# The two adjacent sides of a parallelogram are $2\hat i - 4\hat j - 5\hat k$ and $2\hat i + 2\hat j + 3\hat k$. Find the two-unit vectors parallel to its diagonals. Using the diagonals vectors, find the area of the parallelogram.

Last updated date: 21st Mar 2023

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Answer

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Hint: Area of Parallelogram,$A = \dfrac{1}{2}\left| {{{\vec P}_1} \times {{\vec P}_2}} \right|$.

According to the question, we have two adjacent sides of parallelogram, which is $2\hat i - 4\hat j - 5\hat k$ and $2\hat i + 2\hat j + 3\hat k$

Now first we will assume the given value: -

$

\Rightarrow \vec a = 2\hat i - 4\hat j - 5\hat k \\

\Rightarrow \vec b = 2\hat i + 2\hat j + 3\hat k \\

$

And we know that any one diagonal of a parallelogram is given as

$

\vec P = \vec a + \vec b \\

\Rightarrow 2\hat i - 4\hat j - 5\hat k + 2\hat i + 2\hat j + 3\hat k \\

\Rightarrow 4\hat i - 2\hat j - 2\hat k \\

$

Therefore, we can calculate the unit vector along the diagonal, that is

$\dfrac{{{{\vec P}_1}}}{{\left| {{{\vec P}_1}} \right|}} = \dfrac{{4\hat i - 2\hat j - 2\hat k}}{{\sqrt {16 + 4 + 4} }} = \dfrac{{4\hat i - 2\hat j - 2\hat k}}{{\sqrt {24} }} = \dfrac{{4\hat i - 2\hat j - 2\hat k}}{{2 \times \sqrt 6 }}$

$ \Rightarrow \dfrac{{2\hat i - \hat j - \hat k}}{{\sqrt 6 }}$

Also, another diagonal of a parallelogram is given by: -

$

\Rightarrow {{\vec P}_2} = \vec b - \vec a \\

\Rightarrow 2\hat i + 2\hat j + 3\hat k - 2\hat i + 4\hat j + \hat k \\

\Rightarrow 6\hat j + 8\hat k \\

$

Therefore, unit vector along the diagonal is given by: -

$

\dfrac{{{{\vec P}_2}}}{{\left| {{{\vec P}_2}} \right|}} = \dfrac{{6\hat j + 8\hat k}}{{\sqrt {36 + 64} }} = \dfrac{{6\hat j + 8\hat k}}{{\sqrt {100} }} = \dfrac{{6\hat j + 8\hat k}}{{10}} \\

\Rightarrow \dfrac{{3\hat j + 4\hat k}}{5} \\

$

Now, we will take the cross product of the two diagonals

$

\Rightarrow {{\vec P}_1} \times {{\vec P}_2} \\

\Rightarrow \left( {\begin{array}{*{20}{c}}

{\hat i}&{\hat j}&{\hat k} \\

4&{ - 2}&{ - 2} \\

0&6&8

\end{array}} \right) \\

$

Further solving and simplify gives

$

\Rightarrow \hat i\left( { - 16 + 12} \right) - \hat j\left( {32 - 0} \right) + \hat k\left( {24 - 0} \right) \\

\Rightarrow - 4\hat i - 32\hat j + 24\hat k \\

$

From here, we will calculate the Area of parallelogram

$A = \dfrac{1}{2}\left| {{{\vec P}_1} \times {{\vec P}_2}} \right| = \dfrac{1}{2} \times \sqrt {16 + 1024 + 57} = \dfrac{{\sqrt {1616} }}{2}$

So, the answer is

$ \Rightarrow \dfrac{{4\sqrt {101} }}{2} = 2\sqrt {101} sq.units$

Note: - Whenever such a type of question is asked Always start with finding the diagonals of a parallelogram. After that find the unit vector along the diagonals one by one. Then use the formula Area of Parallelogram in terms of Diagonals,$A = \dfrac{1}{2}\left| {{{\vec P}_1} \times {{\vec P}_2}} \right|$ where ${\vec P_1}$and${\vec P_2}$are diagonals of a Parallelogram.

According to the question, we have two adjacent sides of parallelogram, which is $2\hat i - 4\hat j - 5\hat k$ and $2\hat i + 2\hat j + 3\hat k$

Now first we will assume the given value: -

$

\Rightarrow \vec a = 2\hat i - 4\hat j - 5\hat k \\

\Rightarrow \vec b = 2\hat i + 2\hat j + 3\hat k \\

$

And we know that any one diagonal of a parallelogram is given as

$

\vec P = \vec a + \vec b \\

\Rightarrow 2\hat i - 4\hat j - 5\hat k + 2\hat i + 2\hat j + 3\hat k \\

\Rightarrow 4\hat i - 2\hat j - 2\hat k \\

$

Therefore, we can calculate the unit vector along the diagonal, that is

$\dfrac{{{{\vec P}_1}}}{{\left| {{{\vec P}_1}} \right|}} = \dfrac{{4\hat i - 2\hat j - 2\hat k}}{{\sqrt {16 + 4 + 4} }} = \dfrac{{4\hat i - 2\hat j - 2\hat k}}{{\sqrt {24} }} = \dfrac{{4\hat i - 2\hat j - 2\hat k}}{{2 \times \sqrt 6 }}$

$ \Rightarrow \dfrac{{2\hat i - \hat j - \hat k}}{{\sqrt 6 }}$

Also, another diagonal of a parallelogram is given by: -

$

\Rightarrow {{\vec P}_2} = \vec b - \vec a \\

\Rightarrow 2\hat i + 2\hat j + 3\hat k - 2\hat i + 4\hat j + \hat k \\

\Rightarrow 6\hat j + 8\hat k \\

$

Therefore, unit vector along the diagonal is given by: -

$

\dfrac{{{{\vec P}_2}}}{{\left| {{{\vec P}_2}} \right|}} = \dfrac{{6\hat j + 8\hat k}}{{\sqrt {36 + 64} }} = \dfrac{{6\hat j + 8\hat k}}{{\sqrt {100} }} = \dfrac{{6\hat j + 8\hat k}}{{10}} \\

\Rightarrow \dfrac{{3\hat j + 4\hat k}}{5} \\

$

Now, we will take the cross product of the two diagonals

$

\Rightarrow {{\vec P}_1} \times {{\vec P}_2} \\

\Rightarrow \left( {\begin{array}{*{20}{c}}

{\hat i}&{\hat j}&{\hat k} \\

4&{ - 2}&{ - 2} \\

0&6&8

\end{array}} \right) \\

$

Further solving and simplify gives

$

\Rightarrow \hat i\left( { - 16 + 12} \right) - \hat j\left( {32 - 0} \right) + \hat k\left( {24 - 0} \right) \\

\Rightarrow - 4\hat i - 32\hat j + 24\hat k \\

$

From here, we will calculate the Area of parallelogram

$A = \dfrac{1}{2}\left| {{{\vec P}_1} \times {{\vec P}_2}} \right| = \dfrac{1}{2} \times \sqrt {16 + 1024 + 57} = \dfrac{{\sqrt {1616} }}{2}$

So, the answer is

$ \Rightarrow \dfrac{{4\sqrt {101} }}{2} = 2\sqrt {101} sq.units$

Note: - Whenever such a type of question is asked Always start with finding the diagonals of a parallelogram. After that find the unit vector along the diagonals one by one. Then use the formula Area of Parallelogram in terms of Diagonals,$A = \dfrac{1}{2}\left| {{{\vec P}_1} \times {{\vec P}_2}} \right|$ where ${\vec P_1}$and${\vec P_2}$are diagonals of a Parallelogram.

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