A vector `vec v` in the first octant is inclined to the `x` -axis at `60^(@)`,to the `y` -axis at `45^(@)` and to the `z` -axis at an acute angle.If a plane passing through the points `(sqrt(2),-1,1)` and `(a,b,c)` is normal to `vec v`,then

Options

A. `sqrt(2)a-b+c=1`
B. `a+sqrt(2)a+c=1`
C. `sqrt(2)a+b+c=1`
D. `a+b+sqrt(2)a=1`

Answer

To solve the problem, we need to find the vector \(\vec{v}\) that is inclined to the \(x\)-axis at \(60^\circ\), to the \(y\)-axis at \(45^\circ\), and to the \(z\)-axis at an acute angle. We will then use this vector to determine the condition for the plane passing through the points \((\sqrt{2}, -1, 1)\) and \((a, b, c)\) being normal to \(\vec{v}\).

### Step 1: Determine the direction ratios of the vector \(\vec{v}\)

The direction ratios \(L\), \(M\), and \(N\) of the vector \(\vec{v}\) can be determined using the angles given:

1. **For the \(x\)-axis**:
\[ L = \cos(60^\circ) = \frac{1}{2} \]

2. **For the \(y\)-axis**:
\[ M = \cos(45^\circ) = \frac{1}{\sqrt{2}} \]

3. **For the \(z\)-axis**:
Let \(N = \cos(\theta)\) where \(\theta\) is the acute angle.

### Step 2: Use the normalization condition

Since \(\vec{v}\) is a unit vector, we have:
\[ L^2 + M^2 + N^2 = 1 \]

Substituting the values of \(L\) and \(M\):
\[ \left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + N^2 = 1 \]
\[ \frac{1}{4} + \frac{1}{2} + N^2 = 1 \]
\[ \frac{1}{4} + \frac{2}{4} + N^2 = 1 \]
\[ \frac{3}{4} + N^2 = 1 \]
\[ N^2 = 1 - \frac{3}{4} = \frac{1}{4} \]
\[ N = \frac{1}{2} \quad (\text{since } N \text{ is acute}) \]

### Step 3: Write the vector \(\vec{v}\)

The vector \(\vec{v}\) can now be expressed as:
\[ \vec{v} = \left(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}\right) \]

### Step 4: Find the normal vector to the plane

The normal vector to the plane can be represented as:
\[ \vec{n} = (1, \sqrt{2}, 1) \]

### Step 5: Set up the equation for the plane

The equation of the plane can be expressed using the normal vector and the point \((\sqrt{2}, -1, 1)\):
\[ \vec{n} \cdot \left((a, b, c) - (\sqrt{2}, -1, 1)\right) = 0 \]

Expanding this gives:
\[ 1(a - \sqrt{2}) + \sqrt{2}(b + 1) + 1(c - 1) = 0 \]
\[ a - \sqrt{2} + \sqrt{2}b + \sqrt{2} + c - 1 = 0 \]
\[ a + \sqrt{2}b + c - \sqrt{2} - 1 = 0 \]
\[ a + \sqrt{2}b + c = \sqrt{2} + 1 \]

### Conclusion

Thus, the condition for the plane is:
\[ a + \sqrt{2}b + c = \sqrt{2} + 1 \]