Find all real values of ‘t‘ so that angle between the vectors u = (t − 2, 6 − t, −4) and v = (−4, t − 2, 6 − t) is 120◦ .
Accepted Solution
A:
Answer:for all values Step-by-step explanation:u = (t - 2, 6 - t, - 4)v = ( - 4, t - 2, 6 - t) Angle between them, θ = 120°Use the concept of dot product of two vectors[tex]\overrightarrow{A}.\overrightarrow{B}=A B Cos\theta[/tex]Magnitude of u = [tex]\sqrt{(t-2)^{2}+(6-t)^{2}+(-4)^{2}}[/tex] = [tex]\sqrt{2t^{2}-16t+56}[/tex]Magnitude of v = [tex]\sqrt{(t-2)^{2}+(6-t)^{2}+(-4)^{2}}[/tex] = [tex]\sqrt{2t^{2}-16t+56}[/tex][tex]\overrightarrow{u}.\overrightarrow{v}=-4(t-2)+(6-t)(t-2)-4(6-t)=-t^{2}+8t-28[/tex]By the formula of dot product of two vectors[tex]Cos120 = \frac{-t^{2}+8t-28}{\sqrt{2t^{2}-16t+56}\times \sqrt{2t^{2}-16t+56}}[/tex][tex]-0.5\times {2t^{2}-16t+56} = {-t^{2}+8t-28}}[/tex][tex]{-t^{2}+8t-28}} = {-t^{2}+8t-28}}[/tex]So, for all values of t the angle between these two vectors be 120.