Question

Find a point on the line \(\frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2}\) at a distance of \(\sqrt{2}\) units from the point \((1, 2, 3)\).

Hint:

Always ensure your line equation has coefficients of x, y, and z as +1 before parametrizing.
Distance formula in 3D: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\).

Answer 1

Concept:
• Express the general point on a line in terms of a parameter \(\lambda\).
• Use the distance formula between the general point and the given point.

Step 1: Rewrite the line equation in standard form
The given equation is \(\frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2}\).
To standardise, multiply the second term by -1 in both numerator and denominator:
\[ \frac{x - 2}{3} = \frac{y - 1}{-2} = \frac{z - 3}{2} = \lambda \]

Step 2: Write the general point on the line
Any point \(P\) on the line is given by:
\(x = 3\lambda + 2\)
\(y = -2\lambda + 1\)
\(z = 2\lambda + 3\)

Step 3: Apply the distance formula
The distance from \(P(3\lambda + 2, -2\lambda + 1, 2\lambda + 3)\) to \(Q(1, 2, 3)\) is \(\sqrt{2}\):
\[ \sqrt{(3\lambda + 2 - 1)^2 + (-2\lambda + 1 - 2)^2 + (2\lambda + 3 - 3)^2} = \sqrt{2} \] Squaring both sides:
\[ (3\lambda + 1)^2 + (-2\lambda - 1)^2 + (2\lambda)^2 = 2 \] \[ (9\lambda^2 + 6\lambda + 1) + (4\lambda^2 + 4\lambda + 1) + 4\lambda^2 = 2 \] \[ 17\lambda^2 + 10\lambda + 2 = 2 \] \[ 17\lambda^2 + 10\lambda = 0 \] \[ \lambda(17\lambda + 10) = 0 \] This gives \(\lambda = 0\) or \(\lambda = -10/17\).

Step 4: Find the point
For \(\lambda = 0\):
\(x = 3(0) + 2 = 2, y = -2(0) + 1 = 1, z = 2(0) + 3 = 3\).
One such point is \((2, 1, 3)\).