Question

How do you find a vector orthogonal to both $i+j$ and $i+k$?

Answer 1

Hint: We can let the unit vector orthogonal to both $i+j$ and $i+k$ as $r=ai+bj+ck$. Unit vector means its magnitude must be one, which will give ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1$. Also, since the vector $r$ is orthogonal to both $i+j$ and $i+k$, its dot product with both of these vectors will be equal to zero. On putting these dot product equal to zero, we will get two equations in terms of $a$, $b$ and $c$ which on solving with ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1$ will solve for the unit vector $r$.

Complete step by step solution:
Let us consider the two vectors given in the above question as $u$ and $v$ so that we can write the vector equations
$\Rightarrow u=i+j.......\left( i \right)$
And
\[\Rightarrow v=i+k.......\left( ii \right)\]
Now, let the unit vector orthogonal to both of the vectors $u$ and $v$ be $r$ such that
$\Rightarrow r=\left( ai+bj+ck \right)........\left( iii \right)$
Now, since $r$ is a unit vector, we can write
$\Rightarrow \left| r \right|=1$
From (iii) we can write
$\Rightarrow \sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}=1$
On squaring both the sides, we get
$\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1.......\left( iv \right)$
Now, since $r$ is orthogonal to $u$, we can write
$\Rightarrow r\cdot u=0$
Putting (i) and (iii) in the above equation, we get
\[\begin{align}
  & \Rightarrow \left( ai+bj+ck \right)\cdot \left( i+j \right)=0 \\
 & \Rightarrow a\left( 1 \right)+b\left( 1 \right)+c\left( 0 \right)=0 \\
 & \Rightarrow a+b=0.......\left( v \right) \\
\end{align}\]
Also, since $r$ is orthogonal to $v$, we can write
$\Rightarrow r\cdot v=0$
Putting (ii) and (iii) in the above equation, we get
\[\begin{align}
  & \Rightarrow \left( ai+bj+ck \right)\cdot \left( i+k \right)=0 \\
 & \Rightarrow a\left( 1 \right)+b\left( 0 \right)+c\left( 1 \right)=0 \\
 & \Rightarrow a+c=0.......\left( vi \right) \\
\end{align}\]
Subtracting (vi) from (v) we get
$\begin{align}
  & \Rightarrow a+b-\left( a+c \right)=0 \\
 & \Rightarrow a+b-a-c=0 \\
 & \Rightarrow b-c=0 \\
 & \Rightarrow b=c........\left( vii \right) \\
\end{align}$
Also, subtracting $b$ from both sides of the equation (v), we get
$\begin{align}
  & \Rightarrow a+b-b=0-b \\
 & \Rightarrow a=-b.......\left( viii \right) \\
\end{align}$
From (vii) and (viii) we can write
$\begin{align}
  & \Rightarrow -a=b=c=\lambda \left( say \right) \\
 & \Rightarrow a=-\lambda ,b=\lambda ,c=\lambda .......\left( ix \right) \\
\end{align}$
Therefore, the vector $r$ becomes
$\begin{align}
  & \Rightarrow r=-\lambda i+\lambda j+\lambda k \\
 & \Rightarrow r=\lambda \left( -i+j+k \right).......\left( x \right) \\
\end{align}$
Putting (ix) in (iv) we get
\[\begin{align}
  & \Rightarrow {{\left( -\lambda \right)}^{2}}+{{\lambda }^{2}}+{{\lambda }^{2}}=1 \\
 & \Rightarrow {{\lambda }^{2}}+{{\lambda }^{2}}+{{\lambda }^{2}}=1 \\
 & \Rightarrow 3{{\lambda }^{2}}=1 \\
 & \Rightarrow {{\lambda }^{2}}=\dfrac{1}{3} \\
\end{align}\]
On solving the above equation, we get
\[\begin{align}
  & \Rightarrow \lambda =\pm \dfrac{1}{\sqrt{3}} \\
 & \Rightarrow \lambda =\dfrac{1}{\sqrt{3}},\lambda =-\dfrac{1}{\sqrt{3}} \\
\end{align}\]
Putting $\lambda =\dfrac{1}{\sqrt{3}}$ in (x) we get
$\Rightarrow r=\dfrac{1}{\sqrt{3}}\left( -i+j+k \right)$
Similarly, putting $\lambda =-\dfrac{1}{\sqrt{3}}$ we get
$\Rightarrow r=-\dfrac{1}{\sqrt{3}}\left( -i+j+k \right)$

Hence, the unit vectors perpendicular to the given vectors are $\dfrac{1}{\sqrt{3}}\left( -i+j+k \right)$ and $-\dfrac{1}{\sqrt{3}}\left( -i+j+k \right)$.

Note: We can also use the concept of cross product to obtain the unit vector perpendicular to $u$ and $v$. Since the unit vector $r$ is orthogonal to both $u$ and $v$, it must be normal to the plane formed by $u$ and $v$, which means that it must be parallel to the cross product $u\times v$. So the unit vector $r$ can be written as $r=\pm \dfrac{u\times v}{\left| u\times v \right|}$.