3D Geometry
Distance of a Point from a Plane
Grade 12

Question:

<p>The distance of the point \((1, -5, 9)\) from the plane \(x - y + z = 5\) measured along the line \(x = y = z\) is</p>
<p>\(\dfrac{20}{3}\)</p>
<p>\(3\sqrt{10}\)</p>
<p>\(10\sqrt{3}\)</p>
<p>\(\dfrac{10}{\sqrt{3}}\)</p>

Step-by-Step Solution

Key Concept: The distance along a specific direction equals the distance traveled along that direction's line until it intersects the plane. Parametrize the line in direction (1,1,1) through the given point, find where it meets the plane, then calculate the distance.
Step 1: The line x=y=z has direction vector d = (1,1,1). The line through point P(1,-5,9) parallel to this direction is: x = 1+t, y = -5+t, z = 9+t Step 2: Find intersection with plane x - y + z = 5: (1+t) - (-5+t) + (9+t) = 5 1 + t + 5 - t + 9 + t = 5 15 + t = 5 t = -10 Step 3: The distance along the direction (1,1,1) is |t|√(1^2+1^2+1^2) = |-10|√3 = 10√3 ∴ Answer: C
Correct Answer: C

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