Scalar product question

[Mahmoud Ibrahim]

Vector_Scalar_Product.png (9.29 KiB) Viewed 2284 times

Please , I think that there is a mistake in the question. If we delete the word ( perpendicular ) from the question, then the question will be correct . because there is no unite vectors perpendiculars. Because unite vectors are ( 1i, 1j, 1k), then it is impossible to be 2 unite vectors perpendiculars. After deleting the word ( perpendiculars) , The solution will be in the attached file ( The_Sol.png).
The book said that the answer is (-7).
Please correct my answer and the solution.
Thank you.

[Mahmoud Ibrahim]

Sorry, i studed the vectors again . The unite vector is the vector whose LENGTH ( MAGNITUDE) = unit. Not (1i, 1j, 1k) .
Therefore : there are infinite number of unite vectors and infinite number of them are perpendiculars.
So the book is correct ( the question and the answer)
The attachment is a file of my solution . Now my answer is same as the book answer.
I hope my solution is correct.
Sorry and thanks.

[Baltuilhe]

Hi!

Data:
[tex]\begin{cases}\|\vec{A}\|=1\\\|\vec{B}\|=1\end{cases}[/tex]

Remember that:
[tex]\vec{A}\perp\vec{B}\Rightarrow\vec{A}\cdot\vec{B}=0[/tex]

So:
[tex]\left(\vec{A}-2\vec{B}\right)\cdot\left(3\vec{A}+5\vec{B}\right)\\\\3\vec{A}\cdot\vec{A}+5\vec{A}\cdot\vec{B}-6\vec{A}\cdot\vec{B}-10\vec{B}\cdot\vec{B}\\\\3\overbrace{\|\vec{A}\|^2}^{1}-\overbrace{\vec{A}\cdot\vec{B}}^{0}-10\overbrace{\|\vec{B}\|^2}^{1}\\\\3-0-10=-7[/tex]

I hope I have helped!