The magnitude of two vectors p and q differ by 1. The magnitude of their resultant makes an angle of tan inverse ( 3 / 4 ) with p. The angle between p and q is

To find the angle between vectors p and q, we can use the dot product formula:

p·q = |p| |q| cosθ

Given that the magnitude of p and q differ by 1, we can assume |p| = |q| + 1.

Also, the magnitude of the resultant vector (let's call it r) makes an angle of tan^(-1)(3/4) with p. This means the dot product of r and p will be |r| |p| cosα, where α is the angle between r and p.

Now, let's solve the problem step by step:

1. Start by using the dot product formula for r and p.
r·p = |r| |p| cosα

2. We know that |r| cosα = |q|, so the equation becomes:
|q| = |r| |p| cosα

3. Rewrite |q| as |p| - 1 (since |p| = |q| + 1):
|p| - 1 = |r| |p| cosα

4. Rearrange the equation to isolate cosα:
cosα = (|p| - 1) / (|r| |p|)

5. We can use the Pythagorean identity to find sinα:
sinα = √(1 - cos^2α)

6. Finally, the angle between p and q can be found using the formula:
θ = tan^(-1)(sinα / cosα)

Now, by substituting the given value tan^(-1)(3/4) for α and solving the equation, we will get the angle between p and q.

Let's assume the magnitude of vector p is represented as |p| and the magnitude of vector q is |q|. Given that the magnitude of vectors p and q differ by 1, we can express this as:

|p| = |q| + 1 (Equation 1)

We are also given that the magnitude of the resultant (R) of vectors p and q makes an angle of tan inverse (3/4) with p. Let's denote the angle between p and q as θ.

Now, let's analyze the given information and use it to find the angle θ.

Using trigonometry, we have:

tan(θ) = (|R| sin(θ)) / (|R| cos(θ))
tan(θ) = (|q| sin(θ) + 1) / (|q| cos(θ))

Since tan(θ) = 3/4, we have:

3/4 = (|q| sin(θ) + 1) / (|q| cos(θ))

Now, let's solve this equation to find the value of θ.

Cross-multiply:

4(|q| sin(θ) + 1) = 3(|q| cos(θ))

4|q| sin(θ) + 4 = 3|q| cos(θ)

Divide both sides by |q|:

4 sin(θ) + 4/|q| = 3 cos(θ)

Rearrange the equation:

4 sin(θ) - 3 cos(θ) = -4/|q|

Using the Pythagorean identity (sin²(θ) + cos²(θ) = 1), we can square the equation:

(4 sin(θ) - 3 cos(θ))^2 = (-4/|q|)^2
16 sin²(θ) - 24 sin(θ) cos(θ) + 9 cos²(θ) = 16/|q|^2

Now, we can substitute tan²(θ) = 9/16 (from the given tan inverse(3/4)):

16(1 - tan²(θ)) - 24 tan(θ) = 16/|q|^2

16 - 16 tan²(θ) - 24 tan(θ) = 16/|q|^2

16 (1 - tan²(θ) - 24 tan(θ)) = 16/|q|^2

Simplifying further:

1 - tan²(θ) - 24 tan(θ) = 1/|q|^2

tan²(θ) + 24 tan(θ) + 1/|q|^2 - 1 = 0

Let's substitute tan(θ) = 3/4:

(3/4)^2 + 24 (3/4) + 1/|q|^2 - 1 = 0

9/16 + 72/16 + 1/|q|^2 - 1 = 0

81/16 + 1/|q|^2 - 1 = 0

81/16 + 1/|q|^2 = 1

81 + 16/|q|^2 = 16

16/|q|^2 = 16 - 81 = -65

This resulting equation is not possible, as we cannot have a negative value for the magnitude of a vector. Therefore, the angle θ between vectors p and q cannot be determined with the given information.

hmmm...

If θ is the angle between p+q and p, cosθ = 4/5
If Ø is the angle between p+q and q, then the angle between p and q is θ+Ø

Let's let |q| = a, so |p| = a+1

p•q = a(a+1)cos(θ+Ø)
(p+q)•p = |p+q|*|p| cosθ = (4/5)(a+1)|p+q|
(p+q)•q = |p+q|*|q| cosØ = |p+q|(a) cosØ
|p+q| = pcosθ + qcosØ = (4/5)(a+1) + acosØ

Somewhere in those equations you can surely solve for a and Ø, and you want θ+Ø.

Play around with that, while I do the same.

Or, there may be a handy formula I have forgotten.