Given that `vecA+vecB+vecC=0`, out of three vectors two are equal in magnitude and the magnitude of third vector is `sqrt2` times that of either of two having equal magnitude. Then angle between vectors are given by

Options

A. `30^(@),60^(@),90^(@)`
B. `45^(@),45^(@),90^(@)`
C. `90^(@),135^(@),45^(@)`
D. `90^(@),135^(@),135^(@)`

Explanation

To solve the problem, we need to analyze the given vectors and their relationships. Here’s a step-by-step solution:

### Step 1: Understand the Given Information
We have three vectors \( \vec{A} \), \( \vec{B} \), and \( \vec{C} \) such that:
\[ \vec{A} + \vec{B} + \vec{C} = 0 \]
This implies that the vectors form a closed triangle when represented graphically.

### Step 2: Define the Magnitudes
Let the magnitudes of vectors \( \vec{A} \) and \( \vec{B} \) be \( x \). According to the problem, the magnitude of vector \( \vec{C} \) is:
\[ |\vec{C}| = \sqrt{2} \cdot x \]

### Step 3: Visualize the Vectors
Since the vectors form a triangle, we can visualize this as follows:
- Vectors \( \vec{A} \) and \( \vec{B} \) are equal in magnitude and can be represented as two sides of a triangle.
- Vector \( \vec{C} \) is the third side, which is longer than the other two.

### Step 4: Apply the Law of Cosines
To find the angles between the vectors, we can use the Law of Cosines. For triangle formed by these vectors, we can express the relationship as:
\[ |\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos(\theta) \]
Where \( \theta \) is the angle between vectors \( \vec{A} \) and \( \vec{B} \).

### Step 5: Substitute the Magnitudes
Substituting the magnitudes into the equation:
\[ (\sqrt{2} x)^2 = x^2 + x^2 - 2x^2 \cos(\theta) \]
This simplifies to:
\[ 2x^2 = 2x^2(1 - \cos(\theta)) \]

### Step 6: Simplify the Equation
Dividing both sides by \( 2x^2 \) (assuming \( x \neq 0 \)):
\[ 1 = 1 - \cos(\theta) \]
This implies:
\[ \cos(\theta) = 0 \]

### Step 7: Find the Angle
The angle \( \theta \) for which \( \cos(\theta) = 0 \) is:
\[ \theta = 90^\circ \]

### Step 8: Determine Other Angles
Since \( \vec{A} \) and \( \vec{B} \) are equal in magnitude and the triangle is closed, the angles opposite to equal sides are also equal. Thus, we can denote the angles opposite \( \vec{A} \) and \( \vec{B} \) as \( \alpha \) and \( \beta \) respectively.

Using the triangle angle sum property:
\[ \alpha + \beta + 90^\circ = 180^\circ \]
This leads to:
\[ \alpha + \beta = 90^\circ \]

Since \( \alpha = \beta \), we can set:
\[ 2\alpha = 90^\circ \]
Thus:
\[ \alpha = 45^\circ \]
And:
\[ \beta = 45^\circ \]

### Final Result
The angles between the vectors are:
- \( \angle A = 90^\circ \)
- \( \angle B = 45^\circ \)
- \( \angle C = 45^\circ \)

### Conclusion
The angles between the vectors are \( 90^\circ \), \( 45^\circ \), and \( 45^\circ \). The correct option is \( 4 \).

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