Question

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◦ .

Answer 1

for all values

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

`\overrightarrow{A}.\overrightarrow{B}=A B Cos\theta`

Magnitude of u =
`\sqrt{(t-2)^(2)+(6-t)^(2)+(-4)^(2)}`

=
`\sqrt{2t^(2)-16t+56}`

Magnitude of v =
`\sqrt{(t-2)^(2)+(6-t)^(2)+(-4)^(2)}`

=
`\sqrt{2t^(2)-16t+56}`

`\overrightarrow{u}.\overrightarrow{v}=-4(t-2)+(6-t)(t-2)-4(6-t)=-t^(2)+8t-28`

By the formula of dot product of two vectors

`Cos120 = \frac{-t^(2)+8t-28}{\sqrt{2t^(2)-16t+56}* \sqrt{2t^(2)-16t+56}}`

`-0.5* {2t^(2)-16t+56} = {-t^(2)+8t-28}}`

`{-t^(2)+8t-28}} = {-t^(2)+8t-28}}`

So, for all values of t the angle between these two vectors be 120.