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## Sagot :

**Answer:**

** (a) √(160/207) ≈ 0.87917**

** (b) √(68/77) ≈ 0.93974**

**Step-by-step explanation:**

You want the **cosine of angle PAQ** for the following sets of points:

**(a) P(1,2,-1), A(-2, 1,5), Q(2,-1,0)****(b) P(0,2,-3), A(2,-1,5), Q(-2,3,-1)**.

### Dot product

The dot product of vectors AP and AQ is ...

AP•AQ = |AP|·|AQ|·cos(θ)

where θ is the angle between the vectors. Solving for cos(θ), we have ...

[tex]\cos(\theta)=\dfrac{AP\cdot AQ}{|AP|\times|AQ|}[/tex]

### (a) P(1, 2, -1)

The vectors are ...

AP = P -A = (1, 2, -1) -(-2, 1, 5) = (1+2, 2-1, -1-5) = (3, 1, -6)

AQ = Q -A = (2, -1, 0) -(-2, 1, 5) = (2+2, -1-1, 0-5) = (4, -2, -5)

And their magnitudes are ...

|AP| = √(3² +1² +(-6)²) = √46

|AQ| = √(4² +(-2)² +(-5)²) = 3√5

Then the cosine of the angle is found as ...

AP•AQ = 3·4 +1(-2) -6(-5) = 12 -2 +30 = 40

[tex]\cos(\theta)=\dfrac{40}{\sqrt{46}\cdot3\sqrt{5}}\\\\\\\boxed{\cos(\theta)=\sqrt{\dfrac{160}{207}}}[/tex]

### (b) P(0, 2, -3)

The vectors are ...

AP = P -A = (0, 2, -3) -(2, -1, 5) = (0-2, 2+1, -3-5) = (-2, 3, -8)

AQ = Q -A = (-2, 3, -1) -(2, -1, 5) = (-2-2, 3+1, -1-5) = (-4, 4, -6)

And their magnitudes are ...

|AP| = √((-2)² +3² +(-8)²) = √77

|AQ| = √((-4)² +4² +(-6)²) = √68

Then the cosine of the angle is found as ...

AP•AQ = (-2)(-4) +(3)(4) +(-8)(-6) = 8 +12+48 = 68

[tex]\cos(\theta)=\dfrac{68}{\sqrt{77}\cdot \sqrt{68}}\\\\\\\boxed{\cos(\theta)=\sqrt{\dfrac{68}{77}}}[/tex]

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*Additional comment*

You can also use the law of cosines, but that requires the additional computation of vector PQ and its magnitude.