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# Adjacent sides of a parallelogram are given by the vector $2\hat i - \hat j + 2\hat k$ and $\hat i + 5\hat j + \hat k$. Find a unit vector in the direction of its diagonal. Also, find the area of a parallelogram.

Last updated date: 13th Jun 2024
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Hint: By parallelogram theory of vector addition, the diagonal of the parallelogram is the vector sum of the 2 sides of the parallelogram. After finding the diagonal, we can find the unit vector by dividing the vector with its magnitude. The area can be found by taking the magnitude of the cross product of the 2 vectors.

Complete step by step answer:

We have the vectors $\vec a = 2\hat i - \hat j + 2\hat k$and $\vec b = \hat i + 5\hat j + \hat k$ as sides of a parallelogram. By parallelogram law of vector addition, we get the diagonal vector as the sum these vectors. So, the diagonal vector is given by,
$\vec c = \vec a + \vec b = \left( {2\hat i - \hat j + 2\hat k} \right) + \left( {\hat i + 5\hat j + \hat k} \right)$
We can add 2 vectors by adding their vector components. So, we get,
$\vec c = \left( {2 + 1} \right)\hat i + \left( {5 - 1} \right)\hat j + \left( {2 + 1} \right)\hat k \\ = 3\hat i + 4\hat j + 3\hat k \\$
Now we can find the magnitude of the diagonal vector
$\left| {\vec c} \right| = \sqrt {{3^2} + {4^2} + {3^2}} = \sqrt {9 + 16 + 9} \\ = \sqrt {34} \\$
The unit vector is given by,
$\hat c = \dfrac{{\vec c}}{{\left| c \right|}} = \dfrac{{3\hat i + 4\hat j + 3\hat k}}{{\sqrt {34} }} = \dfrac{3}{{\sqrt {34} }}\hat i + \dfrac{4}{{\sqrt {34} }}\hat j + \dfrac{3}{{\sqrt {34} }}\hat k$
Therefore, the required unit vector is $\dfrac{3}{{\sqrt {34} }}\hat i + \dfrac{4}{{\sqrt {34} }}\hat j + \dfrac{3}{{\sqrt {34} }}\hat k$
To find the area of the parallelogram we take the cross product of $\vec a = 2\hat i - \hat j + 2\hat k$and $\vec b = \hat i + 5\hat j + \hat k$
$\vec a \times \vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 2&{ - 1}&2 \\ 1&5&1 \end{array}} \right| \\ = \left( { - 1 - 10} \right)\hat i - \left( {2 - 2} \right)\hat j + \left( {10 + 1} \right)\hat k \\ = - 11\hat i + 11\hat k \\$
Area is given by magnitude of the cross product
$A = \left| {\vec a \times \vec b} \right| = \sqrt {{{11}^2} + {{11}^2}} = \sqrt {2 \times {{11}^2}} = 11\sqrt 2 {\text{ }}sq.{\text{ }}units$
Therefore, area of the parallelogram is $11\sqrt 2 {\text{ }}sq.{\text{ }}units$.

Note: As vector addition is commutative, we can add the vector in any order. Cross product or the vector product of two vectors gives a vector that is perpendicular to both the vectors. The magnitude of the cross product of 2 vectors gives the area of the parallelogram formed by the vectors. Cross product is calculated using the determinant method.