Porisma

Discussion questions · Section 2 of 2

Questions 7–10 in this collection

Equate absolute normal distances to two intersecting planes · Separate a unique triple intersection from a common line · +2 more

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  1. 7

    The planes p1:r⋅(2,−2,1)=1p_1:\mathbf r\cdot(2,-2,1)=1 and p2:r⋅(−6,3,2)=−1p_2:\mathbf r\cdot(-6,3,2)=-1 meet in line l.

    (i) Find the acute angle between the planes.

    (ii) Find a vector equation for l.

    (iii) A(4,3,c)A(4,3,c) is equidistant from the two planes. Calculate both possible values of c.

    [2013/II/4]

    CJC · Discussion · Question 7

  2. 8

    The planes p1,p2,p3p_1,p_2,p_3 are 2x−5y+3z=32x-5y+3z=3, 3x+2y−5z=−53x+2y-5z=-5, 5x+λy+17z=μ5x+\lambda y+17z=\mu, respectively. Here λ,μ\lambda,\mu are constants. When λ=−20.9\lambda=-20.9, μ=16.6\mu=16.6, find x,y,z satisfying all three equations.

    The first two planes meet in a line l.

    (i) Find a vector equation of l.

    (ii) If p3p_3 contains l, find λ,μ\lambda,\mu.

    (iii) If instead p3p_3 does not intersect l, what can be said about λ,μ\lambda,\mu?

    (iv) Find a Cartesian equation of the plane containing l and (1,−1,3)(1,-1,3).

    [2008/I/11(modified)]

    CJC · Discussion · Question 8

  3. 9

    (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

  4. 10

    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. marks

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

    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. marks

    A mirror is placed on plane VBC.

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

    Expand diagram

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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

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