   Chapter 12.3, Problem 26E

Chapter
Section
Textbook Problem

# Find the values of x such that the angle between the vectors ⟨2, 1,−1⟩, and ⟨1, x, 0⟩ is 45°.

To determine

To find: The values of x .

Explanation

Given:

a=2,1,1 , b=1,x,0 and θ=45° .

Formula:

Write the expression to find ab in terms of θ .

ab=|a||b|cosθ (1)

Here,

|a| is the magnitude of a vector,

|b| is the magnitude of b vector, and

θ is the angle between vectors a and b.

Consider a general expression to find dot product between two three-dimensional vectors.

ab=a1,a2,a3b1,b2,b3

ab=a1b1+a2b2+a3b3 (2)

Consider a general expression to find magnitude of a three dimensional vector that is a=a1,a2,a3 .

|a|=a12+a22+a32 (3)

Similarly, Consider a general expression to find magnitude of a three dimensional vector that is b=b1,b2,b3 .

|b|=b12+b22+b32 (4)

In equation (2), substitute 2 for a1 , 1 for a2 , –1 for a3 , 1 for b1 , x for b2 and 0 for b3 .

ab=(2)(1)+(1)(x)+(1)(0)=2+x+0=2+x

In equation (3), substitute 2 for a1 , 1 for a2 and –1 for a3 .

|a|=(2)2+(1)2+(1)2=4+1+1=6

In equation (4), substitute 1 for b1 , x for b2 and 0 for b3 .

|b|=(1)2+(x)2+(0)2=1+x2

In equation (1), substitute 2+x for ab , 6 for |a| , 1+x2 for |b| and 45° for θ

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