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.