Porisma

Worked solution

Interpret a plane equation and derive the line-intersection parameter

(a) Points P,Q,R have position vectors p,q,r\mathbf p,\mathbf q,\mathbf r. P and Q are fixed; R varies. Given r=(x,y,z)\mathbf r=(x,y,z), p=(−1,2,4)\mathbf p=(-1,2,4), q=(3,−5,2)\mathbf q=(3,-5,2) and (r−p)⋅q=0(\mathbf r-\mathbf p)\cdot\mathbf q=0, find the relation between x,y,z and describe the locus of R.

(b) Interpret r⋅n=d\mathbf r\cdot\mathbf n=d geometrically, where n\mathbf n is a constant unit vector and d is a constant scalar. State what d represents.

(c) Given b⋅n≠0\mathbf b\cdot\mathbf n\ne0, solve r=a+tb\mathbf r=\mathbf a+t\mathbf b and r⋅n=d\mathbf r\cdot\mathbf n=d for t in terms of a,b,n,d\mathbf a,\mathbf b,\mathbf n,d. Interpret the solution geometrically.

[2017&2020/I/(modified)]

CJC · Discussion · Question 9

Answer

(a) 3x−5y+2z=−53x-5y+2z=-5; (b) signed normal displacement d; (c) t=(d−a⋅n)/(b⋅n)t=(d-\mathbf a\cdot\mathbf n)/(\mathbf b\cdot\mathbf n).

(a) Expand the displacement-normal condition

(x+1,y−2,z−4)⋅(3,−5,2)=0(x+1,y-2,z-4)\cdot(3,-5,2)=0 gives 3x+3−5y+10+2z−8=03x+3-5y+10+2z-8=0, or 3x−5y+2z=−53x-5y+2z=-5. R lies on the plane through P with normal (3,−5,2)(3,-5,2).

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(b) Use the unit-normal convention

r⋅n=d\mathbf r\cdot\mathbf n=d is a plane normal to the unit vector n\mathbf n. The scalar d is its signed perpendicular displacement from the origin measured in the direction of that normal. The unsigned distance is ∣d∣|d|; the unit-length condition is essential to that direct interpretation.

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(c) Substitute the line into the plane

(a+tb)⋅n=d(\mathbf a+t\mathbf b)\cdot\mathbf n=d gives a⋅n+t(b⋅n)=d\mathbf a\cdot\mathbf n+t(\mathbf b\cdot\mathbf n)=d. Since the denominator is nonzero, t=(d−a⋅n)/(b⋅n)t=(d-\mathbf a\cdot\mathbf n)/(\mathbf b\cdot\mathbf n). Thus the unique intersection point has position vector r=a+[(d−a⋅n)/(b⋅n)]b\mathbf r=\mathbf a+[(d-\mathbf a\cdot\mathbf n)/(\mathbf b\cdot\mathbf n)]\mathbf b. The line is not parallel to the plane.

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Common mistakes

Signed displacement d and distance |d| have different meanings.

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