Let $\overline{a}={{\alpha }_{1}}\widehat{i}+{{\alpha }_{2}}\widehat{j}+{{\alpha }_{3}}\widehat{k},\overline{b}={{\beta }_{1}}\widehat{i}+{{\beta }_{2}}\widehat{j}+{{\beta }_{3}}\widehat{k}$ and $\overline{c}={{\gamma }_{1}}\widehat{i}+{{\gamma }_{2}}\widehat{j}+{{\gamma }_{3}}\widehat{k},\left| \overline{a} \right|=2\sqrt{2},\overline{a}$makes an angle $\dfrac{\pi }{3}$ with the plane of $\overline{b},\overline{c}$ and angle between $\overline{b},\overline{c}$ is$\dfrac{\pi }{6}$, then ${{\left| \begin{matrix}
{{\alpha }_{1}} & {{\alpha }_{2}} & {{\alpha }_{3}} \\
{{\beta }_{1}} & {{\beta }_{2}} & {{\beta }_{3}} \\
{{\gamma }_{1}} & {{\gamma }_{2}} & {{\gamma }_{3}} \\
\end{matrix} \right|}^{n}}$is equal to (n is even natural number)
A) ${{\left[ \dfrac{\left| \overline{b} \right|\left| \overline{a} \right|}{6} \right]}^{\dfrac{n}{2}}}$
B) ${{\left[ \dfrac{\sqrt{3}\left| \overline{b} \right|\left| \overline{c} \right|}{\sqrt{2}} \right]}^{n}}$
C) $\dfrac{\left| \overline{b} \right|{{\left| \overline{c} \right|}^{\dfrac{n}{2}}}}{\sqrt{3}{{2}^{n}}}$
D) ${{\left[ \dfrac{\sqrt{2}\left| \overline{b} \right|\left| \overline{a} \right|}{\sqrt{3}} \right]}^{n}}$
Answer
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Hint: For solving this question we will simplify the given ${{\left| \begin{matrix}
{{\alpha }_{1}} & {{\alpha }_{2}} & {{\alpha }_{3}} \\
{{\beta }_{1}} & {{\beta }_{2}} & {{\beta }_{3}} \\
{{\gamma }_{1}} & {{\gamma }_{2}} & {{\gamma }_{3}} \\
\end{matrix} \right|}^{n}}$using the basic concept that $\left| \begin{matrix}
{{\alpha }_{1}} & {{\alpha }_{2}} & {{\alpha }_{3}} \\
{{\beta }_{1}} & {{\beta }_{2}} & {{\beta }_{3}} \\
{{\gamma }_{1}} & {{\gamma }_{2}} & {{\gamma }_{3}} \\
\end{matrix} \right|=\left[ \begin{matrix}
\overline{a} & \overline{b} & \overline{c} \\
\end{matrix} \right]$and $\begin{align}
& \left[ \begin{matrix}
\overline{a} & \overline{b} & \overline{c} \\
\end{matrix} \right]=\left( \overline{a}\times \overline{b} \right)\cdot \overline{c} \\
& \Rightarrow \overline{a}\cdot \left( \overline{b}\times \overline{c} \right)
\end{align}$ and conclude to the answer.
Complete step by step answer:
Given in the question that $\overline{a}={{\alpha }_{1}}\widehat{i}+{{\alpha }_{2}}\widehat{j}+{{\alpha }_{3}}\widehat{k},\overline{b}={{\beta }_{1}}\widehat{i}+{{\beta }_{2}}\widehat{j}+{{\beta }_{3}}\widehat{k}$ and $\overline{c}={{\gamma }_{1}}\widehat{i}+{{\gamma }_{2}}\widehat{j}+{{\gamma }_{3}}\widehat{k},\left| \overline{a} \right|=2\sqrt{2},\overline{a}$makes an angle $\dfrac{\pi }{3}$ with the plane of $\overline{b},\overline{c}$ and angle between $\overline{b},\overline{c}$ is$\dfrac{\pi }{6}$ where $\overline{a},\overline{b},\overline{c}$ are 3 vectors.
${{\left| \begin{matrix}
{{\alpha }_{1}} & {{\alpha }_{2}} & {{\alpha }_{3}} \\
{{\beta }_{1}} & {{\beta }_{2}} & {{\beta }_{3}} \\
{{\gamma }_{1}} & {{\gamma }_{2}} & {{\gamma }_{3}} \\
\end{matrix} \right|}^{n}}$can be written as ${{\left[ \begin{matrix}
\overline{a} & \overline{b} & \overline{c} \\
\end{matrix} \right]}^{n}}$
As we know from the basic concept that $\begin{align}
& \left[ \begin{matrix}
\overline{a} & \overline{b} & \overline{c} \\
\end{matrix} \right]=\left( \overline{a}\times \overline{b} \right)\cdot \overline{c} \\
& \Rightarrow \overline{a}\cdot \left( \overline{b}\times \overline{c} \right)
\end{align}$
From the concept as we know that the normal of the plane containing $\overline{b},\overline{c}$ is $\overline{b}\times \overline{c}$.
From the question it is given that $\overline{a}$ makes an angle $\dfrac{\pi }{3}$ with the plane of $\overline{b},\overline{c}$
So $\overline{a}$will make an angle $\dfrac{\pi }{6}$ with the normal of the plane containing $\overline{b},\overline{c}$ that is $\overline{b}\times \overline{c}$.
As we know that $\overline{a}\cdot \overline{b}=\left| \overline{a} \right|\left| \overline{b} \right|\cos \theta $where $\theta $is the angle between the normal of the plane containing of$\overline{b}$and $\overline{a}$.
We also know that $\overline{a}\times \overline{b}=\left| \overline{a} \right|\left| \overline{b} \right|\sin \theta \left| \widehat{n} \right|$where $\theta $is the angle between$\overline{a},\overline{b}$ and $\left| \widehat{n} \right|$ is the magnitude of the unit vector along the normal which will be equal to 1.
Hence the simplified form of ${{\left| \begin{matrix}
{{\alpha }_{1}} & {{\alpha }_{2}} & {{\alpha }_{3}} \\
{{\beta }_{1}} & {{\beta }_{2}} & {{\beta }_{3}} \\
{{\gamma }_{1}} & {{\gamma }_{2}} & {{\gamma }_{3}} \\
\end{matrix} \right|}^{n}}$ will be
$\begin{align}
& \Rightarrow {{\left[ \begin{matrix}
\overline{a} & \overline{b} & \overline{c} \\
\end{matrix} \right]}^{n}} \\
& \Rightarrow {{\left( \overline{a}\cdot \left( \overline{b}\times \overline{c} \right) \right)}^{n}} \\
& \Rightarrow {{\left( \left| \overline{a} \right|.\left| \overline{b}\times \overline{c} \right|\cos \dfrac{\pi }{6} \right)}^{n}}
\end{align}$
Since, $\overline{a}$makes an angle $\dfrac{\pi }{6}$ with the normal of the plane containing $\overline{b},\overline{c}$ that is $\overline{b}\times \overline{c}$.
$\begin{align}
& {{\left( \left| \overline{a} \right|.\left| \overline{b}\times \overline{c} \right|\cos \dfrac{\pi }{6} \right)}^{n}} \\
& \Rightarrow {{\left( 2\sqrt{2}\left| \overline{b}\times \overline{c} \right|\dfrac{\sqrt{3}}{2} \right)}^{n}} \\
\end{align}$
Since,$\left| \overline{a} \right|=2\sqrt{2}$is given in the question and we know that $\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$
This can be further simplified as
$\begin{align}
& {{\left( 2\sqrt{2}\left| \overline{b}\times \overline{c} \right|\dfrac{\sqrt{3}}{2} \right)}^{^{{}}}} \\
& \Rightarrow {{\left( \sqrt{6}\left| \overline{b}\times \overline{c} \right| \right)}^{n}} \\
& \Rightarrow {{\left( \sqrt{6}\left| \overline{b} \right|\left| \overline{c} \right|\sin \dfrac{\pi }{6}\left| \widehat{n} \right| \right)}^{n}} \\
\end{align}$
Since, we know that $\left| \widehat{n} \right|=1$and $\sin \dfrac{\pi }{6}=\dfrac{1}{2}$
$\begin{align}
& {{\left( \sqrt{6}\left| \overline{b} \right|\left| \overline{c} \right|\sin \dfrac{\pi }{6}\left| \widehat{n} \right| \right)}^{n}} \\
& \Rightarrow {{\left( \sqrt{6}\left| \overline{b} \right|\left| \overline{c} \right|\dfrac{1}{2} \right)}^{n}} \\
& \Rightarrow {{\left( \dfrac{\sqrt{3}}{\sqrt{2}}\left| \overline{b} \right|\left| \overline{c} \right| \right)}^{n}} \\
\end{align}$
We can conclude that${{\left| \begin{matrix}
{{\alpha }_{1}} & {{\alpha }_{2}} & {{\alpha }_{3}} \\
{{\beta }_{1}} & {{\beta }_{2}} & {{\beta }_{3}} \\
{{\gamma }_{1}} & {{\gamma }_{2}} & {{\gamma }_{3}} \\
\end{matrix} \right|}^{n}}$= ${{\left[ \dfrac{\sqrt{3}\left| \overline{b} \right|\left| \overline{c} \right|}{\sqrt{2}} \right]}^{n}}$
Hence, option (B) is correct.
Note:
Here we should take care that $\overline{a}\cdot \overline{b}=\left| \overline{a} \right|\left| \overline{b} \right|\cos \theta $where $\theta $is the angle between the normal of the plane containing of$\overline{b}$and $\overline{a}$ and $\overline{a}\times \overline{b}=\left| \overline{a} \right|\left| \overline{b} \right|\sin \theta \left| \widehat{n} \right|$where $\theta $is the angle between$\overline{a},\overline{b}$ and $\left| \widehat{n} \right|$ is the magnitude of the unit vector along the normal which will be equal to 1 if we by mistake take the interchange $\cos \theta $with $\sin \theta $the entire result will be different and we will end up concluding a wrong answer.
{{\alpha }_{1}} & {{\alpha }_{2}} & {{\alpha }_{3}} \\
{{\beta }_{1}} & {{\beta }_{2}} & {{\beta }_{3}} \\
{{\gamma }_{1}} & {{\gamma }_{2}} & {{\gamma }_{3}} \\
\end{matrix} \right|}^{n}}$using the basic concept that $\left| \begin{matrix}
{{\alpha }_{1}} & {{\alpha }_{2}} & {{\alpha }_{3}} \\
{{\beta }_{1}} & {{\beta }_{2}} & {{\beta }_{3}} \\
{{\gamma }_{1}} & {{\gamma }_{2}} & {{\gamma }_{3}} \\
\end{matrix} \right|=\left[ \begin{matrix}
\overline{a} & \overline{b} & \overline{c} \\
\end{matrix} \right]$and $\begin{align}
& \left[ \begin{matrix}
\overline{a} & \overline{b} & \overline{c} \\
\end{matrix} \right]=\left( \overline{a}\times \overline{b} \right)\cdot \overline{c} \\
& \Rightarrow \overline{a}\cdot \left( \overline{b}\times \overline{c} \right)
\end{align}$ and conclude to the answer.
Complete step by step answer:
Given in the question that $\overline{a}={{\alpha }_{1}}\widehat{i}+{{\alpha }_{2}}\widehat{j}+{{\alpha }_{3}}\widehat{k},\overline{b}={{\beta }_{1}}\widehat{i}+{{\beta }_{2}}\widehat{j}+{{\beta }_{3}}\widehat{k}$ and $\overline{c}={{\gamma }_{1}}\widehat{i}+{{\gamma }_{2}}\widehat{j}+{{\gamma }_{3}}\widehat{k},\left| \overline{a} \right|=2\sqrt{2},\overline{a}$makes an angle $\dfrac{\pi }{3}$ with the plane of $\overline{b},\overline{c}$ and angle between $\overline{b},\overline{c}$ is$\dfrac{\pi }{6}$ where $\overline{a},\overline{b},\overline{c}$ are 3 vectors.
${{\left| \begin{matrix}
{{\alpha }_{1}} & {{\alpha }_{2}} & {{\alpha }_{3}} \\
{{\beta }_{1}} & {{\beta }_{2}} & {{\beta }_{3}} \\
{{\gamma }_{1}} & {{\gamma }_{2}} & {{\gamma }_{3}} \\
\end{matrix} \right|}^{n}}$can be written as ${{\left[ \begin{matrix}
\overline{a} & \overline{b} & \overline{c} \\
\end{matrix} \right]}^{n}}$
As we know from the basic concept that $\begin{align}
& \left[ \begin{matrix}
\overline{a} & \overline{b} & \overline{c} \\
\end{matrix} \right]=\left( \overline{a}\times \overline{b} \right)\cdot \overline{c} \\
& \Rightarrow \overline{a}\cdot \left( \overline{b}\times \overline{c} \right)
\end{align}$
From the concept as we know that the normal of the plane containing $\overline{b},\overline{c}$ is $\overline{b}\times \overline{c}$.
From the question it is given that $\overline{a}$ makes an angle $\dfrac{\pi }{3}$ with the plane of $\overline{b},\overline{c}$
So $\overline{a}$will make an angle $\dfrac{\pi }{6}$ with the normal of the plane containing $\overline{b},\overline{c}$ that is $\overline{b}\times \overline{c}$.
As we know that $\overline{a}\cdot \overline{b}=\left| \overline{a} \right|\left| \overline{b} \right|\cos \theta $where $\theta $is the angle between the normal of the plane containing of$\overline{b}$and $\overline{a}$.
We also know that $\overline{a}\times \overline{b}=\left| \overline{a} \right|\left| \overline{b} \right|\sin \theta \left| \widehat{n} \right|$where $\theta $is the angle between$\overline{a},\overline{b}$ and $\left| \widehat{n} \right|$ is the magnitude of the unit vector along the normal which will be equal to 1.
Hence the simplified form of ${{\left| \begin{matrix}
{{\alpha }_{1}} & {{\alpha }_{2}} & {{\alpha }_{3}} \\
{{\beta }_{1}} & {{\beta }_{2}} & {{\beta }_{3}} \\
{{\gamma }_{1}} & {{\gamma }_{2}} & {{\gamma }_{3}} \\
\end{matrix} \right|}^{n}}$ will be
$\begin{align}
& \Rightarrow {{\left[ \begin{matrix}
\overline{a} & \overline{b} & \overline{c} \\
\end{matrix} \right]}^{n}} \\
& \Rightarrow {{\left( \overline{a}\cdot \left( \overline{b}\times \overline{c} \right) \right)}^{n}} \\
& \Rightarrow {{\left( \left| \overline{a} \right|.\left| \overline{b}\times \overline{c} \right|\cos \dfrac{\pi }{6} \right)}^{n}}
\end{align}$
Since, $\overline{a}$makes an angle $\dfrac{\pi }{6}$ with the normal of the plane containing $\overline{b},\overline{c}$ that is $\overline{b}\times \overline{c}$.
$\begin{align}
& {{\left( \left| \overline{a} \right|.\left| \overline{b}\times \overline{c} \right|\cos \dfrac{\pi }{6} \right)}^{n}} \\
& \Rightarrow {{\left( 2\sqrt{2}\left| \overline{b}\times \overline{c} \right|\dfrac{\sqrt{3}}{2} \right)}^{n}} \\
\end{align}$
Since,$\left| \overline{a} \right|=2\sqrt{2}$is given in the question and we know that $\cos \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$
This can be further simplified as
$\begin{align}
& {{\left( 2\sqrt{2}\left| \overline{b}\times \overline{c} \right|\dfrac{\sqrt{3}}{2} \right)}^{^{{}}}} \\
& \Rightarrow {{\left( \sqrt{6}\left| \overline{b}\times \overline{c} \right| \right)}^{n}} \\
& \Rightarrow {{\left( \sqrt{6}\left| \overline{b} \right|\left| \overline{c} \right|\sin \dfrac{\pi }{6}\left| \widehat{n} \right| \right)}^{n}} \\
\end{align}$
Since, we know that $\left| \widehat{n} \right|=1$and $\sin \dfrac{\pi }{6}=\dfrac{1}{2}$
$\begin{align}
& {{\left( \sqrt{6}\left| \overline{b} \right|\left| \overline{c} \right|\sin \dfrac{\pi }{6}\left| \widehat{n} \right| \right)}^{n}} \\
& \Rightarrow {{\left( \sqrt{6}\left| \overline{b} \right|\left| \overline{c} \right|\dfrac{1}{2} \right)}^{n}} \\
& \Rightarrow {{\left( \dfrac{\sqrt{3}}{\sqrt{2}}\left| \overline{b} \right|\left| \overline{c} \right| \right)}^{n}} \\
\end{align}$
We can conclude that${{\left| \begin{matrix}
{{\alpha }_{1}} & {{\alpha }_{2}} & {{\alpha }_{3}} \\
{{\beta }_{1}} & {{\beta }_{2}} & {{\beta }_{3}} \\
{{\gamma }_{1}} & {{\gamma }_{2}} & {{\gamma }_{3}} \\
\end{matrix} \right|}^{n}}$= ${{\left[ \dfrac{\sqrt{3}\left| \overline{b} \right|\left| \overline{c} \right|}{\sqrt{2}} \right]}^{n}}$
Hence, option (B) is correct.
Note:
Here we should take care that $\overline{a}\cdot \overline{b}=\left| \overline{a} \right|\left| \overline{b} \right|\cos \theta $where $\theta $is the angle between the normal of the plane containing of$\overline{b}$and $\overline{a}$ and $\overline{a}\times \overline{b}=\left| \overline{a} \right|\left| \overline{b} \right|\sin \theta \left| \widehat{n} \right|$where $\theta $is the angle between$\overline{a},\overline{b}$ and $\left| \widehat{n} \right|$ is the magnitude of the unit vector along the normal which will be equal to 1 if we by mistake take the interchange $\cos \theta $with $\sin \theta $the entire result will be different and we will end up concluding a wrong answer.
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