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(a) In unit-vector notation, what is r=a+b+cif a=5.0i^+4.0j^-6.0k^, ,and b=2.0i^+2.0j^-3.0k^ ? (b) Calculate the angle between rand the positive z axis. (c) What is the component ofa along the direction of b? (d) What is the component of bperpendicular to the direction of bbut in the plane of aand b ? (Hint: For (c), see Eq.3-20 and Fig. 3-18)

Short Answer

Expert verified
  1. The value of vector r=11i^+5.0j^-7.0k^

  2. Angle between rand positive z axis is 120°.

  3. The component of aalong the direction of bis -4.9.

  4. Component of aperpendicular to the direction of bbut in the plane of aand bis 7.3

Step by step solution

01

Given data

The vectors a,band care,

a=5.0i^+4.0j^+6.0k^

role="math" localid="1654745877724" b=2.0i^+2.0j^+3.0k^

c=4.0i^+3.0j^+2.0k^

02

Understanding the scalar product of vectors

This problem is based on dot product which is nothing but the scalar product. The scalar product of two vectors gives a scalar quantity.

The expression for the scalar product is as follows:

a.b=axbx+ayby+azbz=abcosθ … (i)

The expression for the magnitude of a vector is as follows:

A=Ax2+Ay2+Az2 … (ii)

03

(a) Determination of r→

The value of vector ris calculated as:

role="math" localid="1654746771591" r=a+b+c=(5.0i^+4.0j^+6.0k^)-(2.0i^+2.0j^+3.0k^)+(4.0i^+3.0j^+2.0k^)=11i^+5.0j^-7.0k^

Thus, the value of vector role="math" localid="1654746992113" r=11i^+5.0j^-7.0k^.

04

(b) Determination of the angle between r→ and positive z axis

Equation (i) can be used to calculate the angle between rand positive z axis.


r=11i^+5.0j^+7.0k^

The unit vector along positive z-axis is,

k^=1.0k^

Therefore, the dot product,

r.k^=(r=11i^+5.0j^-7.0k^).(1.0k^)=-7.0

The magnitude of rand k^is given by,

|r|=112+52+72=14

|k^|=1

The angle between rand positive z axis is calculated as follows:

r.k^cosθ(-7.0)=(14)(1)cosθcosθ=-714θ=120θ

Thus, the angle between r and the positive z axis is 120°.

05

(c) Determination of the component of a→ along the direction of b→ 

Component of aalong the direction of bis given by acosθwhere θis the angle between aandb.

The magnitude of vector aand bis,

|a|=52+42+62=8.77

|b|=22+22+32=4.12

The dot product of vector aand bis,

a.b=(5.0i^+4.0j^-6.0k^).(2.0i^+2.0j^+3.0k^)=-20

The angle between vector aand bis,

(-20)=(8.77)(4.12)cosθcosθ=-2036.13θ=123.6°

Now, the component of aalong the direction of bis given by,

acosθ=(8.77)cos(123.6)=-4.9

Thus, the component of aalong the direction of bis -4.9.

06

(d) Determination of the component of a→ perpendicular to the direction of b→ but in the plane of a→ and b→

Component of aalong the direction perpendicular to band in the plane of aand bis given by asinθ, where θis the angle between aand b.

Therefeore,

localid="1654749201788" asinθ=(8.77)sin(123.6)=7.3

Thus, the component of aalong the direction perpendicular to band in the plane of aand bis 7.3.

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