Porisma

Worked solution

Use a diagonal direction for a line and a parallelogram area

Relative to O, A,B,C have position vectors i+6j−2k\mathbf i+6\mathbf j-2\mathbf k, −3i+2j−5k-3\mathbf i+2\mathbf j-5\mathbf k, 3i−2j+2k3\mathbf i-2\mathbf j+2\mathbf k. L passes through A,C.

(i) Find a vector equation of L. marks

(ii) Find the exact length of projection of AB onto L. marks

(iii) Hence or otherwise find the shortest distance from B to L in exact form. marks

D is such that ABCD is a parallelogram. Use a vector product to find its exact area. marks

[2012/IJC/Promo/10]

CJC · Review problems · Question 3

Answer

L:r=(1,6,−2)+λ(1,−4,2)L:r=(1,6,-2)+\lambda(1,-4,2); projection 6/216/\sqrt{21}; distance 511/75\sqrt{11/7}; area 103310\sqrt{33}.

(i) Choose a nonzero line direction

AC=(2,−8,4)=2(1,−4,2)AC=(2,-8,4)=2(1,-4,2). Hence L:r=(1,6,−2)+λ(1,−4,2)L:\mathbf r=(1,6,-2)+\lambda(1,-4,2), λ∈R\lambda\in\mathbb R.

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(ii) Project the displacement AB

AB=(−4,−4,−3)AB=(-4,-4,-3). The projection length AN is ∣AB⋅(1,−4,2)∣/21=∣−4+16−6∣/21=6/21|AB\cdot(1,-4,2)|/\sqrt{21}=|-4+16-6|/\sqrt{21}=6/\sqrt{21}.

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(iii) Resolve the right triangle

∣AB∣=16+16+9=41|AB|=\sqrt{16+16+9}=\sqrt{41}, so BN=AB2−AN2=41−36/21=275/7=511/7BN=\sqrt{AB^2-AN^2}=\sqrt{41-36/21}=\sqrt{275/7}=5\sqrt{11/7}.

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Parallelogram area using the printed vector product

The source uses ∣BA×CA∣=∣(4,4,3)×(−2,8,−4)∣=∣(−40,10,40)∣=10∣(−4,1,4)∣=1033|BA\times CA|=|(4,4,3)\times(-2,8,-4)|=|(-40,10,40)|=10|(-4,1,4)|=10\sqrt{33}. CA is a diagonal; its cross product with the adjacent-side vector BA still gives the full parallelogram area because the parallel side component contributes zero.

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Common mistakes

The last ask is the parallelogram area: do not insert the triangle factor 1/2.

Solution notes

Every printed ask, source direction, projection/Pythagoras equations and the BA×CA area method retained. No figure.

Read the hint