
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.
Answer
485.1k+ views
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.
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.
Recently Updated Pages
What percentage of the area in India is covered by class 10 social science CBSE

The area of a 6m wide road outside a garden in all class 10 maths CBSE

What is the electric flux through a cube of side 1 class 10 physics CBSE

If one root of x2 x k 0 maybe the square of the other class 10 maths CBSE

The radius and height of a cylinder are in the ratio class 10 maths CBSE

An almirah is sold for 5400 Rs after allowing a discount class 10 maths CBSE

Trending doubts
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Why is there a time difference of about 5 hours between class 10 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

Write an application to the principal requesting five class 10 english CBSE
