Question

Please I need it now​

Answer 1

1. A vector $\vec x$ is a unit vector if its magnitude is 1. Given

$\vec A = \dfrac{\sqrt3}2\,\vec\imath - \dfrac12\,\vec\jmath \\\\ \vec B = -\dfrac{\sqrt2}2\,\vec\imath - \dfrac{\sqrt2}2\,\vec\jmath \\\\ \vec C = \dfrac12\,\vec\imath - \dfrac{\sqrt3}2\,\vec\jmath - \dfrac12\,\vec k$

$\vec A$ and $\vec B$ are unit vectors, while $\vec C$ is not, since

$\|\vec A\| = \sqrt{\left(\dfrac{\sqrt3}2\right)^2 + \left(-\dfrac12\right)^2} = 1 \\\\ \|\vec B\| = \sqrt{\left(-\dfrac{\sqrt2}2\right)^2 + \left(-\dfrac{\sqrt2}2\right)^2} = 1 \\\\ \|\vec C\| = \sqrt{\left(\dfrac12\right)^2+\left(-\dfrac{\sqrt3}2\right)^2+\left(-\dfrac12\right)^2} =\dfrac{\sqrt5}2 \neq 1$

2. Given some vector $\vec x$, you can get the unit vector in the same direction as

If

$\vec A = -12\,\vec\imath + 9\,\vec\jmath \\\\ \vec B = 15\,\vec\imath-20\,\vec\jmath$

then the unit vectors in the direction of $\vec A$ and $\vec B$, respectively, are

$\dfrac{\vec A}{\|\vec A\|} = \dfrac{-12\,\vec\imath+9\,\vec\jmath}{\sqrt{(-12)^2+9^2}} = \dfrac{-12\,\vec\imath+9\,\vec\jmath}{15} = -\dfrac45\,\vec\imath + \dfrac35\,\vec\jmath \\\\ \dfrac{\vec B}{\|\vec B\|} = \dfrac{15\,\vec\imath-20\,\vec\jmath}{\sqrt{15^2+(-20)^2}} = \dfrac{15\,\vec\imath-20\,\vec\jmath}{25} = \dfrac35\,\vec\imath - \dfrac45\,\vec\jmath$

3.

$\vec A = \|\vec A\| \left(\cos(180^\circ-\theta)\,\vec\imath + \sin(180^\circ-\theta)\,\vec\jmath\right) \\\\ \vec A = 25 \left(-\sin(37^\circ)\,\vec\imath + \cos(37^\circ)\,\vec\jmath\right) \\\\ \vec A = -15\,\vec\imath + 20\,\vec\jmath$

$\vec B = \|\vec B\| \left(\cos(180^\circ + \alpha)\,\vec\imath + \sin(180^\circ+\alpha)\,\vec\jmath\right) \\\\ \vec B = 75 \left(-\sin(53^\circ)\,\vec\imath-\cos(53^\circ)\,\vec\jmath\right) \\\\ \vec B = -60\,\vec\imath - 45\,\vec\jmath$