Respuesta :
Angle between [tex]u = -5i-4j , v=-4i-3j[/tex] is [tex]x =0[/tex]° .
Step-by-step explanation:
We have , two vectors u = <-5, -4>, v = <-4, -3> or , [tex]u = -5i-4j , v=-4i-3j[/tex]
We need to find angle between these two vectors . Let's find out:
We know that dot product of two vectors is defined as :
[tex]u.v =|u|(|v|)cosx[/tex] , where x is angle between u & v !
⇒ [tex]u.v =|u|(|v|)cosx[/tex]
⇒ [tex]cosx =\frac{u.v}{|u|(|v|)}[/tex]
Now , [tex]u.v = (-5i-4j)(-4i-3j)[/tex]
⇒ [tex]u.v = (-5i-4j)(-4i)-(-5i-4j)(3j)[/tex]
⇒ [tex]u.v = 20+12[/tex] { [tex]i(j) = j(i) =0[/tex] }
⇒ [tex]u.v = 32[/tex]
Now , Modulus of any vector [tex]r = xi+yj[/tex] is [tex]|r| = \sqrt{x^{2}+y^{2}}[/tex] So ,
[tex]|u| = \sqrt{(-5)^{2}+(-4)^{2}} = \sqrt{25+16} = \sqrt{41} \\\\|v| = \sqrt{(-4)^{2}+(-3)^{2}} = \sqrt{16+9} = \sqrt{25} = 5[/tex]
Putting all these values in equation [tex]cosx =\frac{u.v}{|u|(|v|)}[/tex] we get:
⇒ [tex]cosx =\frac{32}{5(\sqrt{41})}[/tex]
⇒ [tex]cos^{-1}(cosx) =cos^{-1}(\frac{32}{5(6.4)})[/tex]
⇒ [tex]x =cos^{-1}(1)[/tex] { [tex]cos0 = 1[/tex] }
⇒ [tex]x =0[/tex]°
Therefore , Angle between [tex]u = -5i-4j , v=-4i-3j[/tex] is [tex]x =0[/tex]° .
