The two adjacent sides of a parallelogram are 2\(\hat{i}\)-4\(\hat j\)+5\(\hat k\)and \(\hat{i}\)-2\(\hat j\)-3\(\hat k\). Find the unit vector parallel to its diagonal. Also, find its area.
The two adjacent sides of a parallelogram are 2\(\hat{i}\)-4\(\hat j\)+5\(\hat k\)and \(\hat{i}\)-2\(\hat j\)-3\(\hat k\). Find the unit vector parallel to its diagonal. Also, find its area.
Solution and Explanation
Adjacent sides of a parallelogram are given as \(\vec a\)=2\(\hat{i}\)-4\(\hat j\)+5\(\hat k\)and b→=\(\hat{i}\)-2\(\hat j\)-3\(\hat k\)
Then,the diagonal of a parallelogram is given by \(\vec a\)+\(\vec b\).
\(\vec a\)+\(\vec b\)=(2+1)\(\hat{i}\)+(-4-2)\(\hat j\)+(5-3)\(\hat k\)=3\(\hat{i}\)-6\(\hat j\)+2\(\hat k\)
Thus, the unit vector parallel to the diagonal is
\(\frac{\vec a+\vec b}{|\vec a+\vec b|}\)=\(\frac{3\hat i-6\hat j+2\hat k}{\sqrt{32}}\)+(-6)2+22=3\(\hat{i}\)-6\(\hat j\)+2\(\hat k\)\(\sqrt{9+36+4}\)=\(\frac{3\hat i-6\hat j+2\hat k}{7}\)=\(\frac{3}{7}\hat i-\frac{6}{7}\hat j+\frac{2}{7}\hat k\)
∴Area of parallelogram ABCD =\(\hat{i}\)(12+10)-\(\hat j\)(-6-5)+\(\hat k\)(-4+4)
=22\(\hat i\)+11\(\hat j\)
=11(2\(\hat i\)+\(\hat j\))
∴|\(\vec a\times \vec b\)|=11\(\sqrt{22}\)+12=11\(\sqrt{5}\)
Hence, the area of a parallelogram is 11\(\sqrt{5}\) square units.
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Concepts Used:
Vector Algebra
A vector is an object which has both magnitudes and direction. It is usually represented by an arrow which shows the direction(→) and its length shows the magnitude. The arrow which indicates the vector has an arrowhead and its opposite end is the tail. It is denoted as
The magnitude of the vector is represented as |V|. Two vectors are said to be equal if they have equal magnitudes and equal direction.
Vector Algebra Operations:
Arithmetic operations such as addition, subtraction, multiplication on vectors. However, in the case of multiplication, vectors have two terminologies, such as dot product and cross product.