Question

Prove that vector A + vector b dot to vector minus 3 vector b is equals to 2 a square minus A B cos theta minus 3 b square

Answer 1

(

A

+2

B

).(2

A

−3

B

)

=

A

.2

A

−

A

.3

B

+2

B

.2

A

−2

B

.3

B

=2∣A∣

2

−3

A

.

B

+4

B

A−6∣B∣

2

=2∣A∣

2

+∣A∣∗∣B∣ Cos ϕ−6∣B∣

2

we use the following in the above proof:

\begin{lgathered}\vec{A}.\vec{B}=|A|*|B|* Cos\ \phi,\ where\ \phi=angle\ between\ vectors\ \vec{A}\ and\ \vec{B}\\\vec{A}.\vec{B}=\vec{B}.\vec{A}\\\vec{nA}=n*\vec{A}\end{lgathered}

A

.

B

=∣A∣∗∣B∣∗Cos ϕ, where ϕ=angle between vectors

A

and

B

A

.

B

=

B

.

A

nA

=n∗

A