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Original question & working

A parallel line and a perpendicular plane · source reference

[2010/JJC/Prelim/I/7] · Question pp. 3 · Working pp. 13, 14

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Question

Line l passes through A(−1,2,3)(-1,2,3) and B(5,14,11)(5,14,11). Plane p is 2x+3y−6z=−72x+3y-6z=-7.

(i) Show l is parallel to p but not contained in it. [3]

(ii) Find the distance from l to p. [3]

(iii) Find a Cartesian equation of the plane containing l and perpendicular to p. [2]

[2010/JJC/Prelim/I/7]

Source working

(i) Direction AB→=(5,14,11)−(−1,2,3)=(6,12,8)=2(3,6,4)\overrightarrow{AB}=(5,14,11)-(-1,2,3)=(6,12,8)=2(3,6,4).

Plane p: r⋅(2,3,−6)=−7\mathbf r\cdot(2,3,-6)=-7.

d⋅n=(3,6,4)⋅(2,3,−6)=6+18−24=0,\mathbf d\cdot\mathbf n=(3,6,4)\cdot(2,3,-6)=6+18-24=0,

so l is parallel to p.

OA→⋅n=(−1,2,3)⋅(2,3,−6)=−2+6−18=−14≠−7.\overrightarrow{OA}\cdot\mathbf n=(-1,2,3)\cdot(2,3,-6)=-2+6-18=-14\ne-7.

A lies on l but not p, so l is parallel to but not contained in p.

(ii) C(−7/2,0,0)(-7/2,0,0) lies on p since (−7/2,0,0)⋅(2,3,−6)=−7(-7/2,0,0)\cdot(2,3,-6)=-7.

CA→=(−1,2,3)−(−7/2,0,0)=(5/2,2,3).\overrightarrow{CA}=(-1,2,3)-(-7/2,0,0)=(5/2,2,3). distance=∣CA→⋅(2,3,−6)∣∣(2,3,−6)∣=14+9+36∣(5/2,2,3)⋅(2,3,−6)∣=∣5+6−18∣7=1 unit.\begin{aligned}\text{distance}&=\frac{|\overrightarrow{CA}\cdot(2,3,-6)|}{|(2,3,-6)|}\\&=\frac1{\sqrt{4+9+36}}|(5/2,2,3)\cdot(2,3,-6)|\\&=\frac{|5+6-18|}7=1\text{ unit}.\end{aligned}

(iii) Let required plane be p1p_1. A normal is (3,6,4)×(2,3,−6)=(−48,26,−3)(3,6,4)\times(2,3,-6)=(-48,26,-3).

A lies on p1p_1, so

r⋅(−48,26,−3)=(−1,2,3)⋅(−48,26,−3)=91,\mathbf r\cdot(-48,26,-3)=(-1,2,3)\cdot(-48,26,-3)=91, −48x+26y−3z=91.-48x+26y-3z=91.

Teacher notes

l∥pl\parallel p means the line direction is perpendicular to the plane normal: d⋅n=0\mathbf d\cdot\mathbf n=0.

To show A does not lie on p, show its coordinates fail p’s equation. Pick any point C satisfying p’s equation to calculate distance.

A normal of p1p_1 is u×v\mathbf u\times\mathbf v, where u,v are parallel to p1p_1: choose the normal of p and direction of l.

Native diagrams

Distance panel: plane p, C in it, l through A(−1,2,3)(-1,2,3) above, a red perpendicular distance and normal (2,3,−6)(2,3,-6). Perpendicular-plane panel: p and p1p_1 intersect, l lies in p1p_1, A on l, direction (3,6,4)(3,6,4) and normal (2,3,−6)(2,3,-6) lie in p1p_1, with their right-angle relations.

Line l parallel to plane p at distance 1 Plane p₁ contains l and is perpendicular to p

Printed answer key

(ii) 11 unit; (iii) −48x+26y−3z=91-48x+26y-3z=91.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question document pages 3; solution document pages 13,14. Original page numbers are retained as text metadata.