Question

The acute angle that the vector $2\hat{i}-2\hat{j}+\hat{k}$ makes with the plane contained by the two vectors $2\hat{i}+3\hat{j}-\hat{k}$ and $\hat{i}-\hat{j}+2\hat{k}$ is given by: -
(a) ${{\cos }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$
(b) ${{\sin }^{-1}}\left( \dfrac{1}{\sqrt{5}} \right)$
(c) ${{\tan }^{-1}}\left( \sqrt{2} \right)$
(d) ${{\cot }^{-1}}\left( \sqrt{2} \right)$

Answer 1

Hint: First of all find a plane that is perpendicular to the two vectors given by taking the cross product using the determinant formula given as \[\vec{n}=\vec{a}\times \vec{b}=\left( \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   2 & 3 & -1 \\
   1 & -1 & 2 \\
\end{matrix} \right)\], where $\vec{n}$ is the perpendicular plane, $\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+2\hat{k}$. Now, assume that the acute angle between the plane that contain these two vectors and the given vector $2\hat{i}-2\hat{j}+\hat{k}$ is $\theta $, using which consider the angle between the perpendicular plane $\vec{n}$ and the vector $2\hat{i}-2\hat{j}+\hat{k}$ as $\left( \dfrac{\pi }{2}-\theta \right)$. Take the dot product of $\vec{n}$ with the vector $2\hat{i}-2\hat{j}+\hat{k}$ using the formula $\vec{n}.\vec{a}=n\times a\times \cos \left( \dfrac{\pi }{2}-\theta \right)$ and find the value of $\theta $.

Complete step by step solution:
Here we have been provided with a plane that contains two vectors $2\hat{i}+3\hat{j}-\hat{k}$ and $\hat{i}-\hat{j}+2\hat{k}$. We have been asked to determine the acute angle between this plane and the vector $2\hat{i}-2\hat{j}+\hat{k}$.
Let us assume the given two vectors that lie in a plane as $\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+2\hat{k}$. Now, to solve the question first we need to determine a plane that will be perpendicular to both the vectors a and b. We know that the cross product of two vectors result in a vector that is perpendicular to both the given vectors. So let us consider the vector plane that is perpendicular to both $\vec{a}$ and $\vec{b}$ is $\vec{n}$. Therefore taking the cross product of $\vec{a}$ and $\vec{b}$ we get,
\[\Rightarrow \vec{n}=\vec{a}\times \vec{b}=\left( \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   2 & 3 & -1 \\
   1 & -1 & 2 \\
\end{matrix} \right)\]
Expanding the determinant we get,
\[\begin{align}
  & \Rightarrow \vec{n}=\hat{i}\left( 6-1 \right)-\hat{j}\left( 4-\left( -1 \right) \right)+\hat{k}\left( -2-3 \right) \\
 & \Rightarrow \vec{n}=5\hat{i}-5\hat{j}-5\hat{k} \\
\end{align}\]
Now, let us assume that the acute angle between the plane containing $\vec{a}$ and $\vec{b}$and the vector $2\hat{i}-2\hat{j}+\hat{k}$ is $\theta $, so the angle between $\vec{n}$ and $2\hat{i}-2\hat{j}+\hat{k}$ will be $\theta $ as shown in the figure.

Considering the dot product of $\vec{n}$ and $2\hat{i}-2\hat{j}+\hat{k}$ which will be the product of magnitude of these vectors and the cosine of the angle between the two, we get,
\[\begin{align}
  & \Rightarrow \left( 5\hat{i}-5\hat{j}-5\hat{k} \right).\left( 2\hat{i}-2\hat{j}+\hat{k} \right)=\sqrt{{{5}^{2}}+{{\left( -5 \right)}^{2}}+{{\left( -5 \right)}^{2}}}\times \sqrt{{{2}^{2}}+{{\left( -2 \right)}^{2}}+{{\left( 1 \right)}^{2}}}\times \cos \left( \dfrac{\pi }{2}-\theta \right) \\
 & \Rightarrow 7\times 2+\left( -3 \right)\times \left( -2 \right)+\left( -5 \right)\times 1=\sqrt{75}\times \sqrt{9}\cos \left( \dfrac{\pi }{2}-\theta \right) \\
 & \Rightarrow 15=3\times 5\sqrt{3}\cos \left( \dfrac{\pi }{2}-\theta \right) \\
\end{align}\]
We know that $\cos \left( \dfrac{\pi }{2}-\theta \right)=\sin \theta $, so on simplifying the above relation we get,
\[\begin{align}
  & \Rightarrow \sin \theta =\left( \dfrac{1}{\sqrt{3}} \right) \\
 & \Rightarrow \theta ={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right) \\
\end{align}\]
We know that $\sin \theta =\dfrac{p}{h}$ where p = perpendicular and h = hypotenuse so we get \[\theta ={{\sin }^{-1}}\left( \dfrac{p}{h} \right)\]. On comparing the \[{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)\] with \[{{\sin }^{-1}}\left( \dfrac{p}{h} \right)\] we get $p=1$ and $h=\sqrt{3}$. Using the Pythagoras theorem \[h=\sqrt{{{b}^{2}}+{{p}^{2}}}\] to calculate the value of base (b) we get,
\[\begin{align}
  & \Rightarrow b=\sqrt{{{h}^{2}}-{{p}^{2}}} \\
 & \Rightarrow b=\sqrt{{{\left( \sqrt{3} \right)}^{2}}-{{1}^{2}}} \\
 & \Rightarrow b=\sqrt{2} \\
\end{align}\]
We know that $\tan \theta =\dfrac{p}{b}$, $\cot \theta =\dfrac{b}{p}$ and $\cos \theta =\dfrac{b}{h}$, so we can write \[{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)\] as:
\[\Rightarrow {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)={{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{\sqrt{3}} \right)={{\tan }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)={{\cot }^{-1}}\left( \sqrt{2} \right)\]
Hence option (d) is the correct answer.

Note: We do not have any method to directly determine the equation of a plane containing two vectors and that is why we have found the perpendicular plane first. If we would have been provided with a point $\vec{k}$ through which the plane would have been passing then we can directly use the formula $\left( \vec{r}-\vec{k} \right).\left( \vec{a}\times \vec{b} \right)$ to get the equation of the plane. Here, in the options there are four different inverse functions and that is why we have found the value of $\theta $ in terms of those four functions to check the correct one.