Question

A cube is 4.4 cm on a side, with one corner at the origin. Part 1 (a) What is the unit vector pointing from the origin to the diagonally opposite corner at location < 4.4, 4.4, 4.4 > cm? (Express your answer in vector form.) r^= < , , > By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor. Attempts: 0 of 10 used Save for later Submit Answer Part 2 (b) What is the angle in degrees from this diagonal to one of the adjacent edges of the cube? °

Answer 1

Answer:

(a) $\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

(b) $\theta = 85.44^{\circ}$

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:

$\hat{A} = \frac{\vec{A}}{|A|}$

$\hat{A} = \frac{4.4\hat{i} + 4.4\hat{j} + 4.4\hat{k}}{\sqrt{()4.4}^{2} + (4.4)^{2} + (4.4)^{2}}$

$\hat{A} = \frac{4.4(\hat{i} + \hat{j} + \hat{k})}{4.4\sqrt{3}}$

$\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

(b) To calculate the angle between the two vectors say A and A' is given by:

$\vec{A}\cdot \vec{A'} = \vec{A}\vec{A'}cos\theta$                      

$\theta = cos^{- 1}(\frac{\vec{A}\cdot \vec{A'}}{\vec{A}\vec{A'}})$        (1)

Now,

The coordinates of the diagonally opposite corner, A' is <0, 0, 1> cm

Thus

$\vec{A'} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}$

Now,

Using equation (1) :

$\theta = cos^{- 1}(\frac{(\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}})\cdot \hat{k}}{|A||A'|})$

$|A||A'| = (\sqrt{4.4^{2} +4.4^{2} + 4.4^{2}})(\sqrt{0^{2} + 0^{2} + 0^{2}}) = 7.261$

Thus

$\theta = cos^{- 1}(\frac{\frac{1}{\sqrt{3}}}{7.261})$

$\theta = cos^{- 1}(0.07946) = 85.44^{\circ}$