Build coordinates from a cuboid

Q14 cuboid with A as origin and basis i j kABCDPQRSijkAB=5, AD=3, AP=2 units
Question diagram 2
Q14 cuboid with A as origin and basis i j kABCDPQRSijkAB=5, AD=3, AP=2 units

Native cuboid ABCD PQRS with basis i,j,k

The diagram above shows a cuboid with horizontal rectangular base ABCDABCD, where AB=5AB=5 units, AD=3AD=3 units and AP=2AP=2 units. The edges AP,BQ,CR,DSAP,BQ,CR,DS are vertical, and PQRSPQRS is the top of the cuboid. The point AA is taken as the origin and vectors i,j,k\mathbf i,\mathbf j,\mathbf k, each of length 1 unit, are taken along AB,AD,APAB,AD,AP respectively.

(i) Show that the line DQDQ has equation r=3j+λ(5i−3j+2k)\mathbf r=3\mathbf j+\lambda(5\mathbf i-3\mathbf j+2\mathbf k), where λ\lambda is a parameter. Find an equation of the line BSBS. marks

(ii) Hence calculate the acute angle between the lines DQ,BSDQ,BS, giving your answer to the nearest 0.1∘0.1^\circ. marks

The point XX lies on RCRC such that RX=23RCRX=\frac23RC, and the point MM is the midpoint of BDBD. The plane pp contains the points R,B,MR,B,M.

(iii) Find, in scalar product form, an equation of the plane pp. marks

(iv) Find the exact value of the shortest distance from XX to the plane pp. marks

[2009/IJC/Prelim/I/13]

Total marks: 13

Solution

(i)

Lines in a cuboid

AA is the origin, i,j,k\mathbf i,\mathbf j,\mathbf k along AB,AD,APAB,AD,AP with AB=5AB=5, AD=3AD=3, AP=2AP=2. Find equations of DQDQ and BSBS.

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Hint

  1. Write down the coordinates of every vertex first.

Answer · (i)

DQ:r=(0,3,0)+λ(5,−3,2)DQ:r=(0,3,0)+\lambda(5,-3,2); BS:r=(5,0,0)+t(−5,3,2)BS:r=(5,0,0)+t(-5,3,2).

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The cuboid and the diagonals DQ and BS

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(ii)

Angle between lines

Calculate the acute angle between DQDQ and BSBS to the nearest 0.1∘0.1^\circ.

Hint

  1. Use the direction vectors (5,−3,2)(5,-3,2) and (−5,3,2)(-5,3,2).

Answer · (ii)

37.9∘37.9^\circ.

(iii)

Plane through three points

XX is on RCRC with RX=23RCRX=\frac23RC; MM is the midpoint of BDBD. Find the equation of plane pp through RR, BB, MM in scalar product form.

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Hint

  1. Normal =RB→×RM→=\overrightarrow{RB}\times\overrightarrow{RM}.

Answer · (iii)

6x+10y−15z=306x+10y-15z=30.

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The plane through R, B and M

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(iv)

Distance from a point to a plane

Find the exact shortest distance from XX to pp.

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Hint

  1. X=(5,3,23)X=\left(5,3,\frac23\right).

  2. Distance =∣n⋅X−d∣∣n∣=\frac{|\mathbf n\cdot X-d|}{|\mathbf n|}.

Answer · (iv)

20/1920/19.

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X and its foot on the plane

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