A vector `vecn` is inclined to `x`-axis at `45^(@)`, to y-axis at `60^(@)` and at an angle to z-axis. If `vecn` is a normal to the plane passing through the point `(sqrt(2),-1,1)`, then the equation of plane is

Options

A. `3sqrt(2)x-4y-3z=7`
B. `4sqrt(2)x+7y+z=2`
C. `sqrt(2)x+y+z=2`
D. `sqrt(2)x-y-z=2`

Explanation

To find the equation of the plane given the vector \(\vec{n}\) that is inclined to the x-axis at \(45^\circ\), to the y-axis at \(60^\circ\), and at an angle \(\theta\) to the z-axis, we can follow these steps:

### Step 1: Determine the direction cosines of the vector \(\vec{n}\)

The direction cosines \(l\), \(m\), and \(n\) can be expressed as:
- \(l = \cos(45^\circ) = \frac{1}{\sqrt{2}}\)
- \(m = \cos(60^\circ) = \frac{1}{2}\)
- \(n = \cos(\theta)\)

Using the identity:
\[ l^2 + m^2 + n^2 = 1 \]
we can substitute the known values:
\[ \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + n^2 = 1 \]

### Step 2: Solve for \(n^2\)

Calculating the squares:
\[ \frac{1}{2} + \frac{1}{4} + n^2 = 1 \]
\[ \frac{2}{4} + \frac{1}{4} + n^2 = 1 \]
\[ \frac{3}{4} + n^2 = 1 \]
\[ n^2 = 1 - \frac{3}{4} = \frac{1}{4} \]
Thus,
\[ n = \frac{1}{2} \]

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

The normal vector \(\vec{n}\) can be expressed as:
\[ \vec{n} = l \hat{i} + m \hat{j} + n \hat{k} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{2} \hat{j} + \frac{1}{2} \hat{k} \]

### Step 4: Use the point-normal form of the plane equation

The equation of a plane can be given by:
\[ \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \]
where \(\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}\) and \(\vec{a}\) is the point through which the plane passes, which is \((\sqrt{2}, -1, 1)\).

Calculating \(\vec{a} \cdot \vec{n}\):
\[ \vec{a} \cdot \vec{n} = \left(\sqrt{2} \cdot \frac{1}{\sqrt{2}} + (-1) \cdot \frac{1}{2} + 1 \cdot \frac{1}{2}\right) \]
\[ = 1 - \frac{1}{2} + \frac{1}{2} = 1 \]

### Step 5: Substitute into the plane equation

Now substituting into the plane equation:
\[ \vec{r} \cdot \vec{n} = 1 \]
This gives:
\[ \left(x \cdot \frac{1}{\sqrt{2}} + y \cdot \frac{1}{2} + z \cdot \frac{1}{2}\right) = 1 \]
Multiplying through by \(2\sqrt{2}\) to eliminate the fractions:
\[ 2x + \sqrt{2}y + \sqrt{2}z = 2\sqrt{2} \]

### Step 6: Rearranging the equation

Rearranging gives:
\[ \sqrt{2}x + \frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = \sqrt{2} \]
To match the standard form, we can multiply through by \(\sqrt{2}\):
\[ 2x + y + z = 2 \]

### Final Equation of the Plane

Thus, the equation of the plane is:
\[ \sqrt{2}x + y + z = 2 \]