3D Geometry
Distance from a Point to a Line
Grade 12
Question:
<p>The given line is \(x = 4y+5,\; z = 3y-6\). A point on the line is \((4\lambda+5,\;\lambda,\;3\lambda-6)\). The distance between the point \((4\lambda+5,\;\lambda,\;3\lambda-6)\) and \((5,3,-6)\) is 3 units. Find the point on the line closest to \((5,3,-6)\).</p>
<p>\((5, 0, -6)\)</p>
<p>\((5, 3, -6)\)</p>
<p>\((1, -1, -3)\)</p>
<p>\((0, 0, 0)\)</p>
Step-by-Step Solution
Key Concept: Convert the line to parametric form, then use the distance formula to find λ values where distance equals 3. The closest point occurs where the line from the given point to the line is perpendicular to the line's direction vector.
Step 1: Identify the line in parametric form. Given x = 4y + 5 and z = 3y - 6, let y = λ. Then the line is: L(λ) = (4λ + 5, λ, 3λ - 6) with direction vector d = (4, 1, 3). Step 2: For the closest point on the line to P(5, 3, -6), the vector from P to the point on the line must be perpendicular to d . Step 3: Vector from P to L(λ): v = (4λ + 5 - 5, λ - 3, 3λ - 6 - (-6)) = (4λ, λ - 3, 3λ) Step 4: Apply perpendicularity condition: v · d = 0
(4λ)(4) + (λ - 3)(1) + (3λ)(3) = 0
16λ + λ - 3 + 9λ = 0
26λ = 3
λ = 3/26 Step 5: Substitute λ = 3/26 into the parametric equation:
x = 4(3/26) + 5 = 12/26 + 130/26 = 142/26 = 71/13
y = 3/26
z = 3(3/26) - 6 = 9/26 - 156/26 = -147/26 ∴ The closest point is (71/13, 3/26, -147/26) or equivalently (71/13, 3/26, -147/26)
Correct Answer: A