehat{j}$ is orthogonal to the vector $-3\widehat{i}+4\widehat{j}$.
The dot product between any two orthogonal vectors will always give the zero value.
So, $\left( -3\widehat{i}+4\widehat{j} \right)\left( a\widehat{i}+b\widehat{j} \right)=0$. In dot product the same axes multiplication gives 1 and different axes multiplication gives 0.
Therefore, $-3a+4b=0$ which on simplification gives $b=\dfrac{3a}{4}$.
We replace the value and get that the vector $a\widehat{i}+\dfrac{3a}{4}\widehat{j}$ is orthogonal to the given vector.
Now we find the magnitude which is $\sqrt{{{a}^{2}}+{{\left( \dfrac{3a}{4} \right)}^{2}}}=\dfrac{5a}{4}$.
The unit vector orthogonal to the given vector $-3\widehat{i}+4\widehat{j}$ is $\dfrac{a\widehat{i}+\dfrac{3a}{4}\widehat{j}}{\pm \dfrac{5a}{4}}=\pm \dfrac{4}{5a}\left( a\widehat{i}+\dfrac{3a}{4}\widehat{j} \right)=\pm \left( \dfrac{4}{5}\widehat{i}+\dfrac{3}{5}\widehat{j} \right)$.
These vectors are on the X-Y plane.
As the given vector is on the X-Y plane, any vector perpendicular to the X-Y plane is orthogonal to $-3\widehat{i}+4\widehat{j}$.
Those unit vectors are $\pm \widehat{k}$.
The vectors which are orthogonal to $\left( -3,4 \right)$ are $\pm \left( \dfrac{4}{5}\widehat{i}+\dfrac{3}{5}\widehat{j} \right),\pm \widehat{k}$.

Note: The other planes are automatically perpendicular to the X-Y plane. That’s why we broke the orthogonal vectors into two parts. The unit vectors are also perpendicular to $\left( -3,4 \right)$.