Projection onto a plane and a reflected triangle

Plane π1\pi_1 and line l are r⋅(2,1,−2)=6\mathbf r\cdot(2,1,-2)=6 and (x−1)/3=(y+4)/2=z−5(x-1)/3=(y+4)/2=z-5 respectively. A has position vector 3i−j+4k3\mathbf i-\mathbf j+4\mathbf k.

(i) Find the shortest distance from A to π1\pi_1. marks

(ii) B satisfies AB→=−5j−2k\overrightarrow{AB}=-5\mathbf j-2\mathbf k. Find the length of projection of AB onto π1\pi_1. marks

(iii) Use (i),(ii) to find area ABC, where C is A reflected in π1\pi_1. marks

(iv) Find the Cartesian equation of π2\pi_2 containing l and O. Hence find the intersection of π1,π2\pi_1,\pi_2. marks

[2010/AJC/Prelim/I/13]

Total marks: 10

Solution

(i)

Distance from a point to a plane

π1:r⋅(2,1,−2)=6\pi_1:\mathbf r\cdot(2,1,-2)=6 and A=(3,−1,4)A=(3,-1,4). Find the shortest distance from AA to π1\pi_1.

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Hint

  1. Take a point PP on π1\pi_1 and project AP→\overrightarrow{AP} onto the normal.

Answer · (i)

33.

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A point and its perpendicular foot

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(ii)

Projection onto a plane

AB→=−5j−2k\overrightarrow{AB}=-5\mathbf j-2\mathbf k. Find the length of the projection of ABAB onto π1\pi_1.

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Hint

  1. The component of AB→\overrightarrow{AB} along the plane has length ∣AB→×n∣∣n∣\frac{|\overrightarrow{AB}\times\mathbf n|}{|\mathbf n|}.

Answer · (ii)

265/32\sqrt{65}/3.

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A segment and its shadow on the plane

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(iii)

Area of a triangle

CC is AA reflected in π1\pi_1. Find the area of triangle ABCABC.

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Hint

  1. AC=2×3=6AC=2\times3=6 and AC⊥π1AC\perp\pi_1.

  2. The base is ACAC; the height is the projection of ABAB onto the plane.

Answer · (iii)

2652\sqrt{65}.

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Move A: its shadow and reflection move together

Explore the perpendicular foot, the projected segment and triangle ABC in one scene. Reset returns to A = (3, −1, 4). B and the plane stay fixed.

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(iv)

Plane through a line and the origin

l:x−13=y+42=z−5l:\frac{x-1}3=\frac{y+4}2=z-5. Find the equation of π2\pi_2 containing ll and OO, and the line where π1\pi_1 and π2\pi_2 meet.

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Hint

  1. π2\pi_2 contains the vectors (1,−4,5)(1,-4,5) and (3,2,1)(3,2,1).

  2. Intersection: solve the two plane equations.

Answer · (iv)

π2:−x+y+z=0\pi_2:-x+y+z=0; intersection r=(2,2,0)+t(1,0,1)r=(2,2,0)+t(1,0,1).

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Two planes and their line

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