Discussion Forum

The length of longer diagonal of the parallelogram constructed on 5a+2b and a-3b , if it is given that $|a|= 2\sqrt{2}$.|b|=3 and the angle between a and b is $\frac{\pi}{4}$ , is 

Answer:

Explanation:

Given : |a|=$2 \sqrt{2}$,|b|=3

One diagonal is 5a+2b+a-3b=6a-b

 Length of one diagonal

=|6a-b|

=$\sqrt{36a^{2}+b^{2}-2 \times 6|a|.|b| \cos 45^{0}}$

 =$\sqrt{36\times 8+9-12\times2\sqrt{2}\times3\times\frac{1}{\sqrt{2}}}$

=$\sqrt{288+9-12\times6}=\sqrt{225}=15$

other diagonal is (5a+2b)-(a-3b)=4a+5b

Its length is 

=$\sqrt{(4a)^{2}+(5b)^{2}+2\times |4a||5b| \cos 45^{0}}$

=$\sqrt{16\times8+25\times9+40\times6}=\sqrt{593}$