Question

How do you find all unit vectors orthogonal to $v=i+j+k$?

Answer 1

Hint: To solve this question, we need to assume a vector orthogonal to the given vector $v=i+j+k$ in the form of three unknown parameters. Since the vectors are orthogonal to each other, their dot product will be equal to zero. From this, we will get an equation relating the three unknown parameters. With the help of that equation, we will be able to reduce our vector in terms of only two unknown parameters. Since we need a unit vector, we will have to divide the vector with its magnitude. The unit vector which we will obtain will be the set of infinite possible unit vectors.

Complete step by step answer:
Let us assume a vector $u$ orthogonal to the given vector $v$, and that vector be given as
$u=xi+yj+zk..........(i)$
According to the question, the vector $v$ is given as
$v=i+j+k..........(ii)$
Now, the term orthogonal means perpendicular. This means that the assumed vector $u$ is perpendicular to the given vector $v$. We know that the dot product of two perpendicular vectors is equal to zero. So we can write
$\Rightarrow u\cdot v=0$
Putting the equations (i) and (ii) in the above equation, we get
$\Rightarrow \left( xi+yj+zk \right)\cdot \left( i+j+k \right)=0$
$\Rightarrow x+y+z=0$
Subtracting $\left( y+z \right)$ from both the sides, we get
\[\begin{align}
  & \Rightarrow x+y+z-\left( y+z \right)=-\left( y+z \right) \\
 & \Rightarrow x=-\left( y+z \right).........(iii) \\
\end{align}\]
Putting the equation (iii) in the equation (i), we get
\[\Rightarrow u=-\left( y+z \right)i+yj+zk..........(iv)\]
Now, according to the question, we have to determine the unit vectors orthogonal to the given vector. So we need to divide the vector $u$ by its magnitude, in order to obtain the unit vector parallel to its direction.
We know that the magnitude of a vector $ai+bj+ck$ is equal to $\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$.
From (iii) we can say that $a=-\left( y+z \right)$, $b=y$, and $c=z$. Therefore we get the magnitude of the vector $u$ as
\[\begin{align}
  & \Rightarrow \left| u \right|=\sqrt{{{\left( -\left( y+z \right) \right)}^{2}}+{{y}^{2}}+{{z}^{2}}} \\
 & \Rightarrow \left| u \right|=\sqrt{{{\left( y+z \right)}^{2}}+{{y}^{2}}+{{z}^{2}}} \\
\end{align}\]
Now, we know that $\left( a+b \right)={{a}^{2}}+2ab+{{b}^{2}}$. Applying this identity to expand ${{\left( y+z \right)}^{2}}$ in the above equation, we get
\[\begin{align}
  & \Rightarrow \left| u \right|=\sqrt{\left( {{y}^{2}}+2yz+{{z}^{2}} \right)+{{y}^{2}}+{{z}^{2}}} \\
 & \Rightarrow \left| u \right|=\sqrt{2\left( {{y}^{2}}+{{z}^{2}} \right)+2yz} \\
 & \Rightarrow \left| u \right|=\sqrt{2\left( {{y}^{2}}+{{z}^{2}}+yz \right)}........(v) \\
\end{align}\]
Dividing (iv) by (v) we get
\[\begin{align}
  & \Rightarrow \dfrac{u}{\left| u \right|}=\dfrac{-\left( y+z \right)i+yj+zk}{\sqrt{2\left( {{y}^{2}}+{{z}^{2}}+yz \right)}} \\
 & \Rightarrow \dfrac{u}{\left| u \right|}=\dfrac{1}{\sqrt{2\left( {{y}^{2}}+{{z}^{2}}+yz \right)}}\left[ -\left( y+z \right)i+yj+zk \right] \\
\end{align}\]
Thus, we have obtained the unit vector orthogonal to the given vector. As we can see that this unit vector is in the form of two independent parameters $y$ and $z$ which can take infinite values. Therefore, there will be infinitely many such unit vectors possible.

Note: We have to determine the unit vector orthogonal to the given vector, so do not forget to divide the vector by its magnitude. Also, the unit vector which we have obtained is a set of infinite possible unit vectors orthogonal to the given vector $v$. For any set of values of $\left( y,z \right)$ we will get a unique unit vector orthogonal to the vector $v$.