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(a) In unit-vector notation, what is r→=a→+b→+c→if a→=5.0i^+4.0j^-6.0k^, ,and b→=2.0i^+2.0j^-3.0k^ ? (b) Calculate the angle between r→and the positive z axis. (c) What is the component ofa→ along the direction of b→? (d) What is the component of b→perpendicular to the direction of b→but in the plane of a→and 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 r→and positive z axis is 120°.

  3. The component of a→along the direction of b→is -4.9.

  4. Component of a→perpendicular to the direction of b→but in the plane of a→and b→is 7.3

Step by step solution

01

Given data

The vectors a→,b→and c→are,

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=ab³¦´Ç²õθ … (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 r→is 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 r→and 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 r→and k^is given by,

|r→|=112+52+72=14

|k^|=1

The angle between r→and 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 a→along the direction of b→is given by acosθwhere θis the angle between a→andb→.

The magnitude of vector a→and b→is,

|a→|=52+42+62=8.77

|b→|=22+22+32=4.12

The dot product of vector a→and b→is,

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

The angle between vector a→and b→is,

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

Now, the component of a→along the direction of b→is given by,

a³¦´Ç²õθ=(8.77)cos(123.6)=-4.9

Thus, the component of a→along the direction of b→is -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 a→along the direction perpendicular to b→and in the plane of a→and b→is given by asinθ, where θis the angle between a→and b→.

Therefeore,

localid="1654749201788" a²õ¾±²Ôθ=(8.77)sin(123.6)=7.3

Thus, the component of a→along the direction perpendicular to b→and in the plane of a→and b→is 7.3.

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Most popular questions from this chapter

Vector a→lies in the yz plane 63.0°from the positive direction of the y axis, has a positive z component, and has magnitude 3.20units. Vector b→lies in xz the plane 48.0from the positive direction of the x axis, has a positive z component, and has magnitude1.40°units. Find (a)role="math" localid="1661144136421" a→·b→, (b)a→×b→, and (c) the angle betweena→andb→.

A wheel with a radius of 45.0 cmrolls without slipping along a horizontal floor (Fig). At time t, the dot P painted on the rim of the wheel is at the point of contact between the wheel and the floor. At a later time t2, the wheel has rolled through one-half of a revolution. What are (a) the magnitude and (b) the angle (relative to the floor) of the displacement of P?

Rock faults are ruptures along which opposite faces of rock have slid past each other. In Fig. 3-35, points A and B coincided before the rock in the foreground slid down to the right. The net displacement AB→is along the plane of the fault.The horizontal component of is the strike-slip AC. The component AB→of that is directed down the plane of the fault is the dip-slip AD.(a) What is the magnitude of the net displacementAB→if the strike-slip is 22.0 mand the dip-slip is 17.0 m? (b) If the plane of the fault is inclined at angleϕ=52.0°to the horizontal, what is the vertical component ofAB→?

Consider a→in the positive direction of x, b→in the positive direction of y, and a scalar d. What is the direction of b→/dif d is

(a) positive and

(b) negative? What is the magnitude of

(c)a→⋅b→and (d)a→⋅b→/d?

What is the direction of the vector resulting from (e)a→×b→and (f)b→×a→?

(g) What is the magnitude of the vector product in (e)?

(h) What is the magnitude of the vector product in (f)? What are

(i) the magnitude and

(j) the direction of a→×b→/dif d is positive?

Threevectorsaregivenbya⃗=3.0i^+3.0j^-2.0k^a→=3.0i^+3.0j^-2.0k^a→=3.0i^+3.0j^-2.0k^,b→=-1.0i^-4.0j^+2.0k^,c→=2.0i^+2.0j^+1.0k.^Find(a)a→.(a→×c→),(b)a→.(b→+c→),(c)a→×(b→+c→).

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