3D Geometry
Points on Lines
Grade 12
Question:
<p><strong>Ex. 63 (A):</strong> The coordinates of a point on the line \(x = 4y + 5, z = 3y - 6\) at a distance 3 from the point \((5, 3, -6)\) is/are</p>
<p>(p) \((-1, -2, 0)\)</p>
<p>(q) \((5, 0, -6)\)</p>
<p>(r) \((2, 5, 7)\)</p>
<p>(s) Other</p>
Step-by-Step Solution
Key Concept: Parametrize the line and use the distance formula to find points at a specific distance from a given point.
Solution: The line can be parametrized. Let \(y = t\), then \(x = 4t + 5, z = 3t - 6\). A general point on the line is \((4t+5, t, 3t-6)\). Distance from \((5, 3, -6)\) is 3: \((4t+5-5)^2 + (t-3)^2 + (3t-6-(-6))^2 = 9\) \((4t)^2 + (t-3)^2 + (3t)^2 = 9\) \(16t^2 + t^2 - 6t + 9 + 9t^2 = 9\) \(26t^2 - 6t = 0\) \(t(26t - 6) = 0 \Rightarrow t = 0 \text{ or } t = \frac{3}{13}\) For \(t = 0\): Point is \((5, 0, -6)\) For \(t = \frac{3}{13}\): Point is \(\left(5 + \frac{12}{13}, \frac{3}{13}, -6 + \frac{9}{13}\right)\) ∴ Answer is (q) \((5, 0, -6)\)
Correct Answer: A