Determining Orthogonal Vectors

[nycmathguy]

Section 6.4

Can you do 56 in detail? Thanks.

 

[MathLover1]

56.

two vectors are perpendicular to each other if the dot product of the two vectors is zero

u= <cos(θ),sin(θ)>
v= <sin(θ),−cos(θ)>

Calculate

u*v
= <cos(θ),sin(θ)> *<sin(θ),−cos(θ)>
=<cos(θ)*sin(θ) +sin(θ)(−cos(θ))>
=<cos(θ)*sin(θ) -sin(θ)(cos(θ))>
=0

=> vectors u and v are orthogonal

 

[nycmathguy]

56.

two vectors are perpendicular to each other if the dot product of the two vectors is zero

u= <cos(θ),sin(θ)>
v= <sin(θ),−cos(θ)>

Calculate

u*v
= <cos(θ),sin(θ)> *<sin(θ),−cos(θ)>
=<cos(θ)*sin(θ) +sin(θ)(−cos(θ))>
=<cos(θ)*sin(θ) -sin(θ)(cos(θ))>
=0

=> vectors u and v are orthogonal

Thanks. I will do the rest this weekend.

 

[nycmathguy]

56.

two vectors are perpendicular to each other if the dot product of the two vectors is zero

u= <cos(θ),sin(θ)>
v= <sin(θ),−cos(θ)>

Calculate

u*v
= <cos(θ),sin(θ)> *<sin(θ),−cos(θ)>
=<cos(θ)*sin(θ) +sin(θ)(−cos(θ))>
=<cos(θ)*sin(θ) -sin(θ)(cos(θ))>
=0

=> vectors u and v are orthogonal

Please visit the link below. How is 44 done using vectors not the law of cosines as suggested by Country Boy?

Find Angles in a Triangle

 

[nycmathguy]

56.

two vectors are perpendicular to each other if the dot product of the two vectors is zero

u= <cos(θ),sin(θ)>
v= <sin(θ),−cos(θ)>

Calculate

u*v
= <cos(θ),sin(θ)> *<sin(θ),−cos(θ)>
=<cos(θ)*sin(θ) +sin(θ)(−cos(θ))>
=<cos(θ)*sin(θ) -sin(θ)(cos(θ))>
=0

=> vectors u and v are orthogonal