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# Let $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ be three non – zero vectors such that no two of them are collinear and $\left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \overrightarrow{c}=\dfrac{1}{3}\left| \overrightarrow{b} \right|\left| \overrightarrow{c} \right|\overrightarrow{a}$, if $\theta$ is the angle between vectors $\overrightarrow{b}$ and $\overrightarrow{c}$, then a value of sin$\theta$ is(a) $\dfrac{2\sqrt{2}}{3}$ (b) $\dfrac{-\sqrt{2}}{3}$ (c) $\dfrac{2}{3}$ (d) $\dfrac{-2\sqrt{3}}{3}$

Hint: To solve this question, we will use the BAC – CAB identity which states the triple cross product of three vectors. We will apply BAC – CAB on the left hand side. Then we will equate the coefficients of vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$. We have to keep in mind that the dot product of two vectors is defined as the product of their magnitudes and cos of the angle between them. Thus, we will try to find the value of either sine or the cosine of the angle $\theta$. Once we get the value of either one of them, we can find the value of sin$\theta$.

It is given to us that $\left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \overrightarrow{c}=\dfrac{1}{3}\left| \overrightarrow{b} \right|\left| \overrightarrow{c} \right|\overrightarrow{a}$, where $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ be three non – zero vectors such that no two of them are collinear.
According to BAC – CAB identity, the triple cross product is given as:
\begin{align} & A\times \left( B\times C \right)=B\left( A.C \right)-C\left( A.B \right)......\left( 1 \right) \\ & \left( A\times B \right)\times C=-C\times \left( A\times B \right) \\ & =-A\left( B.C \right)+B\left( A.C \right)......\left( 2 \right) \end{align}
We will use the second identity on the left hand side.
\begin{align} & \Rightarrow \left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \overrightarrow{c}=\dfrac{1}{3}\left| \overrightarrow{b} \right|\left| \overrightarrow{c} \right|\overrightarrow{a} \\ & \Rightarrow \left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{b}.\overrightarrow{c} \right)\overrightarrow{a}=\dfrac{1}{3}\left| \overrightarrow{b} \right|\left| \overrightarrow{c} \right|\overrightarrow{a} \\ \end{align}
It is given to use that $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ be three non – zero vectors such that no two of them are collinear. Thus, we can equate the coefficients of $\overrightarrow{a}$ and $\overrightarrow{b}$ from both sides.
$\Rightarrow \left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{b}.\overrightarrow{c} \right)\overrightarrow{a}=\dfrac{1}{3}\left| \overrightarrow{b} \right|\left| \overrightarrow{c} \right|\overrightarrow{a}+\left( 0 \right)\overrightarrow{b}$
The coefficient of $\overrightarrow{b}$ on left hand side is $\overrightarrow{a}.\overrightarrow{c}$ and on left hand side is 0.
$\Rightarrow \overrightarrow{a}.\overrightarrow{c}=0$
Similarly, coefficient of $\overrightarrow{a}$ on left hand side is $-\left( \overrightarrow{b}.\overrightarrow{c} \right)$ and on left hand side is $\dfrac{1}{3}\left| \overrightarrow{b} \right|\left| \overrightarrow{c} \right|$.
$\Rightarrow -\left( \overrightarrow{b}.\overrightarrow{c} \right)=\dfrac{1}{3}\left| \overrightarrow{b} \right|\left| \overrightarrow{c} \right|$
But we know that the dot product is defined as the product of the magnitudes of the two vectors and cosine of the angle between them. We also know from the question that $\theta$ is the angle between vectors $\overrightarrow{b}$ and $\overrightarrow{c}$.
\begin{align} & \Rightarrow -\left| \overrightarrow{b} \right|\left| \overrightarrow{c} \right|\cos \theta =\dfrac{1}{3}\left| \overrightarrow{b} \right|\left| \overrightarrow{c} \right| \\ & \Rightarrow \cos \theta =-\dfrac{1}{3} \\ \end{align}
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
$\Rightarrow \sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$
We will substitute $\cos \theta =-\dfrac{1}{3}$
\begin{align} & \Rightarrow \sin \theta =\sqrt{1-\dfrac{1}{9}} \\ & \Rightarrow \sin \theta =\sqrt{\dfrac{8}{9}} \\ & \Rightarrow \sin \theta =\dfrac{2\sqrt{2}}{3} \\ \end{align}

So, the correct answer is “Option A”.

Note: It is to be noted that the dot product of two vectors is a scalar and cross product of two vectors is a vector. Hence, when we equate the coefficients, we can include dot products but not cross products.