Respuesta :
Answer:
(a) [tex]\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}[/tex]
(b) [tex]\theta = 85.44^{\circ}[/tex]
Solution:
As per the question:
Side of the cube, a = 4.4 cm
Coordinates of the diagonally opposite corner, A = <4.4, 4.4, 4.4> cm
Now,
(a) To calculate the unit vector:
[tex]\hat{A} = \frac{\vec{A}}{|A|}[/tex]
[tex]\hat{A} = \frac{4.4\hat{i} + 4.4\hat{j} + 4.4\hat{k}}{\sqrt{()4.4}^{2} + (4.4)^{2} + (4.4)^{2}}[/tex]
[tex]\hat{A} = \frac{4.4(\hat{i} + \hat{j} + \hat{k})}{4.4\sqrt{3}}[/tex]
[tex]\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}[/tex]
(b) To calculate the angle between the two vectors say A and A' is given by:
[tex]\vec{A}\cdot \vec{A'} = \vec{A}\vec{A'}cos\theta[/tex]
[tex]\theta = cos^{- 1}(\frac{\vec{A}\cdot \vec{A'}}{\vec{A}\vec{A'}})[/tex] (1)
Now,
The coordinates of the diagonally opposite corner, A' is <0, 0, 1> cm
Thus
[tex]\vec{A'} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}[/tex]
Now,
Using equation (1) :
[tex]\theta = cos^{- 1}(\frac{(\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}})\cdot \hat{k}}{|A||A'|})[/tex]
[tex]|A||A'| = (\sqrt{4.4^{2} +4.4^{2} + 4.4^{2}})(\sqrt{0^{2} + 0^{2} + 0^{2}}) = 7.261[/tex]
Thus
[tex]\theta = cos^{- 1}(\frac{\frac{1}{\sqrt{3}}}{7.261})[/tex]
[tex]\theta = cos^{- 1}(0.07946) = 85.44^{\circ}[/tex]
