3D Geometry
Projection of line segment on a plane
Grade None
Question:
<p>The length of the projection of the line segment joining the points (5, –1, 4) and (4, –1, 3) on the plane, <em>x</em> + <em>y</em> + <em>z</em> = 7 is</p>
<p>\(\dfrac{2}{3}\)</p>
<p>\(\dfrac{1}{3}\)</p>
<p>\(\sqrt{\dfrac{2}{3}}\)</p>
<p>\(\dfrac{2}{\sqrt{3}}\)</p>
Step-by-Step Solution
Key Concept: The projection length of a line segment onto a plane equals the distance between the projections of its endpoints, which can be found using the formula: √(d² - h²) where d is the segment length and h is the perpendicular distance between the segment and plane.
Step 1: Find the length of the line segment AB where A(5, -1, 4) and B(4, -1, 3). d = √[(5-4)^2 + (-1-(-1))^2 + (4-3)^2] = √[1 + 0 + 1] = √2 Step 2: Find the perpendicular distance from point A to the plane x + y + z = 7. h_1 = |5 + (-1) + 4 - 7|/√(1^2 + 1^2 + 1^2) = |1|/√3 = 1/√3 Step 3: Find the perpendicular distance from point B to the plane x + y + z = 7. h_2 = |4 + (-1) + 3 - 7|/√3 = |-1|/√3 = 1/√3 Step 4: Since both points are at the same perpendicular distance from the plane, the segment is parallel to the plane. The projection length equals the segment length itself. Projection = √2 Step 5: Alternatively, use the formula: Projection = d·cos(θ) where θ is the angle between the segment and plane's normal. The normal vector is n = (1, 1, 1). Direction vector: v = (-1, 0, -1), |v| = √2 cos(θ) = |v·n|/(|v||n|) = |(-1)(1) + (0)(1) + (-1)(1)|/(√2·√3) = 2/(√6) Projection = √2 · √(1 - 4/6) = √2 · √(1/3) = √(2/3) ∴ Answer: C
Correct Answer: C