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Chapter 1: Q1.3P (page 7)

Find the angle between the body diagonals of a cube.

Short Answer

Expert verified

The angle between the digonals is 70.52∘.

Step by step solution

01

Explain the concept and draw the cube using given information

To find the angle between the diagonals, the vector diagonals must be evaluated and then the dot formula must be used.

The cube has 1 unit side and one of the corner is coinciding with the origin. using this information, the cube is drawn as follows:

02

Assume the vectors

The principle diagonals are OM→and BA→. The corrdinates of points O, M, A and B are O=0,0,0,M=1,1,1,B=0,1,0 and A=1,0,1.

Find the vectors OM→and BA→.

OM→=1-0i+1-0j+1-0k=i+j+k

Solve the vector BA→.

BA→=(1-0)i+(0-1)j+(1-0)k=i-j+k

03

Find the dot product between OM→ and AB→ .

The formula of the dot product of the vectors OM→and BA→is

OM→BA→=OMBAcosθ,θis the angle between the vectos OM→and BA→.

Find the dot product of the vectors OM→and BA→.

OM→.BA→=OMBAcosθ2=3cosθcosθ=23θ=70.52∘

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Most popular questions from this chapter

(a) Check product rule (iv) (by calculating each term separately) for the functions

A=xx^+2yy^+3zz^B=3xx^-2xy^

(b) Do the same for product rule (ii).

(c) Do the same for rule (vi).

Compute the divergence of the function v=(rcosθ)r^+(rsinθ)θ^+rsinθcosϕϕ^

Check the divergence theorem for this function, using as your volume the inverted hemispherical bowl of radius R,resting on the xyplane and centered at the origin (Fig. 1.40).

A uniform current density J=J0z^fills a slab straddling the yz plane, fromx=-atox=+a to . A magnetic dipolem=m0x is situated at the origin.

(a) Find the force on the dipole, using Eq. 6.3.

(b) Do the same for a dipole pointing in the direction:m=m0y .

(c) In the electrostatic case, the expressionsF=∇(p.E) and F=(p.∇)Eare equivalent (prove it), but this is not the case for the magnetic analogs (explain why). As an example, calculate(m.∇) for the configurations in (a) and (b).

Question: Evaluate the following integrals:

(a)∫-22(2x+3)δ(3x)dx

(b)∫02(x3+3x+2)δ(1-x)dx

(c)∫-419x2δ(3x+1)dx

(d)∫-∞aδ(x-b)dx

Check the fundamental theorem for gradients, using T=x2+4xy+2yz3the points a=(0,0,0),b=(1,1,1)and the three paths in Fig. 1.28.

(a)=(0.0.0)→(1,0,0)→(1,1,0)→(1,1,1).(b)=(0.0.0)→(0,0,1)→(0,1,1)→(1,1,1).

(c) The parabolic path z=x2,y=x
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