If `veca and vecb ` are two unit vectors perpendicular to each other and `vecc=lamda_1veca+lamda_2vecb+lamda_3(vecaxxvecb)` then the following is (are) true

Options

A. (a)`lambda_1=a*c`
B. (b)`lambda_2=|atimesb|`
C. (c)`lambda_3=|(atimesb)timesc|`
D. (d)`lambda_1+lambda_2+lambda_3=(a+b+atimesb)*c`

Answer

To solve the problem, we will analyze the given information step by step.

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

### Step 2: Define Vector \(\vec{c}\)
The vector \(\vec{c}\) is defined as:
\[ \vec{c} = \lambda_1 \vec{a} + \lambda_2 \vec{b} + \lambda_3 (\vec{a} \times \vec{b}) \]

### Step 3: Calculate \(\vec{a} \cdot \vec{c}\)
To find \(\lambda_1\), we compute:
\[ \vec{a} \cdot \vec{c} = \vec{a} \cdot (\lambda_1 \vec{a} + \lambda_2 \vec{b} + \lambda_3 (\vec{a} \times \vec{b})) \]
Using the properties of the dot product:
\[ \vec{a} \cdot \vec{c} = \lambda_1 (\vec{a} \cdot \vec{a}) + \lambda_2 (\vec{a} \cdot \vec{b}) + \lambda_3 (\vec{a} \cdot (\vec{a} \times \vec{b})) \]
Since \(\vec{a} \cdot \vec{a} = 1\), \(\vec{a} \cdot \vec{b} = 0\), and \(\vec{a} \cdot (\vec{a} \times \vec{b}) = 0\):
\[ \vec{a} \cdot \vec{c} = \lambda_1 \cdot 1 + \lambda_2 \cdot 0 + \lambda_3 \cdot 0 = \lambda_1 \]
Thus, we have:
\[ \lambda_1 = \vec{a} \cdot \vec{c} \]

### Step 4: Calculate \(\vec{b} \cdot \vec{c}\)
Next, we calculate \(\lambda_2\):
\[ \vec{b} \cdot \vec{c} = \vec{b} \cdot (\lambda_1 \vec{a} + \lambda_2 \vec{b} + \lambda_3 (\vec{a} \times \vec{b})) \]
Using the properties of the dot product:
\[ \vec{b} \cdot \vec{c} = \lambda_1 (\vec{b} \cdot \vec{a}) + \lambda_2 (\vec{b} \cdot \vec{b}) + \lambda_3 (\vec{b} \cdot (\vec{a} \times \vec{b})) \]
Since \(\vec{b} \cdot \vec{a} = 0\), \(\vec{b} \cdot \vec{b} = 1\), and \(\vec{b} \cdot (\vec{a} \times \vec{b}) = 0\):
\[ \vec{b} \cdot \vec{c} = \lambda_1 \cdot 0 + \lambda_2 \cdot 1 + \lambda_3 \cdot 0 = \lambda_2 \]
Thus, we have:
\[ \lambda_2 = \vec{b} \cdot \vec{c} \]

### Step 5: Calculate \(\vec{a} \times \vec{b}\)
Now, we calculate \(\lambda_3\):
\[ \lambda_3 = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|} = |\vec{a} \times \vec{b}| \]
Since \(\vec{a}\) and \(\vec{b}\) are unit vectors and perpendicular, we have:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(90^\circ) = 1 \cdot 1 \cdot 1 = 1 \]
Thus:
\[ \lambda_3 = 1 \]

### Step 6: Summarize the Results
We have:
- \(\lambda_1 = \vec{a} \cdot \vec{c}\)
- \(\lambda_2 = \vec{b} \cdot \vec{c}\)
- \(\lambda_3 = 1\)

### Conclusion
Now we can check the options provided in the question to see which statements are true based on our calculations.