In a rectangular prism, `M` is the midpoint of `AD`.
Use vector methods to find
- `angle HBD` to 1 decimal place (2 marks)
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- `angle HBM` to 1 decimal place (2 marks)
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In a rectangular prism, `M` is the midpoint of `AD`.
Use vector methods to find
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i. `text{Consider the position vectors of points using}\ B\ text{as origin:}`
`vec(BH)=((0),(9),(6)), \ abs(vec(BH))=sqrt(9^2+6^2)=sqrt117`
`vec(BD)=((4),(9),(0)), \ abs(vec(BD))=sqrt(4^2+9^2)=sqrt97`
`cos angleHBD` | `=(vec(BH)*vec(BD))/(abs(vec(BH))*abs(vec(BD)))` | |
`=(0xx4+9xx9+6xx0)/(sqrt117 xx sqrt97)` | ||
`angleHBD` | `=cos^(-1)(81/(sqrt117 xx sqrt97))` | |
`=40.5°` |
ii. `vec(BM)=((4),(9/2),(0)), \ abs(vec(BM))=sqrt(4^2+(9/2)^2)=sqrt(145/4)`
`cos angleHBM` | `=(vec(BH) * vec(BM))/(abs(vec(BH))*abs(vec(BM)))` | |
`=(0xx4+9xx9/2+6xx0)/(sqrt99 xx sqrt(145/4))` | ||
`angleHBM` | `=cos^(-1)(40.5/(sqrt99 xx sqrt(145/4)))` | |
`=47.5°` |
A parallelogram is formed by joining the points `P(-2,1,4), Q(1,4,5), R(0,2,3)` and `S(a,b,c)`.
Use vector methods to find `a,b` and `c`. (2 marks)
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`a=-3, b=1, c=2`
`text{Opposite sides of a parallelogram are equal and parallel.}`
`=> vec(PQ)=vec(SR)`
`vec(PQ)=((1+2),(4-3),(5-4))=((3),(1),(1))`
`vec(SR)=((-a),(2-b),(3-c))`
`text{Equating coordinates:}`
`-a=3\ \ =>\ \ a=-3`
`2-b=1\ \ =>\ \ b=1`
`3-c=1\ \ =>\ \ c=2`
Classify the triangle formed by joining the points `A(3,1,0), B(-2,4,3)` and `C(3,3,-2)`. (4 marks)
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`text{ΔABC is a right-angled (scalene) triangle with a right-angle at B.}`
`text{(See Worked Solutions)}`
`text{Calculating the side lengths:}`
`abs(AB)=sqrt((3+2)^2+(1-4)^2+(0-3)^2)=sqrt43`
`abs(BC)=sqrt((-2-3)^2+(4-3)^2+(3+2)^2)=sqrt51`
`abs(AC)=sqrt((3-3)^2+(1-3)^2+(0+2)^2)=sqrt8`
`=>\ text{Triangle is scalene.}`
`text{From above,}`
`abs(BC)^2=abs(AB)^2+abs(AC)^2\ \ (angleBAC=90°)`
`vec(AB)=((3),(1),(0))+lambda((-2-3),(4-1),(3-0))=((3),(1),(0))+lambda((-5),(3),(3))`
`vec(AB)*vec(AC)=((-5),(3),(3))((0),(2),(-2))=0+6-6=0`
`:.vec(AB) ⊥ vec(AC)`
`:.\ text{ΔABC is a right-angled (scalene) triangle with a right-angle at A.}`
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The diagram below shows the tetrahedron with vertices `A, B, C` and `S`.
The point `K` is defined by `vec(SK)=(1)/(4) vec(SB)+(1)/(3) vec(SC)`, as shown in the diagram.
The point `L` is the point of intersection of the straight lines `S K` and `B C`.
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i. | `lambda vecu+mu vecv` | `=0` |
`lambda vecu` | `=-mu vecv` |
`lambda=0\ \ text{or}\ \ vecu=-(mu/lambda)vecv=k vecv\ \ (kinRR)`
`text{S}text{ince}\ \ vecu and vecv\ \ text{are not parallel}`
`=> lambda=mu=0`
ii. `\lambda_1 \vec{u}+\mu_1 \vec{v}=\lambda_2 \vec{u}+\mu_2 \vec{v}`
`\lambda_1 \vec{u}-\lambda_2 \vec{u}+\mu_1 \vec{v}-\mu_2 \vec{v}` | `=vec0` | |
`(\lambda_1-\lambda_2)\vec{u}+(\mu_1-\mu_2)\vec{v}` | `=vec0` |
`text{Using part (i):}`
`(\lambda_1-\lambda_2)=0 and (\mu_1-\mu_2)=0`
`:.\lambda_1=\lambda_2 and \mu_1=\mu_2\ …\ text{as required}`
iii. | `vec(BL)` | `=lambda vec(BC)` |
`=lambda(vec(BS)+vec(SC))\ \ \ …\ (1)` |
`vec(BL)` | `=vec(BS)+mu vec(SK)` | |
`=-vec(SB)+mu(1/4vec(SB)+1/3vec(SC))` | ||
`=-vec(SB)+mu/4vec(SB)+mu/3vec(SC)` | ||
`=mu/3vec(SC)+(mu/4-1)vec(SB)\ \ \ …\ (2)` |
`text{Using}\ \ (1) = (2):`
`lambda vec(BS)+ lambda vec(SC)=mu/3vec(SC)+(mu/4-1)vec(SB)`
`mu/3=lambda, \ \ 1-mu/4=lambda`
`mu/3` | `=1-mu/4` | |
`(4mu+3mu)/12` | `=1` | |
`(7mu)/12` | `=1` | |
`mu` | `=12/7` |
`lambda=(12/7)/3=4/7`
`:.vec(BL)=(4)/(7) vec(BC)\ \ text{… as required}`
iv. `vec(AP)=-6 vec(AB)-8 vec(AC)`
`text{If}\ P\ text{lies on}\ AL, ∃k\ text{such that}\ \ vec(AP)=k vec(AL)`
`-6 vec(AB)-8 vec(AC)` | `=k vec(AL)` | |
`-6 vec(AB)-8 (vec(AB)+vec(BC))` | `=k(vec(AB)+vec(BL))` | |
`-14 vec(AB)-8vec(BC)` | `=k(vec(AB)+4/7vec(BC))\ \ text{(see part (iii))}` | |
`-14 vec(AB)-8vec(BC)` | `=kvec(AB)+(4k)/7vec(BC)` |
`k=-14, \ \ (4k)/7=-8\ \ =>\ \ k=-14`
`:.\ P\ text{lies on}\ \ AL.`
The diagram shows the pyramid `ABCDS` where `ABCD` is a square. The diagonals of the square bisect each other at `H`.
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Let `G` be the point such that `overset->{GA} + overset->{GB} + overset->{GC} + overset->{GD} + overset->{GS} = underset~0`.
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i. `text{S} text{ince diagonal} \ overset->{AC} \ text{is bisected by H:}`
`overset->{HA} =- overset->{HC}`
`text{Similarly for diagonal} \ overset->{BD}`
`overset->{HB} = – overset->{HD}`
`:. \ overset->{HA} + overset->{HB} + overset->{HC} + overset->{HD}` | `= – overset->{HC} – overset->{HD} + overset->{HC} + overset->{HD}` |
`= underset~0` |
`4 overset->{GH} + overset->{HS} + overset->{GH}` | `= underset~0` |
`5 overset->{GH}` | `= overset->{SH}` |
`overset->{GH}` | `= 1/5 overset->{SH}` |
`overset->{HG}` | `= 1/5 overset->{HS}` |
`:. \ λ = 1/5`
Four cubes are placed in a line as shown on the diagram.
Which of the following vectors is equal to `overset->{AB} + overset->{CQ}`
`B`
`overset->{AB} \ + \ overset->{CQ}` | `= overset->{CD} + overset->{DP}` | |
`= overset->{CP}` |
`=>\ B`
The base of a pyramid is the parallelogram `ABCD` with vertices at points `A(2,−1,3), B(4,−2,1), C(a,b,c)` and `D(4,3,−1)`. The apex (top) of the pyramid is located at `P(4,−4,9)`.
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i. | `overset(->)(AB)` | `= (4 – 2)underset~i + (−2 + 1)underset~j + (1 – 3)underset~k` |
`= 2underset~i – underset~j – 2underset~k` |
`text(S)text(ince)\ ABCD\ text(is a parallelogram)\ => \ overset(->)(AB)= overset(->)(DC)`
`overset(->)(DC) = (a – 4)underset~i + (b – 3)underset~j + (c + 1)underset~k`
`a – 4 = 2 \ => \ a = 6`
`b – 3 = −1 \ => \ b = 2`
`c + 1 = −2 \ => \ c = −3`
ii. `overset(->)(AB) = 2underset~i – underset~j – 2underset~k`
`overset(->)(AD) = 2underset~i + 4underset~j – 4underset~k`
`cos angleBAD` | `= (overset(->)(AB) · overset(->)(AD))/(|overset(->)(AB)| · |overset(->)(AD)|)` |
`= (4 – 4 + 8)/(sqrt(4 + 1 + 4) · sqrt(4 + 16 + 16))` | |
`= 4/9` |
iii. | `1/2 xx text(Area)_(ABCD)` | `= 1/2 ab sin c` |
`text(Area)_(ABCD)` | `= |overset(->)(AB)| · |overset(->)(AD)| *sin(cos^(−1)\ 4/9)` |
`:. text(Area)_(ABCD)` | `= 3 · 6 · sqrt65/9` |
`= 2sqrt65\ text(u²)` |
Let `OABCD` be a right square pyramid where `underset ~a = vec(OA),\ underset ~b = vec(OB),\ underset ~c = vec(OC)` and `underset ~d = vec(OD)`.
Show that `underset~a + underset~c = underset~b + underset~d`. (3 marks)
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`text(See Worked Solution)`
`text(Let)\ \ A=(p,–p,–k),`
`underset ~a` | `= overset(->)(OA) = punderset~i – punderset~j – qunderset~k` |
`underset ~b` | `= overset(->)(OB) = punderset~i + punderset~j – qunderset~k` |
`underset ~c` | `= overset(->)(OC) = −punderset~i + punderset~j – qunderset~k` |
`underset ~d` | `= overset(->)(OA) = −punderset~i – punderset~j – qunderset~k` |
`underset~a + underset~c` | `= −2qunderset~k` |
`underset ~b + underset ~d` | `= −2qunderset~k` |
`:. underset~a + underset~c = underset~b + underset~d`
A cube with side length 3 units is pictured below.
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i. `A(3, 0 , 0), \ \ G(0, 3, 3)`
`vec(AG)` | `= ((0), (3), (3)) – ((3), (0), (0)) = ((text{−3}), (3), (3))` | |
`|\ vec(AG)\ |` | `= sqrt (9 + 9 + 9)` | |
`= 3 sqrt 3\ text(units)` |
ii. | `H (3, 3, 3)` |
`vec(BH) = ((3), (3), (3))` |
`vec(AG) ⋅ vec(BH)` | `= |\ vec(AG)\ | ⋅ |\ vec(BH)\ |\ cos theta` |
`((text{−3}), (3), (3)) ⋅ ((3), (3), (3))` | `= sqrt (9 + 9 + 9) ⋅ sqrt (9 + 9 + 9) cos theta` |
`-9 + 9 + 9` | `= 27 cos theta` |
`cos theta` | `= 1/3` |
`theta` | `= 70.52…` |
`= 70^@32′` |
`ABCDEFGH` are the vertices of a rectangular prism.
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i. | `A(2, text{−2}, 0),` | `G(text{−2}, 2, 2)` |
`D(2, 2, 0),` | `F (text{−2}, text{−2}, 2)` |
`text(Midpoint)\ AG = ((1/2 (2 – 2)),(1/2 (text{−2} + 2)),(1/2 (0 + 2))) = ((0), (0), (1))`
`text(Midpoint)\ DF = ((1/2 (2 – 2)),(1/2 (2 – 2)),(1/2 (0 + 2))) = ((0), (0), (1))`
`text(S) text(ince midpoints are the same), AG and DF\ text(intersect.)`
ii. | `vec(AG) = ((text{−2}), (2), (2)) – ((2), (text{−2}), (0)) = ((text{−4}), (4), (2))` |
`vec(DF) = ((text{−2}), (text{−2}), (2)) – ((2), (2), (0)) = ((text{−4}), (text{−4}), (2))` |
`vec (AG) ⋅ vec (DF) = |\ vec (AG)\ | ⋅ |\ vec(DF)\ |\ cos theta`
`((text{−4}), (4), (2)) ⋅ ((text{−4}), (text{−4}), (2)) = sqrt 36 sqrt 36 cos theta`
`16 – 16 + 4` | `= 36 cos theta` |
`cos theta` | `= 1/9` |
`theta` | `= 83.62…` |
`= 83^@37′` |