If `veca and vecb` are two unit vectors perpenicualar to each other and `vecc= lambda_(1)veca+ lambda_2 vecb+ lambda_(3)(vecaxx vecb),` then which of the following is (are) true ?

Options

A. `lambda_(1)= veca.vecc`
B. `lambda_(2)= |vecb xx vecc|`
C. `lambda_(3)= |vecaxx vecb|xxvecc|`
D. `lambda_(1)veca + lambda_(2) vecb + lambda_(3) (vecaxxvecb)`

Explanation

To solve the problem, we need to analyze the given vectors and their relationships. Let's break it down step by step.

### Step 1: Understand the Given Information
We have two unit vectors, \(\vec{a}\) and \(\vec{b}\), which are perpendicular to each other. This means:
- \(|\vec{a}| = 1\)
- \(|\vec{b}| = 1\)
- \(\vec{a} \cdot \vec{b} = 0\)

We also have a vector \(\vec{c}\) defined as:
\[ \vec{c} = \lambda_1 \vec{a} + \lambda_2 \vec{b} + \lambda_3 (\vec{a} \times \vec{b}) \]

### Step 2: Find \(\lambda_1\)
To find \(\lambda_1\), we take the dot product of \(\vec{c}\) with \(\vec{a}\):
\[ \lambda_1 = \vec{a} \cdot \vec{c} \]
Substituting \(\vec{c}\):
\[ \lambda_1 = \vec{a} \cdot (\lambda_1 \vec{a} + \lambda_2 \vec{b} + \lambda_3 (\vec{a} \times \vec{b})) \]
Using the properties of dot products:
\[ \lambda_1 = \lambda_1 (\vec{a} \cdot \vec{a}) + \lambda_2 (\vec{a} \cdot \vec{b}) + \lambda_3 (\vec{a} \cdot (\vec{a} \times \vec{b})) \]
Since \(\vec{a} \cdot \vec{b} = 0\) and \(\vec{a} \cdot (\vec{a} \times \vec{b}) = 0\):
\[ \lambda_1 = \lambda_1 \cdot 1 + 0 + 0 \Rightarrow \lambda_1 = \lambda_1 \]
This equation is always true.

### Step 3: Find \(\lambda_2\)
Next, we find \(\lambda_2\) by taking the dot product of \(\vec{c}\) with \(\vec{b}\):
\[ \lambda_2 = \vec{b} \cdot \vec{c} \]
Substituting \(\vec{c}\):
\[ \lambda_2 = \vec{b} \cdot (\lambda_1 \vec{a} + \lambda_2 \vec{b} + \lambda_3 (\vec{a} \times \vec{b})) \]
Using the properties of dot products:
\[ \lambda_2 = \lambda_1 (\vec{b} \cdot \vec{a}) + \lambda_2 (\vec{b} \cdot \vec{b}) + \lambda_3 (\vec{b} \cdot (\vec{a} \times \vec{b})) \]
Since \(\vec{b} \cdot \vec{a} = 0\) and \(\vec{b} \cdot (\vec{a} \times \vec{b}) = 0\):
\[ \lambda_2 = 0 + \lambda_2 \cdot 1 + 0 \Rightarrow \lambda_2 = \lambda_2 \]
This equation is also always true.

### Step 4: Find \(\lambda_3\)
Now, we find \(\lambda_3\) by taking the dot product of \(\vec{c}\) with \((\vec{a} \times \vec{b})\):
\[ \lambda_3 = (\vec{a} \times \vec{b}) \cdot \vec{c} \]
Substituting \(\vec{c}\):
\[ \lambda_3 = (\vec{a} \times \vec{b}) \cdot (\lambda_1 \vec{a} + \lambda_2 \vec{b} + \lambda_3 (\vec{a} \times \vec{b})) \]
Using the properties of dot products:
\[ \lambda_3 = \lambda_1 ((\vec{a} \times \vec{b}) \cdot \vec{a}) + \lambda_2 ((\vec{a} \times \vec{b}) \cdot \vec{b}) + \lambda_3 ((\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b})) \]
Since \((\vec{a} \times \vec{b}) \cdot \vec{a} = 0\) and \((\vec{a} \times \vec{b}) \cdot \vec{b} = 0\):
\[ \lambda_3 = 0 + 0 + \lambda_3 |\vec{a} \times \vec{b}|^2 \]
The magnitude of \(\vec{a} \times \vec{b}\) is:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(90^\circ) = 1 \cdot 1 \cdot 1 = 1 \]
Thus:
\[ \lambda_3 = \lambda_3 \cdot 1 \Rightarrow \lambda_3 = \lambda_3 \]
This equation is always true.

### Step 5: Conclusion
From our analysis, we can conclude:
- \(\lambda_1 = \vec{a} \cdot \vec{c}\)
- \(\lambda_2 = \vec{b} \cdot \vec{c}\)
- \(\lambda_3 = (\vec{a} \times \vec{b}) \cdot \vec{c}\)

Thus, the correct options are:
- Option A: \(\lambda_1 = \vec{a} \cdot \vec{c}\)
- Option D: \(\lambda_1 \vec{a} + \lambda_2 \vec{b} + \lambda_3 (\vec{a} \times \vec{b}) = \vec{c}\)