In vectors given that Pbar+Qbar+Rbaris equal to 0.|Pbar|=|Qbar|&|Rbar|=√2|Pbar|.which of the following can be the angles berween Pbar,Qbar;Qbar,Rbar;and Rbar,Pbar?

To solve the problem involving vectors P, Q, and R, let's break down the information given and analyze the relationships between these vectors step by step.

Understanding the Vector Relationships

We know that the sum of the vectors is zero: P + Q + R = 0. This implies that vector R can be expressed in terms of vectors P and Q as follows:

R = -P - Q

Magnitude Relationships

Next, we have the conditions on the magnitudes of the vectors:

• |P| = |Q|
• |R| = √2 |P|

Since |P| = |Q|, we can denote the common magnitude as |P| = |Q| = a. Therefore, the magnitude of R becomes:

|R| = √2 * a

Finding the Angles Between the Vectors

To find the angles between these vectors, we can use the Law of Cosines, which relates the lengths of the sides of a triangle to the cosine of one of its angles. Considering the triangle formed by the three vectors, we can denote the angles as follows:

• θ₁ between P and Q
• θ₂ between Q and R
• θ₃ between R and P

Applying the Law of Cosines

Using the Law of Cosines for the triangle formed by vectors P, Q, and R, we can express the following:

• |R|² = |P|² + |Q|² - 2|P||Q|cos(θ₁)
• |P|² = |Q|² + |R|² - 2|Q||R|cos(θ₂)
• |Q|² = |R|² + |P|² - 2|R||P|cos(θ₃)

Calculating Each Angle

Substituting the magnitudes we defined earlier:

• |R|² = (√2 * a)² = 2a²
• |P|² = a²
• |Q|² = a²

Now substituting these into the first equation:

2a² = a² + a² - 2a²cos(θ₁)

Solving this gives:

2a² = 2a² - 2a²cos(θ₁) ⟹ 2a²cos(θ₁) = 0 ⟹ cos(θ₁) = 0

This means:

θ₁ = 90°

Using a similar approach for the second angle:

a² = a² + 2a² - 2a(√2a)cos(θ₂)

Solving this gives:

0 = 2a² - 2a²cos(θ₂) ⟹ cos(θ₂) = 1/√2

Thus, θ₂ is:

θ₂ = 45°

Lastly, for θ₃:

a² = 2a² + a² - 2(√2a)(a)cos(θ₃)

Solving similarly leads us to:

θ₃ = 135°

Conclusion on Angles

Thus, the angles between the vectors can be summarized as:

• Angle between P and Q (θ₁) = 90°
• Angle between Q and R (θ₂) = 45°
• Angle between R and P (θ₃) = 135°

This configuration satisfies the conditions imposed on the magnitudes and ensures that the vectors remain in equilibrium, summing to zero. Each angle can be visualized through the geometric relationships in a triangular formation, demonstrating the unique properties of vectors in a plane.