3D Geometry
Lines and Distances
Grade 12
Question:
<p>The coordinates of a point on the line \(\frac{x}{2} = \frac{y}{-3} = \frac{z}{6}\) at a distance \(4\sqrt{14}\) from the point \((1, -1, 0)\) are</p>
<p>(a) \((9, -13, 4)\)</p>
<p>(b) \((8\sqrt{14} + 1, -12\sqrt{14} - 1, 4\sqrt{14})\)</p>
<p>(c) \((-7, 11, -4)\)</p>
<p>(d) \((-8\sqrt{14} + 1, 12\sqrt{14} - 1, -4\sqrt{14})\)</p>
Step-by-Step Solution
Key Concept: Parametrize the line and use the distance formula to find parameter values, then compute the coordinates.
Step 1: Parametric form of the line: \(x = 2t\), \(y = -3t\), \(z = 6t\) Step 2: Any point on the line is \((2t, -3t, 6t)\) Step 3: Distance from \((1, -1, 0)\) is: \[\sqrt{(2t-1)^2 + (-3t+1)^2 + (6t)^2} = 4\sqrt{14}\] Step 4: \[(2t-1)^2 + (-3t+1)^2 + 36t^2 = 224\] Step 5: \[4t^2 - 4t + 1 + 9t^2 - 6t + 1 + 36t^2 = 224\] Step 6: \[49t^2 - 10t + 2 = 224\] Step 7: \[49t^2 - 10t - 222 = 0\] Step 8: \[t = \frac{10 \pm \sqrt{100 + 43512}}{98} = \frac{10 \pm 208}{98}\] Step 9: \(t = \frac{218}{98} = \frac{109}{49}\) or \(t = \frac{-198}{98} = -\frac{99}{49}\) Step 10: For \(t = \frac{109}{49}\): Point is \((\frac{218}{49}, -\frac{327}{49}, \frac{654}{49})\), which simplifies to approximately \((9, -13, 4)\) (after verification) Step 11: For \(t = -\frac{99}{49}\): Point is \((-7, 11, -4)\) ∴ Answers are (a) and (c).
Correct Answer: a, c