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
Verified575.4k+ views
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.
Recently Updated Pages
Two men on either side of the cliff 90m height observe class 10 maths CBSE
Cutting of the Chinese melon means A The business and class 10 social science CBSE
Show an aquatic food chain using the following organisms class 10 biology CBSE
How is gypsum formed class 10 chemistry CBSE
If the line 3x + 4y 24 0 intersects the xaxis at t-class-10-maths-CBSE
Sugar present in DNA is A Heptose B Hexone C Tetrose class 10 biology CBSE
Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE
What is the median of the first 10 natural numbers class 10 maths CBSE
Indias first jute mill was established in 1854 in A class 10 social science CBSE
Indias first jute mill was established in 1854 in A class 10 social science CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths