Question

If the unit vector $\overrightarrow{a}$ makes angle $\dfrac{\pi }{3}$ with $\widehat{i}$, $\dfrac{\pi }{4}$ with $\widehat{j}$ and an acute angle $\theta$ with $\widehat{k}$, then find the value of $\theta$.

Answer 1

To solve this question, we should know the formula of dot product. If $\overrightarrow{a},\overrightarrow{b}$ are two vectors and the angle between them is $\theta$, the dot product of the vectors a, b is defined as $\overrightarrow{a}.\overrightarrow{b}=\left| a \right|\left| b \right|\cos \theta$. Let us assume the vector $\overrightarrow{a}$ as $x\widehat{i}+y\widehat{j}+z\widehat{k}$. We can infer from the question that the angles between the vector $\overrightarrow{a}$ and $\widehat{i}$,$\widehat{j}$ are $\dfrac{\pi }{3}$,$\dfrac{\pi }{4}$ respectively. We can also infer that $\left| a \right|=1$ as $\overrightarrow{a}$ is a unit vector. By using the dot product formula with the two vectors $\widehat{i}$,$\widehat{j}$, we get the values of x, y. We know that $\left| a \right|=1\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}=1$. Using this, we get two values of z. By doing the dot product with $\widehat{k}$ and applying the condition that the angle $\theta$ should be acute, we get unique values of z and $\theta$.

Complete step-by-step solution:
Let us assume the vector $\overrightarrow{a}$ as $x\widehat{i}+y\widehat{j}+z\widehat{k}$.
Let us consider the dot products of the vector $\overrightarrow{a}$ with $\widehat{i}$,$\widehat{j}$.
The dot product of the vectors a, b is defined as $\overrightarrow{a}.\overrightarrow{b}=\left| a \right|\left| b \right|\cos \theta$.
We know that the vectors $\widehat{i},\widehat{j},\widehat{k}$ are representing unit vectors along x, y, z axis respectively.
$\widehat{i}.\widehat{i}=\left| i \right|\left| i \right|\cos \theta$
As $\widehat{i}$ is a unit vector $\left| i \right|=1$ and the angle between$\widehat{i}$ and $\widehat{i}$ is $\theta ={{0}^{\circ }}$
$\widehat{i}.\widehat{i}=1.1.\cos 0=1$
Similarly,
$\begin{aligned} & \widehat{j}.\widehat{j}=1 \\\ & \widehat{k}.\widehat{k}=1 \\\ \end{aligned}$
$\widehat{i}.\widehat{j}=\left| i \right|\left| j \right|\cos \theta$
As $\widehat{i}$, $\widehat{j}$ are unit vectors $\left| i \right|=1,\left| j \right|=1$ and the angle between$\widehat{i}$ and $\widehat{j}$ is $\theta ={{90}^{\circ }}$
$\widehat{i}.\widehat{j}=1.1\cos 90=0$
Similarly, dot product of any two vectors of $\widehat{i},\widehat{j},\widehat{k}$ is zero and dot product of the same unit vectors is 1.

Let us consider $\overrightarrow{a}\text{ }.\text{ }\widehat{i}$,
$\overrightarrow{a}.\widehat{i}=\left| a \right|\left| i \right|\cos \dfrac{\pi }{3}$
We know that $\left| a \right|=1\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}=1$, $\left| i \right|=1$
Using them, we get
$\begin{aligned} & \left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right).\text{ }\widehat{i}=\dfrac{1}{2} \\\ & x\widehat{i}.\widehat{i}+y\widehat{j}.\widehat{i}+z\widehat{k}.\widehat{i}=\dfrac{1}{2} \\\ & x=\dfrac{1}{2} \\\ \end{aligned}$
Let us consider $\overrightarrow{a}\text{ }.\text{ }\widehat{j}$,
$\overrightarrow{a}.\widehat{j}=\left| a \right|\left| j \right|\cos \dfrac{\pi }{4}$
We know that $\left| a \right|=1\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}=1$, $\left| j \right|=1$
Using them, we get
$\begin{aligned} & \left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right).\text{ }\widehat{j}=\dfrac{1}{\sqrt{2}} \\\ & y=\dfrac{1}{\sqrt{2}} \\\ \end{aligned}$
Let us consider $\overrightarrow{a}\text{ }.\text{ }\widehat{k}$,
Let us assume the required angle be $\theta$.
$\overrightarrow{a}.\widehat{k}=\left| a \right|\left| k \right|\cos \theta$
We know that $\left| a \right|=1\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}=1$, $\left| k \right|=1$
Using them, we get
$\begin{aligned} & \left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right).\text{ }\widehat{k}=\cos \theta \\\ & z=\cos \theta \\\ \end{aligned}$
We know that $\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}=1$. Substituting the values of x and y, we get
$\begin{aligned} & \sqrt{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+{{z}^{2}}}=1 \\\ & \sqrt{\dfrac{1}{4}+\dfrac{1}{2}+{{z}^{2}}}=1 \\\ & {{z}^{2}}+\dfrac{3}{4}=1 \\\ & {{z}^{2}}=\dfrac{1}{4} \\\ & z=\pm \dfrac{1}{2} \\\ \end{aligned}$
But we know that $z=\cos \theta$ and $\theta$ is acute. So,
$\cos \theta >0$
So, we get
$\begin{aligned} & z=\cos \theta =\dfrac{1}{2} \\\ & \theta =\dfrac{\pi }{3} \\\ \end{aligned}$
$\therefore$ The angle between $\overrightarrow{a}$ and $\widehat{k}$ is $\dfrac{\pi }{3}$

Note: There is a property in vectors that if a vector $\overrightarrow{a}$ makes angles ${{\theta }_{1}},{{\theta }_{2}},{{\theta }_{3}}$ with $\widehat{\hat{i}},\widehat{j},\widehat{k}$ respectively, then ${{\cos }^{2}}{{\theta }_{1}}+{{\cos }^{2}}{{\theta }_{2}}+{{\cos }^{2}}{{\theta }_{3}}=1$. Using this relation, we can write that $\begin{aligned} & {{\cos }^{2}}\dfrac{\pi }{3}+{{\cos }^{2}}\dfrac{\pi }{4}+{{\cos }^{2}}\theta =1 \\\ & {{\cos }^{2}}\theta =1-\dfrac{1}{2}-\dfrac{1}{4} \\\ & {{\cos }^{2}}\theta =\dfrac{1}{4} \\\ & \cos \theta =\pm \dfrac{1}{2} \\\ \end{aligned}$
As $\theta$ is acute angle, we can write that
$\begin{aligned} & \cos \theta =\dfrac{1}{2} \\\ & \theta =\dfrac{\pi }{3} \\\ \end{aligned}$