Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# 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: 13th Jul 2024
Total views: 453.9k
Views today: 8.53k
Verified
453.9k+ views
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
$A = \dfrac{1}{2}\left| {{{\vec P}_1} \times {{\vec P}_2}} \right| = \dfrac{1}{2} \times \sqrt {16 + 1024 + 57} = \dfrac{{\sqrt {1616} }}{2}$
$\Rightarrow \dfrac{{4\sqrt {101} }}{2} = 2\sqrt {101} sq.units$
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.