Question

The length of the diagonals of a parallelogram constructed on the vectors $\overrightarrow p = 2\overrightarrow a + \overrightarrow b$and $\overrightarrow q = \overrightarrow a - 2\overrightarrow b$, where $\overrightarrow a ,\overrightarrow b$are unit vectors forming an angle of ${60^0}$ are?

Answer 1

First we have to define what the terms we need to solve the problem are.
First, we have to know about the parallelogram which is the simple quadrilateral with two pairs that are parallel sides. Also, from the opposite sides of the facing sides of a parallelogram are at the equivalent length.
Hence the opposite angles of the parallelogram are used to be equal in measure.
Thus, the length of the parallelogram is based on the area and base from the height.

Complete step by step answer:
Since from the given parallelogram the diagonals are given as in the form of $\overrightarrow p = 2\overrightarrow a + \overrightarrow b$and$\overrightarrow q = \overrightarrow a - 2\overrightarrow b$, but the general parallelogram diagonals are $\overrightarrow p + \overrightarrow q$and$\overrightarrow p - \overrightarrow q$.
Thus, the given diagonals can be rewritten as $\overrightarrow p + \overrightarrow q = 2\overrightarrow a + \overrightarrow b - (\overrightarrow a - 2\overrightarrow b )$ (substituting the values of p vector and q vector).
Simplifying we get$\overrightarrow p + \overrightarrow q = 3\overrightarrow a - \overrightarrow b$, similarly we can also able to find the value for$\overrightarrow p - \overrightarrow q = \overrightarrow a + 3\overrightarrow b$.
Now we are going to apply the square of the terms as well the root of the terms so the values will be not affected.
Thus, the modulus of these vectors can be rewritten as $\overrightarrow p + \overrightarrow q = \sqrt {9{a^2} - 6\overrightarrow {a.b} + {b^2}}$while square the terms the vector will be eliminated to whole square common terms.
Also, for $\overrightarrow p - \overrightarrow q = \sqrt {{a^2} + 6\overrightarrow {a.b} + 9{b^2}}$ (by subtraction of the vectors p and q)
Hence a and b are the vectors of the unit and modulus will be becoming as $\sqrt {9 - 6\cos {{60}^0} + 1}$and$\sqrt {1 + 6\cos {{60}^0} + 9}$, where cos sixty degree is one by three; and substitute the value into the root.
Therefore, we get $\sqrt 7 ,\sqrt {13}$is the required answer for the given question.

Note: since cos sixty degrees can be written as the value one, using the trigonometric tables.
Also, a and b are the unit vectors so the value will become one in the above simplification, thus by the vector’s formula, we get the required resultant.