Let `vec(alpha)=hat(i) +2 hat(j) - hat(k), vec(beta) = 2 hat(i) - hat(j) + 3 hat(k) and vec(lambda)=2hat(i) + hat(j) + 6 hat(k)` be three vectors. If `vec(alpha) and vec(beta)` are both perpendicular to the vector `vec(delta) and vec(delta).vec(lambda)=10`, then what is the magnitude of `vec(delta)`?

Options

A. `sqrt(3) ` units
B. `2sqrt(3)` units
C. `(sqrt(3))/(2)` unit
D. `(1)/(sqrt(3))` unit

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the Given Vectors
We have three vectors:
- \(\vec{\alpha} = \hat{i} + 2\hat{j} - \hat{k}\)
- \(\vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k}\)
- \(\vec{\lambda} = 2\hat{i} + \hat{j} + 6\hat{k}\)

### Step 2: Assume the Form of \(\vec{\delta}\)
Let \(\vec{\delta} = a\hat{i} + b\hat{j} + c\hat{k}\).

### Step 3: Use the Perpendicular Condition
Since \(\vec{\alpha}\) is perpendicular to \(\vec{\delta}\), we have:
\[ \vec{\delta} \cdot \vec{\alpha} = 0 \]
Calculating this dot product:
\[ (a\hat{i} + b\hat{j} + c\hat{k}) \cdot (\hat{i} + 2\hat{j} - \hat{k}) = a + 2b - c = 0 \quad \text{(Equation 1)} \]

### Step 4: Apply the Second Perpendicular Condition
Since \(\vec{\beta}\) is also perpendicular to \(\vec{\delta}\), we have:
\[ \vec{\delta} \cdot \vec{\beta} = 0 \]
Calculating this dot product:
\[ (a\hat{i} + b\hat{j} + c\hat{k}) \cdot (2\hat{i} - \hat{j} + 3\hat{k}) = 2a - b + 3c = 0 \quad \text{(Equation 2)} \]

### Step 5: Use the Given Dot Product Condition
We know that:
\[ \vec{\delta} \cdot \vec{\lambda} = 10 \]
Calculating this dot product:
\[ (a\hat{i} + b\hat{j} + c\hat{k}) \cdot (2\hat{i} + \hat{j} + 6\hat{k}) = 2a + b + 6c = 10 \quad \text{(Equation 3)} \]

### Step 6: Solve the System of Equations
Now we have a system of three equations:
1. \(a + 2b - c = 0\) (Equation 1)
2. \(2a - b + 3c = 0\) (Equation 2)
3. \(2a + b + 6c = 10\) (Equation 3)

#### From Equation 1:
Rearranging gives:
\[ c = a + 2b \quad \text{(Equation 4)} \]

#### Substitute Equation 4 into Equations 2 and 3:
Substituting \(c\) in Equation 2:
\[ 2a - b + 3(a + 2b) = 0 \]
\[ 2a - b + 3a + 6b = 0 \implies 5a + 5b = 0 \implies a + b = 0 \implies b = -a \quad \text{(Equation 5)} \]

Substituting \(b\) from Equation 5 into Equation 4:
\[ c = a + 2(-a) = a - 2a = -a \]

Now substituting \(b\) and \(c\) into Equation 3:
\[ 2a + (-a) + 6(-a) = 10 \]
\[ 2a - a - 6a = 10 \implies -5a = 10 \implies a = -2 \]

Now substituting \(a\) back to find \(b\) and \(c\):
\[ b = -(-2) = 2 \]
\[ c = -(-2) = 2 \]

### Step 7: Write \(\vec{\delta}\)
Thus, we have:
\[ \vec{\delta} = -2\hat{i} + 2\hat{j} + 2\hat{k} \]

### Step 8: Calculate the Magnitude of \(\vec{\delta}\)
The magnitude of \(\vec{\delta}\) is given by:
\[ |\vec{\delta}| = \sqrt{(-2)^2 + 2^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} \]

### Final Answer
The magnitude of \(\vec{\delta}\) is \(2\sqrt{3}\).

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