Question

Let $\vec{a},\vec{b}\text{ and }\vec{c}$ three unit vectors, out of which vectors $\vec{b}\text{ and }\vec{c}$ are non-parallel. If $\alpha $ and $\beta $ are the angles which vector $\vec{a}$ makes with vector $\vec{b}\text{ and }\vec{c}$ respectively and $\vec{a}\times \left( \vec{b}\times \vec{c} \right)=\dfrac{1}{2}\vec{b}$ then $\left| \alpha -\beta \right|$ is equal to:
\[\begin{align}
  & A{{.60}^{\circ }} \\
 & B{{.30}^{\circ }} \\
 & C{{.90}^{\circ }} \\
 & D{{.45}^{\circ }} \\
\end{align}\]

Answer 1

Hint: To solve this question, we will use the formula of vector triple product of three vectors which is given as \[\vec{p}\times \left( \vec{q}\times \vec{r} \right)=\left( \vec{p}\cdot \vec{r} \right)\vec{q}-\left( \vec{p}\cdot \vec{q} \right)\vec{r}\] where $\vec{p},\vec{q}\text{ and }\vec{r}$ are there vectors. Then, we will compare the value obtained to $\dfrac{1}{2}\vec{b}$ as we are given $\vec{a}\times \left( \vec{b}\times \vec{c} \right)=\dfrac{1}{2}\vec{b}$ then we will use the formula of dot product of two vector when angle between them is given, $\vec{p}\cdot \vec{q}=\left| {\vec{p}} \right|\left| {\vec{q}} \right|\cos \theta $ where $\vec{p}\text{ and }\vec{q}$ are two vectors and $\theta $ is the angle between them.

Complete step by step answer:
Given that $\vec{a},\vec{b}\text{ and }\vec{c}$ are unit vectors.
Let us first define a unit vector. A unit vector is a vector having magnitude as 1. That is, if $\vec{p}$ is a unit vector than $\left| {\vec{p}} \right|=1$
If $\vec{p}=x\hat{i}+y\hat{i}+z\hat{k}$ then magnitude of $\vec{p}=\left| {\vec{p}} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$
As $\vec{a},\vec{b}\text{ and }\vec{c}$ are unit vector $\left| {\vec{a}} \right|=1,\left| {\vec{b}} \right|=1\text{ and }\left| {\vec{c}} \right|=1$
Given that $\alpha $ is angle between vector $\vec{d}\text{ and }\vec{b}$

And $\beta $ is angle between $\vec{a}\text{ and }\vec{c}$

Now, we will define a vector triple product.
Vector triple product of three vectors $\vec{p},\vec{q}\text{ and }\vec{r}$ is defined as the cross product of the $\vec{p}$ with cross product of $\vec{q}\text{ and }\vec{r}$
It is represented as \[\vec{p}\times \left( \vec{q}\times \vec{r} \right)\]
The formula for vector triple product is
\[\begin{align}
  & \vec{p}\times \left( \vec{q}\times \vec{r} \right)=\left( \vec{p}\cdot \vec{r} \right)\vec{q}-\left( \vec{p}\cdot \vec{q} \right)\vec{r} \\
 & \Rightarrow \vec{p}\times \left( \vec{q}\times \vec{r} \right)=\left( \vec{p}\cdot \vec{r} \right)\vec{q}-\left( \vec{q}\cdot \vec{r} \right)\vec{p} \\
\end{align}\]
In general we have
\[\vec{p}\times \left( \vec{q}\times \vec{r} \right)\ne \left( \vec{p}\times \vec{q} \right)\times \vec{r}\]
We are given that \[\vec{a}\times \left( \vec{b}\times \vec{c} \right)=\dfrac{1}{2}\vec{b}\] which is vector triple product of $\vec{a},\vec{b}\text{ and }\vec{c}$
Using the formula of vector triple product we have:
\[\vec{a}\times \left( \vec{b}\times \vec{c} \right)=\left( \vec{a}\cdot \vec{c} \right)\vec{b}-\left( \vec{a}\cdot \vec{b} \right)\vec{c}\]
We had \[\begin{align}
  & \vec{a}\times \left( \vec{b}\times \vec{c} \right)=\dfrac{1}{2}\vec{b} \\
 & \left( \vec{a}\cdot \vec{c} \right)\vec{b}-\left( \vec{a}\cdot \vec{b} \right)\vec{c}=\dfrac{1}{2}\vec{b}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)} \\
\end{align}\]
Comparing both sides of above equation and coefficient of vector $\vec{b}$ we get that
\[\left( \vec{a}\cdot \vec{c} \right)\vec{b}=\dfrac{1}{2}\vec{b}\text{ and }-\left( \vec{a}\cdot \vec{b} \right)\vec{c}=0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}\]
This is so as there is no term on RHS of (i) having $\vec{c}$
Again comparing terms of \[\left( \vec{a}\cdot \vec{c} \right)\vec{b}=\dfrac{1}{2}\vec{b}\] and cancelling vector $\vec{b}$ we get
\[\left( \vec{a}\cdot \vec{c} \right)=\dfrac{1}{2}\]
We have a formula of dot product of two vectors $\vec{p}\text{ and }\vec{q}$ when angle between them is $\theta $ is given as
\[\vec{p}\cdot \vec{q}=\left| {\vec{p}} \right|\left| {\vec{q}} \right|\cos \theta \]
Where $\vec{p}$ is magnitude of $\vec{p}\text{ and }\left| {\vec{q}} \right|$ is magnitude of $\vec{q}$
Using this formula above in $\vec{a}\cdot \vec{c}=\dfrac{1}{2}$ we get
\[\left| {\vec{a}} \right|\left| {\vec{c}} \right|\cos \beta =\dfrac{1}{2}\]
This is so as we had angle between $\vec{a}\text{ and }\vec{c}$ as $\beta $
Now, $\vec{a}\text{ and }\vec{c}$ are both unit vectors $\left| {\vec{a}} \right|=1\text{ and }\left| {\vec{c}} \right|=1$
Substituting this in above equation we get:
\[\begin{align}
  & 1\cdot 1\cos \beta =\dfrac{1}{2} \\
 & \cos \beta =\dfrac{1}{2} \\
\end{align}\]
Now, we know that value of \[\cos \dfrac{\pi }{3}=\dfrac{1}{2}\Rightarrow \cos \beta =\cos \dfrac{\pi }{3}\]
Applying \[{{\cos }^{-1}}\] both sides we get:
\[\beta =\dfrac{\pi }{3}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)}\]
We have obtained from equation (ii) that
\[\begin{align}
  & 0=-\left( \vec{a}\cdot \vec{b} \right)\cdot \vec{c} \\
 & \left( \vec{a}\cdot \vec{b} \right)\cdot \vec{c}=0 \\
\end{align}\]
Now, as $\vec{c}$ is unit vector $\left| {\vec{c}} \right|=1$ then
\[\begin{align}
  & \left( \vec{a}\cdot \vec{b} \right)\cdot \vec{c}=0 \\
 & \vec{a}\cdot \vec{b}=0 \\
\end{align}\]
Now, applying formula of dot product stated above between $\vec{a}\text{ and }\vec{b}$ we get:
\[\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \alpha =0\]
As angle between $\vec{a}\text{ and }\vec{b}$ is $\alpha $
Now, as $\left| {\vec{a}} \right|=1=\left| {\vec{b}} \right|$ as both $\vec{a}\text{ and }\vec{b}$ are unit vector.
\[\begin{align}
  & 1\cdot 1\cdot \cos \alpha =0 \\
 & \cos \alpha =0 \\
\end{align}\]
We know that \[\cos \dfrac{\pi }{2}=0\Rightarrow \cos \alpha =\cos \dfrac{\pi }{2}\]
Applying ${{\cos }^{-1}}$ both sides of above equation we get
\[\alpha =\dfrac{\pi }{2}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iv)}\]
So from equation (iii) and (iv) we have $\alpha =\dfrac{\pi }{2}\text{ and }\beta =\dfrac{\pi }{3}$
Then value of \[\begin{align}
  & \left| \alpha -\beta \right|=\left| \dfrac{\pi }{2}-\dfrac{\pi }{3} \right| \\
 & \left| \alpha -\beta \right|=\left| \dfrac{3\pi -2\pi }{6} \right|=\left| \dfrac{\pi }{6} \right| \\
\end{align}\]
Value of $\left| \alpha -\beta \right|=\left| \dfrac{\pi }{6} \right|\Rightarrow {{30}^{\circ }}$

So, the correct answer is “Option B”.

Note: The possibility of confusion in this question can be at the point where we have used \[\left( \vec{a}\cdot \vec{b} \right)\cdot \vec{c}=0\Rightarrow \vec{a}\cdot \vec{b}=0\] this is possible as $\vec{c}$ is a unit vector. $\vec{c}$ being unit vector $\left| {\vec{c}} \right|=1$ hence if the magnitude of any vector is 1 or non-zero then it cannot be a 0 vector. So, only possibility left is that \[\vec{a}\cdot \vec{b}=0\]