Question

When to non-zero vectors a and b are perpendicular to each other their magnitude of resultant is R.
When they are opposte to each other their resultant is of magnitude R/√2. Find
Find the value of (a/b + b/a)​

Answer 1

Answer:

4

check the attachment.

#dynamite

Answer 2

Given,

case 1): when two vectors $a,b$ are perpendicular i.e angle between $a\&b$ is

$90$°

case 2): When two vectors $a,b$ are opposite to each other i.e angle between $a\&b$ is $180$°

Now we know that the resultant of two vectors is given by

$R= \sqrt{a^2+b^2+2abcos\theta}$

useful formula $(a-b)^2 = a^2+b^2-2ab$

From case(1)

$\theta = 90$° then,

$R_1= \sqrt{a^2+b^2+2ab(cos90)}$

we know that $cos90 = 0$

$R_1 = \sqrt{a^2+b^2+0} \\a^2+b^2=R_1^2$

given the value of resultant in first case is $R$ then

$a^2+b^2 = R^2\\ R = \sqrt{a^2+b^2} ----- > (A)$

From case(2)

$\theta = 180$° then,

$R= \sqrt{a^2+b^2+(2ab(cos180))}$

we know that $cos 180 =-1$

$R_2 = \sqrt{a^2+b^2+(2ab(-1))} \\R_2 = \sqrt{a^2+b^2-2ab} \\R_2= a-b$

given that resultant of second case is $\frac{R}{\sqrt{2} }$ then,

$\frac{R}{\sqrt{2} } = a-b\\\\R = (\sqrt{2})(a-b) ---- > (B)$

Now, from $(A)=(B)$  we get

$\sqrt{a^2+b^2} =\sqrt{2}(a-b)\\ a^2+b^2=2(a-b)^2\\a^2+b^2 =2[a^2+b^2-2ab]\\a^2+b^2=2a^2+2b^2-4ab\\4ab = a^2+b^2$

Now, divide the above equation with both $a \& b$ we get

$\frac{4ab}{ab} =\frac{a^2}{ab}+\frac{b^2}{ab} \\\\4 = \frac{a}{b}+\frac{b}{a} \\\\\frac{a}{b}+\frac{b}{a} = 4$

∴The value of $\frac{a}{b}+\frac{b}{a} =4$

#SPJ2