Question

Find \[|\vec a|and|\vec b|,\,if(\vec a + \vec b).(\vec a - \vec b) = 8\,and\,\,|\vec a| = 8|\vec b|\].

Answer 1

Hint: We will multiply the vector by using the algebraic identity $(a + b)(a - b) = {a^2} - {b^2}$.
Thereafter we will calculate the value $a$ vector and $b$ vector.


Complete step by step solution:
$(\vec a + \vec b).(\vec a - \vec b) = 8$
Use algebraic identity $(a + b)(a - b) = {a^2} - {b^2}$
Then, $|\vec a{|^2} - |\vec b{|^2} = 8$ …..(1)
Now, \[|\vec a| = 8|\vec b|\] ….(2)
Now, substituting equation (2) in equation (1), we have
${[8|\vec b|]^2} = |\vec b{|^2} = 8$
$64|\vec b{|^2} - |\vec b{|^2} = 8$
$ \Rightarrow 63|\vec b{|^2} = 8$
$ \Rightarrow |\vec b{|^2} = \dfrac{8}{{63}}$
$|b{|^2} = \sqrt {\dfrac{8}{{63}}} $
$|\vec b| = \sqrt {\dfrac{{4 \times 2}}{{7 \times 9}}} $
$|\vec b| = \dfrac{{2\sqrt 2 }}{{3\sqrt 7 }}$
Now,
$|\vec a| = 8|\vec b|$
$|\vec a| = 9 \times \dfrac{{2\sqrt 2 }}{{3\sqrt 7 }}$
$|a| = \dfrac{{16\sqrt 2 }}{{3\sqrt 7 }}$
Additional information: The dot product between two vectors $a$ and $b$ is $a.b = ||a||||b||\cos \theta $,
where $\theta $ is the angle between vectors $a\,\,and\,\,b$.


Note: Students should solve the value of the square root carefully while you convert the square root value into square otherwise you will get the wrong answer.