The shortest distance between the lines `2x + y + z - 1 = 0 = 3x + y + 2z - 2` and `x = y = z`, is

Options

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

Answer

To find the shortest distance between the lines given by the equations of the planes \(2x + y + z - 1 = 0\) and \(3x + y + 2z - 2 = 0\), and the line defined by \(x = y = z\), we can follow these steps:

### Step 1: Identify the planes and their normal vectors
The equations of the planes can be rewritten in the standard form:
- Plane 1: \(2x + y + z - 1 = 0\) has a normal vector \(\mathbf{n_1} = (2, 1, 1)\).
- Plane 2: \(3x + y + 2z - 2 = 0\) has a normal vector \(\mathbf{n_2} = (3, 1, 2)\).

### Step 2: Find a direction vector for the line \(x = y = z\)
The line \(x = y = z\) can be represented by the direction vector \(\mathbf{b} = (1, 1, 1)\).

### Step 3: Find the direction vector perpendicular to both planes
To find the direction vector that is perpendicular to both planes, we can take the cross product of their normal vectors:
\[ \mathbf{b} = \mathbf{n_1} \times \mathbf{n_2} \]

Calculating the cross product:
\[ \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 1 \\ 3 & 1 & 2 \end{vmatrix} = \mathbf{i} \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} \]

Calculating the determinants:
\[ = \mathbf{i} (1 \cdot 2 - 1 \cdot 1) - \mathbf{j} (2 \cdot 2 - 1 \cdot 3) + \mathbf{k} (2 \cdot 1 - 1 \cdot 3) \]
\[ = \mathbf{i} (2 - 1) - \mathbf{j} (4 - 3) + \mathbf{k} (2 - 3) \]
\[ = \mathbf{i} (1) - \mathbf{j} (1) - \mathbf{k} (1) = (1, -1, -1) \]

### Step 4: Find the distance between the two lines
To find the shortest distance \(d\) between the two lines, we can use the formula:
\[ d = \frac{|(\mathbf{a_2} - \mathbf{a_1}) \cdot \mathbf{b}|}{|\mathbf{b}|} \]
where \(\mathbf{a_1}\) and \(\mathbf{a_2}\) are points on the two lines.

Let’s choose \(\mathbf{a_1} = (0, 0, 0)\) (a point on the line \(x = y = z\)) and \(\mathbf{a_2} = (0, 0, 1)\) (a point on the plane \(2x + y + z - 1 = 0\)).

Now, we calculate \(\mathbf{a_2} - \mathbf{a_1} = (0, 0, 1) - (0, 0, 0) = (0, 0, 1)\).

### Step 5: Calculate the dot product and magnitude
Now we compute the dot product:
\[ (\mathbf{a_2} - \mathbf{a_1}) \cdot \mathbf{b} = (0, 0, 1) \cdot (1, -1, -1) = 0 \cdot 1 + 0 \cdot (-1) + 1 \cdot (-1) = -1 \]

Next, we calculate the magnitude of \(\mathbf{b}\):
\[ |\mathbf{b}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 6: Calculate the distance
Now we can find the distance:
\[ d = \frac{| -1 |}{\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Final Answer
Thus, the shortest distance between the lines is \(\frac{1}{\sqrt{3}}\).

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