Which of the following planes is parallel to the line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}?$

Question

Accepted Answer

To determine which plane is parallel to the given line, we need to understand the relationship between the direction vector of the line and the normal vector of the plane. A plane is parallel to a line if the direction vector of the line is orthogonal (perpendicular) to the normal vector of the plane.

The given line is represented in symmetric form as:

$$
\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}
$$

From this, we can extract the direction vector of the line, which is $\mathbf{d} = \langle 3, 4, 5 \rangle$.

For a plane given by the equation $ax + by + cz = d$, the normal vector is $\mathbf{n} = \langle a, b, c \rangle$.

### Explanation for Option 3:

The equation of the plane in Option 3 is:

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

The normal vector for this plane is $\mathbf{n} = \langle 2, 1, -2 \rangle$.

To check if the plane is parallel to the line, we need to verify if the direction vector $\mathbf{d} = \langle 3, 4, 5 \rangle$ is orthogonal to the normal vector $\mathbf{n} = \langle 2, 1, -2 \rangle$. This is done by calculating the dot product:

$$
\mathbf{d} \cdot \mathbf{n} = 3 \times 2 + 4 \times 1 + 5 \times (-2) = 6 + 4 - 10 = 0
$$

Since the dot product is zero, the direction vector is orthogonal to the normal vector, confirming that the plane is parallel to the line. Therefore, Option 3 is correct.

### Explanation for Other Options:

- **Option 1: $2x + 3y + 4z = \sqrt{29}$**

The normal vector is $\langle 2, 3, 4 \rangle$. The dot product with the direction vector is:

$$
  3 \times 2 + 4 \times 3 + 5 \times 4 = 6 + 12 + 20 = 38 \neq 0
  $$

Since the dot product is not zero, the plane is not parallel to the line.

- **Option 2: $3x + 4y + 5z = 5\sqrt{2}$**

The normal vector is $\langle 3, 4, 5 \rangle$. The dot product with the direction vector is:

$$
  3 \times 3 + 4 \times 4 + 5 \times 5 = 9 + 16 + 25 = 50 \neq 0
  $$

Since the dot product is not zero, the plane is not parallel to the line.

- **Option 4: $2x - y + 2z = 3$**

The normal vector is $\langle 2, -1, 2 \rangle$. The dot product with the direction vector is:

$$
  3 \times 2 + 4 \times (-1) + 5 \times 2 = 6 - 4 + 10 = 12 \neq 0
  $$

Since the dot product is not zero, the plane is not parallel to the line.

In conclusion, only Option 3 has a normal vector that is orthogonal to the direction vector of the line, making it the correct choice for a plane parallel to the given line.