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) ${\cos }^{-1}\left({\displaystyle \frac{1}{\sqrt{3}}}\right)$
(b) ${\sin }^{-1}\left({\displaystyle \frac{1}{\sqrt{5}}}\right)$
(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 $$\overrightarrow{n}=\overrightarrow{a}\times \overrightarrow{b}=\left(\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& -1\\ 1& -1& 2\end{array}\right)$$, where $\overrightarrow{n}$ is the perpendicular plane, $\overrightarrow{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{i}-\hat{j}+2\hat{k}$. 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 $\overrightarrow{n}$ and the vector $2\hat{i}-2\hat{j}+\hat{k}$ as $({\displaystyle \frac{\pi }{2}}-\theta )$. Take the dot product of $\overrightarrow{n}$ with the vector $2\hat{i}-2\hat{j}+\hat{k}$ using the formula $\overrightarrow{n}.\overrightarrow{a}=n\times a\times \cos ({\displaystyle \frac{\pi }{2}}-\theta )$ 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 $\overrightarrow{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{i}-\hat{j}+2\hat{k}$. 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 $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\overrightarrow{n}$. Therefore taking the cross product of $\overrightarrow{a}$ and $\overrightarrow{b}$ we get,
$$\Rightarrow \overrightarrow{n}=\overrightarrow{a}\times \overrightarrow{b}=\left(\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& -1\\ 1& -1& 2\end{array}\right)$$
Expanding the determinant we get,
$$\begin{array}{rl}& \Rightarrow \overrightarrow{n}=\hat{i}(6-1)-\hat{j}(4-(-1))+\hat{k}(-2-3)\\ & \Rightarrow \overrightarrow{n}=5\hat{i}-5\hat{j}-5\hat{k}\end{array}$$
Now, let us assume that the acute angle between the plane containing $\overrightarrow{a}$ and $\overrightarrow{b}$and the vector $2\hat{i}-2\hat{j}+\hat{k}$ is $\theta$, so the angle between $\overrightarrow{n}$ and $2\hat{i}-2\hat{j}+\hat{k}$ will be $\theta$ as shown in the figure.

Considering the dot product of $\overrightarrow{n}$ 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,
$$\begin{array}{rl}& \Rightarrow (5\hat{i}-5\hat{j}-5\hat{k}).(2\hat{i}-2\hat{j}+\hat{k})=\sqrt{5^2+{(-5)}^2+{(-5)}^2}\times \sqrt{2^2+{(-2)}^2+{\left(1\right)}^2}\times \cos ({\displaystyle \frac{\pi }{2}}-\theta )\\ & \Rightarrow 7\times 2+(-3)\times (-2)+(-5)\times 1=\sqrt{75}\times \sqrt{9}\cos ({\displaystyle \frac{\pi }{2}}-\theta )\\ & \Rightarrow 15=3\times 5\sqrt{3}\cos ({\displaystyle \frac{\pi }{2}}-\theta )\end{array}$$
We know that $\cos ({\displaystyle \frac{\pi }{2}}-\theta )=\sin \theta$, so on simplifying the above relation we get,
$$\begin{array}{rl}& \Rightarrow \sin \theta =\left({\displaystyle \frac{1}{\sqrt{3}}}\right)\\ & \Rightarrow \theta ={\sin }^{-1}\left({\displaystyle \frac{1}{\sqrt{3}}}\right)\end{array}$$
We know that $\sin \theta ={\displaystyle \frac{p}{h}}$ where p = perpendicular and h = hypotenuse so we get $$\theta ={\sin }^{-1}\left({\displaystyle \frac{p}{h}}\right)$$. On comparing the $${\sin }^{-1}\left({\displaystyle \frac{1}{\sqrt{3}}}\right)$$ with $${\sin }^{-1}\left({\displaystyle \frac{p}{h}}\right)$$ 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,
$$\begin{array}{rl}& \Rightarrow b=\sqrt{h^2-p^2}\\ & \Rightarrow b=\sqrt{{\left(\sqrt{3}\right)}^2-1^2}\\ & \Rightarrow b=\sqrt{2}\end{array}$$
We know that $\tan \theta ={\displaystyle \frac{p}{b}}$, $\cot \theta ={\displaystyle \frac{b}{p}}$ and $\cos \theta ={\displaystyle \frac{b}{h}}$, so we can write $${\sin }^{-1}\left({\displaystyle \frac{1}{\sqrt{3}}}\right)$$ as:
$$\Rightarrow {\sin }^{-1}\left({\displaystyle \frac{1}{\sqrt{3}}}\right)={\cos }^{-1}\left({\displaystyle \frac{\sqrt{2}}{\sqrt{3}}}\right)={\tan }^{-1}\left({\displaystyle \frac{1}{\sqrt{2}}}\right)={\cot }^{-1}\left(\sqrt{2}\right)$$
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 $\overrightarrow{k}$ through which the plane would have been passing then we can directly use the formula $(\overrightarrow{r}-\overrightarrow{k}).(\overrightarrow{a}\times \overrightarrow{b})$ 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.