Porisma

Worked solution

Drop a normal from a point and reflect through its foot

A plane passes through (1,−2,1)(1,-2,1) and (2,−1,−1)(2,-1,-1) and is parallel to the line r=(1,−1,0)+s(3,1,−2)\mathbf r=(1,-1,0)+s(3,1,-2), s∈Rs\in\mathbb R.

(i) Find the foot of the perpendicular from A(1,1,−1)A(1,1,-1) to the plane.

(ii) Hence or otherwise find the shortest distance from A to the plane.

(iii) Find the image of A reflected in the plane.

CJC · Discussion · Question 4

Answer

F=(1,−3/5,−9/5)F=(1,-3/5,-9/5); distance 45/54\sqrt5/5; image (1,−11/5,−13/5)(1,-11/5,-13/5).

(i) Construct the plane and the normal line

Two plane directions are (3,1,−2)(3,1,-2) and (2,−1,−1)−(1,−2,1)=(1,1,−2)(2,-1,-1)-(1,-2,1)=(1,1,-2). Thus r=(1,−2,1)+s(3,1,−2)+t(1,1,−2)\mathbf r=(1,-2,1)+s(3,1,-2)+t(1,1,-2). Their cross product is (0,4,2)=2(0,2,1)(0,4,2)=2(0,2,1). The scalar equation is r⋅(0,2,1)=−3\mathbf r\cdot(0,2,1)=-3, or 2y+z=−32y+z=-3. The normal through A is r=(1,1,−1)+λ(0,2,1)\mathbf r=(1,1,-1)+\lambda(0,2,1).

How did this part go?

(i) Intersect the normal with the plane

At the foot F=(1,1+2λ,−1+λ)F=(1,1+2\lambda,-1+\lambda), 2(1+2λ)−1+λ=−32(1+2\lambda)-1+\lambda=-3, so 5λ=−45\lambda=-4. Therefore F=(1,−3/5,−9/5)F=(1,-3/5,-9/5).

How did this part go?

(ii) Take the length of AF

AF→=(0,−8/5,−4/5)\overrightarrow{AF}=(0,-8/5,-4/5), so AF=16/5=45/5AF=\sqrt{16/5}=4\sqrt5/5 units.

How did this part go?

(iii) Use the foot as the midpoint

OF→=12(OA→+OA′→)\overrightarrow{OF}=\tfrac12(\overrightarrow{OA}+\overrightarrow{OA\prime}), hence A′=2F−A=(1,−11/5,−13/5)A\prime=2F-A=(1,-11/5,-13/5). The normal line passes through A, F and its reflected image, with equal distances on opposite sides of the plane.

How did this part go?

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A plane spanned by (3,1,-2) and (1,1,-2), followed by a normal-section sketch showing A, perpendicular foot F and reflected point A prime equally spaced across the plane; the normal is (0,2,1).
A plane spanned by (3,1,-2) and (1,1,-2), followed by a normal-section sketch showing A, perpendicular foot F and reflected point A prime equally spaced across the plane; the normal is (0,2,1).
A plane spanned by (3,1,-2) and (1,1,-2), followed by a normal-section sketch showing A, perpendicular foot F and reflected point A prime equally spaced across the plane; the normal is (0,2,1).

Common mistakes

The reflected point is twice the foot minus the original point.

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