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

(a)
`\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{√(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}}`

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

`\hat{A} = \frac{4.4(\hat{i} + \hat{j} + \hat{k})}{4.4√(3)}`

`\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{√(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}}{√(3)})\cdot \hat{k}}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)(((1)/(√(3)))/(7.261))`

`\theta = cos^(- 1)(0.07946) = 85.44^(\circ)`