The shortest distance between the z-axis and the line $$x + y + 2z - 3 = 0 = 2x + 3y + 4z - 4$$, is

The problem requires finding the shortest distance between the z-axis and the line defined by the equations $$ x + y + 2z - 3 = 0 $$ and $$ 2x + 3y + 4z - 4 = 0 $$. The z-axis is the line where $$ x = 0 $$ and $$ y = 0 $$, so any point on it can be written as $$ (0, 0, t) $$ for some real number $$ t $$. The given line is the intersection of two planes: $$ x + y + 2z = 3 $$ and $$ 2x + 3y + 4z = 4 $$.

To find the shortest distance between two lines in space, we need their direction vectors and a point on each line. First, find the direction vector of the given line. Since the line is the intersection of two planes, its direction vector is perpendicular to the normal vectors of both planes. The normal vector of the first plane $$ x + y + 2z = 3 $$ is $$ \vec{n_1} = (1, 1, 2) $$, and the normal vector of the second plane $$ 2x + 3y + 4z = 4 $$ is $$ \vec{n_2} = (2, 3, 4) $$. The direction vector $$ \vec{d} $$ of the line is the cross product of $$ \vec{n_1} $$ and $$ \vec{n_2} $$:

$$ \vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 2 & 3 & 4 \end{vmatrix} $$

Expanding the determinant:

$$ \hat{i} (1 \cdot 4 - 2 \cdot 3) - \hat{j} (1 \cdot 4 - 2 \cdot 2) + \hat{k} (1 \cdot 3 - 1 \cdot 2) $$

$$ = \hat{i} (4 - 6) - \hat{j} (4 - 4) + \hat{k} (3 - 2) $$

$$ = \hat{i} (-2) - \hat{j} (0) + \hat{k} (1) $$

$$ = (-2, 0, 1) $$

So, the direction vector of the given line is $$ (-2, 0, 1) $$. The direction vector of the z-axis is $$ (0, 0, 1) $$, since it is parallel to the z-axis.

Next, find a point on the given line by solving the system of equations $$ x + y + 2z = 3 $$ and $$ 2x + 3y + 4z = 4 $$. Multiply the first equation by 2:

$$ 2x + 2y + 4z = 6 $$

Subtract the second equation from this:

$$ (2x + 2y + 4z) - (2x + 3y + 4z) = 6 - 4 $$

$$ 2x + 2y + 4z - 2x - 3y - 4z = 2 $$

$$ -y = 2 $$

$$ y = -2 $$

Substitute $$ y = -2 $$ into the first equation:

$$ x + (-2) + 2z = 3 $$

$$ x + 2z = 5 $$

Let $$ z = t $$, then $$ x = 5 - 2t $$. So, a point on the line is $$ (5 - 2t, -2, t) $$. Choosing $$ t = 0 $$, the point is $$ (5, -2, 0) $$.

Now, for the z-axis, we can take the point $$ (0, 0, 0) $$, as it lies on the axis. So, we have:

• Line 1 (z-axis): Point $$ A(0, 0, 0) $$, direction vector $$ \vec{a} = (0, 0, 1) $$
• Line 2 (given line): Point $$ B(5, -2, 0) $$, direction vector $$ \vec{b} = (-2, 0, 1) $$

The vector connecting point A to point B is $$ \overrightarrow{AB} = B - A = (5 - 0, -2 - 0, 0 - 0) = (5, -2, 0) $$.

The shortest distance between two skew lines is given by the formula:

$$ d = \frac{ | \overrightarrow{AB} \cdot (\vec{a} \times \vec{b}) | }{ | \vec{a} \times \vec{b} | } $$

First, compute the cross product $$ \vec{a} \times \vec{b} $$:

$$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 1 \\ -2 & 0 & 1 \end{vmatrix} $$

Expanding the determinant:

$$ \hat{i} (0 \cdot 1 - 1 \cdot 0) - \hat{j} (0 \cdot 1 - 1 \cdot (-2)) + \hat{k} (0 \cdot 0 - 0 \cdot (-2)) $$

$$ = \hat{i} (0 - 0) - \hat{j} (0 - (-2)) + \hat{k} (0 - 0) $$

$$ = \hat{i} (0) - \hat{j} (2) + \hat{k} (0) $$

$$ = (0, -2, 0) $$

So, $$ \vec{a} \times \vec{b} = (0, -2, 0) $$. The magnitude is $$ | \vec{a} \times \vec{b} | = \sqrt{0^2 + (-2)^2 + 0^2} = \sqrt{4} = 2 $$.

Now, compute the dot product $$ \overrightarrow{AB} \cdot (\vec{a} \times \vec{b}) = (5, -2, 0) \cdot (0, -2, 0) $$:

$$ = 5 \cdot 0 + (-2) \cdot (-2) + 0 \cdot 0 $$

$$ = 0 + 4 + 0 $$

$$ = 4 $$

The absolute value is $$ |4| = 4 $$.

Therefore, the shortest distance is:

$$ d = \frac{4}{2} = 2 $$

To confirm, the lines are skew because they are not parallel (since $$ (-2, 0, 1) $$ is not a scalar multiple of $$ (0, 0, 1) $$) and do not intersect (as the y-coordinate of the given line is always -2, while on the z-axis it is 0). Hence, the formula applies.

So, the shortest distance is 2, which corresponds to Option B.