Question

Find the area of the parallelogram whose diagonals are $3\hat i + 4\hat j$ and $\hat i + \hat j + \hat k$.

Answer 1

Hint: Use the information, area of parallelogram when diagonals are given can be calculated as $\dfrac{1}{2}|{\vec d_1} \times {\vec d_2}|$, where ${\vec d_1}$ and ${\vec d_2}$ are its diagonals.

Complete step-by-step answer:

We know that, area of parallelogram when diagonals are given can be calculated as $\dfrac{1}{2}|{\vec d_1} \times {\vec d_2}|$, where ${\vec d_1}$ and ${\vec d_2}$ are its diagonals. Here, we have given the diagonals as $3\hat i + 4\hat j$ and $\hat i + \hat j + \hat k$. So, the required area is,
\dfrac{1}{2}|{{\vec d}_1} \times {{\vec d}_2}| \\\ = \dfrac{1}{2}|\left( {3\hat i + 4\hat j} \right) \times \left( {\hat i + \hat j + \hat k} \right)| \\\ = \dfrac{1}{2}\left| {\left[ {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\\ 3&4&0 \\\ 1&1&1 \end{array}} \right]} \right| \\\ = \dfrac{1}{2}\left| {\left( {4 - 0} \right)\hat i - \left( {3 - 0} \right)\hat j + \left( {3 - 4} \right)\hat k} \right| \\\ = \dfrac{1}{2}\left| {\left( 4 \right)\hat i - \left( 3 \right)\hat j - \hat k} \right| \to (1) \\\
As, we know, the magnitude of any vector of form $a\hat i + b\hat j + c\hat k$ will be
$\left| {a\hat i + b\hat j + c\hat k} \right| = \sqrt {{a^2} + {b^2} + {c^2}}$.
So, from the above formulae equation (1) can be written as
$= \dfrac{1}{2}\left( {\sqrt {{{(4)}^2} + {{( - 3)}^2} + {{( - 1)}^2}} } \right)\left( {\because a = 4,b = - 3,c = - 1} \right) \\\ = \dfrac{1}{2}\left( {\sqrt {16 + 9 + 1} } \right) \\\ = \dfrac{1}{2}\left( {\sqrt {26} } \right) \\\$
Hence the required area is $\dfrac{1}{2}\sqrt {26}$ square unit.

Note: In vector calculus, one needs to understand the formula in order to apply it. One needs to visualise for the sake of understanding and it is very important to remember the formula for calculation of modulus of vector , keeping the magnitude the same but changing the directions will not change the modulus of any vector.