Equation of a line in 3D

Solution

Option (b) is correct.

Explanation:

To find the perpendicular distance \(d\) from the point \(P(2, 3, 4)\) to the line defined by the symmetric equations \[ \frac{x - 0}{1} = \frac{y - 0}{0} = \frac{z - 0}{0}, \] we can express the line in parametric form as \[ x = t, \quad y = 0, \quad z = 0. \] The direction vector of the line \(\mathbf{d}\) is \((1, 0, 0)\) and a point on the line is \(O(0, 0, 0)\). The vector \(\mathbf{OP}\) from point \(O\) to point \(P\) is \[ \mathbf{OP} = (2 - 0, 3 - 0, 4 - 0) = (2, 3, 4). \] The distance \(d\) from point \(P\) to the line can be calculated using the formula: \[ d = \frac{|\mathbf{OP} \cdot (\mathbf{d} \times \mathbf{OP})|}{|\mathbf{d}|}. \] First, compute \(\mathbf{d} \times \mathbf{OP}\): \[ \mathbf{d} \times \mathbf{OP} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 0 \\ 2 & 3 & 4 \end{vmatrix} = (0 \cdot 4 - 0 \cdot 3)\mathbf{i} - (1 \cdot 4 - 0 \cdot 2)\mathbf{j} + (1 \cdot 3 - 0 \cdot 2)\mathbf{k} = (0, -4, 3). \] Next, calculate the magnitude of \(\mathbf{d} \times \mathbf{OP}\): \[ |\mathbf{d} \times \mathbf{OP}| = \sqrt{0^2 + (-4)^2 + 3^2} = \sqrt{0 + 16 + 9} = \sqrt{25} = 5. \] Now, compute the magnitude of \(\mathbf{d}\): \[ |\mathbf{d}| = \sqrt{1^2 + 0^2 + 0^2} = 1. \] Finally, substitute back into the distance formula: \[ d = \frac{5}{1} = 5 \text{ units}. \] Thus, the perpendicular distance from point \(P(2, 3, 4)\) to the line is 5 units.