Porisma
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Vectors: planes, normals and distances

Practice section 5 of 7

Questions 25–30 in this collection

Practice questions

Discussion practice · 1 of 6

Project a pyramid shadow under two different lighting models

A right opaque pyramid has square base ABCD at ground level, centre O and vertex V. Perpendicular unit vectors i,j,k\mathbf i,\mathbf j,\mathbf k point along AB, AD, OV, respectively. AB has length 8 and OV has length 2h. A point light source is at P(20,−4,0)P(20,-4,0). A screen of height 35 has its base on the ground and lies in r⋅(1,0,0)=α\mathbf r\cdot(1,0,0)=\alpha, where α<−4\alpha<-4, as shown.

(i) Find a vector equation for the line depicting the light ray from P to V, in terms of h. [2]

(ii) Find an inequality between α,h\alpha,h so the pyramid’s shadow does not exceed the screen height. [3]

The point light is now replaced by parallel light rays perpendicular to the screen; also h=10h=10.

(iii) Find the exact length of the shadow cast by edge VB on the screen. [3]

A mirror is placed on plane VBC.

(iv) Find a vector equation of plane VBC and hence the mirror’s inclination to the ground. [4]

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Given setup: square ABCD on the ground centred at O, vertex V vertically above O, i along AB, j along AD, k along OV. P(20,-4,0) is beyond B on the AB ground line. The screen stands on x=alpha less than -4 and has height 35; base AB=8 and OV=2h.
Given setup: square ABCD on the ground centred at O, vertex V vertically above O, i along AB, j along AD, k along OV. P(20,-4,0) is beyond B on the AB ground line. The screen stands on x=alpha less than -4 and has height 35; base AB=8 and OV=2h.
Given setup: square ABCD on the ground centred at O, vertex V vertically above O, i along AB, j along AD, k along OV. P(20,-4,0) is beyond B on the AB ground line. The screen stands on x=alpha less than -4 and has height 35; base AB=8 and OV=2h.

[2017/RI/I/11]

CJC · Discussion · Question 10

Solution

Review practice · 2 of 6

Build an intersection line and a perpendicular plane

Planes π1,π2\pi_1,\pi_2 have equations r⋅(0,1,2)=2\mathbf r\cdot(0,1,2)=2, r⋅(2,−1,4)=12\mathbf r\cdot(2,-1,4)=12. A=(7,2,0), O is the origin.

(i) Verify A lies in both planes. [1]

(ii) They meet in line L. Show that a vector equation of L is r=(7,2,0)+λ(3,2,−1)\mathbf r=(7,2,0)+\lambda(3,2,-1), λ∈R\lambda\in\mathbb R. [2]

(iii) B=(7,−4,9), and N is its perpendicular foot onto π2\pi_2. Find N. [3]

(iv) Find the Cartesian equation of π3\pi_3 containing L and perpendicular to π2\pi_2. [3]

[2008/SRJC/Promo/15]

CJC · Review problems · Question 1

Solution

Review practice · 3 of 6

Use prism geometry to choose line and plane directions

Prism ABCDEF has rectangular base ABEF. Triangles ABC, DEF are congruent and perpendicular to the base. A,C have position vectors (−1,3,−5)(−1,3,−5), (3,7,5)(3,7,5). The unit vector in the direction FE is (5,−3,2)/38(5,−3,2)/\sqrt{38} and BC has equation r=(6,2,2)+λ(−3,5,3)\mathbf r=(6,2,2)+\lambda(-3,5,3), λ real. Find:

(i) a vector equation of AB and the position vector of B; [3]

(ii) an equation of plane ABC in the form r⋅n=D\mathbf r\cdot\mathbf n=D; [3]

(iii) a vector equation of BE; [2]

(iv) the position vector of the perpendicular foot from C onto plane ABEF. [4]

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Prism ABCDEF: base ABEF is rectangular, triangles ABC and DEF are congruent and perpendicular to the base. Corresponding edges AF, BE, CD are parallel. Hidden edges AB, BC, BE are dashed. This is a perspective schematic retaining all source vertex labels and incidences, not a projection of the supplied coordinates.
Prism ABCDEF: base ABEF is rectangular, triangles ABC and DEF are congruent and perpendicular to the base. Corresponding edges AF, BE, CD are parallel. Hidden edges AB, BC, BE are dashed. This is a perspective schematic retaining all source vertex labels and incidences, not a projection of the supplied coordinates.
Prism ABCDEF: base ABEF is rectangular, triangles ABC and DEF are congruent and perpendicular to the base. Corresponding edges AF, BE, CD are parallel. Hidden edges AB, BC, BE are dashed. This is a perspective schematic retaining all source vertex labels and incidences, not a projection of the supplied coordinates.

[2010/MJC/Promo/12]

CJC · Review problems · Question 2

Solution

Review practice · 4 of 6

Find a line-plane angle and reflect a point

Plane Π has equation r⋅(1,2,−3)=5\mathbf r\cdot(1,2,-3)=5. Line l passes through P with position vector (7,4,−6)(7,4,−6) and is parallel to (7,3,−5)(7,3,−5).

(i) Find the perpendicular distance from the origin to Π. [3]

(ii) Find the acute angle between l and Π. [3]

(iii) Find the position vector of the perpendicular foot from P to Π. Hence find the position vector of P’s reflection in Π. [6]

[2007/IJC/I/10(modified)]

CJC · Review problems · Question 3

Solution

Review practice · 5 of 6

Separate plane angles, intersections and orthogonal directions

Planes Π1,Π2\Pi_1,\Pi_2 are r⋅(2,4,1)=10\mathbf r\cdot(2,4,1)=10, r⋅(1,3,1)=8\mathbf r\cdot(1,3,1)=8.

(i) Find their acute angle. [2]

(ii) Obtain a vector equation of their intersection line l1l_1. [2]

Another line l2l_2 is (x−2)/6=(7−z)/3(x-2)/6=(7-z)/3, y=m, where m is real.

(iii) If Π1 and l2 intersect at (6,m,5), find m. [2]

(iv) Show l1 and l2 are perpendicular for all m. [2]

[2010/DHS/II/4(modified)]

CJC · Review problems · Question 4

Solution

Review practice · 6 of 6

Construct a plane parallel to a line and measure their separation

Line l is x−a=y+2=(z+5)/5x-a=y+2=(z+5)/5, and plane p1 is 3x+2y−z=43x+2y-z=4. Plane p1 contains l.

(i) Show a=1. [2]

Plane p2 passes through E=(−2,−5,8) and meets p1 in l.

(ii) Find the Cartesian equation of p2. [3]

(iii) Find N, the perpendicular foot from E to p1. [3]

Plane p3 contains E,N and does not intersect l.

(iv) Find its Cartesian equation. Hence find the distance from l to p3. [4]

[2012/AJC/Promo/10(modified)]

CJC · Review problems · Question 5

Solution

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