3D Geometry
Distance of a Point from a Plane
Grade 12
Question:
<p>The distance of the point \((1, 3, -7)\) from the plane passing through the point \((1, -1, -1)\), having normal perpendicular to both the lines \(\dfrac{x-1}{1} = \dfrac{y+2}{-2} = \dfrac{z-4}{3}\) and \(\dfrac{x-2}{2} = \dfrac{y+1}{-1} = \dfrac{z+7}{-1}\), is</p>
<p>\(\dfrac{10}{\sqrt{83}}\)</p>
<p>\(\dfrac{5}{\sqrt{83}}\)</p>
<p>\(\dfrac{10}{\sqrt{74}}\)</p>
<p>\(\dfrac{20}{\sqrt{74}}\)</p>
Step-by-Step Solution
Key Concept: The normal to the plane is perpendicular to both given lines, so it's parallel to the cross product of their direction vectors. Use this normal and the point on the plane to find the plane equation, then apply the point-to-plane distance formula.
Step 1: Find the normal vector to the plane using cross product of direction vectors of the two lines. Direction vectors: d_1 = (1, -2, 3) and d_2 = (2, -1, -1) n = d_1 × d_2 = | i j k |
|1 -2 3|
|2 -1 -1| = i (2 + 3) - j (-1 - 6) + k (-1 + 4) = i (5) - j (-7) + k (3) = (5, 7, 3) Step 2: Write the plane equation using point (1, -1, -1) and normal (5, 7, 3). 5(x - 1) + 7(y + 1) + 3(z + 1) = 0 5x - 5 + 7y + 7 + 3z + 3 = 0 5x + 7y + 3z + 5 = 0 Step 3: Apply point-to-plane distance formula for point P(1, 3, -7): Distance = |5(1) + 7(3) + 3(-7) + 5| / √(5^2 + 7^2 + 3^2) = |5 + 21 - 21 + 5| / √(25 + 49 + 9) = |10| / √83 = 10/√83 = 10√83/83 ∴ Answer: C
Correct Answer: C