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Vectors, EXT2 V1 2025 SPEC1 2

Consider the following two lines, \(L_1\) and \(L_2\).

\(L_1\) passes through the point \(A_1(2,3,1)\) and has direction  \(\underset{\sim}{u}=\underset{\sim}{i}+2 \underset{\sim}{j}-\underset{\sim}{k}\).

\(L_2\) passes through the point \(A_2(1,3,2)\) and has direction  \(\underset{\sim}{v}=-\underset{\sim}{i}-\underset{\sim}{j}+\underset{\sim}{k}\).

Find the coordinates of the point of intersection of the two lines.   (3 marks)

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\(\text{Intersection at}\ (3,5,0).\)

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\(\underset{\sim}{u}=\underset{\sim}{i}+2 \underset{\sim}{j}-\underset{\sim}{k} \ \ \text {passes through} \ A_1(2,3,1)\)

\(L_1:(2+t) \underset{\sim}{i}+(3+2 t) \underset{\sim}{j}+(1-t) \underset{\sim}{k}\)

\(\underset{\sim}{v}=\underset{\sim}{-i}-\underset{\sim}{j}+\underset{\sim}{k} \ \ \text {passes through}\  A_2(1,3,2)\)

\(L_2:(1-s) \underset{\sim}{i}+(3-s)\underset{\sim}{j}+(2+s) \underset{\sim}{k}\)

\(\text{Equating components for intersection:}\)

\(2+t=1-s \ \ \Rightarrow \ \ s+t=-1\ \ldots\ (1)\)

\(3+2 t=3-s \ \ \Rightarrow \ \ s+2 t=0\ \ldots\ (2)\)

\(\text{Subtract: (2) – (1)}\)

\(t=1 \ \Rightarrow \ s=-2\)

\(\text{Point of intersection:}\)

\((2+1,3+2,1-1) \equiv(3,5,0)\)

Vectors, EXT2 V1 2021 HSC 12c

Two lines are given by  `text(r)_1 = ((-2),(1),(3)) + lambda((1),(0),(2))`  and  `text(r)_2 = ((4),(-2),(q)) + mu ((p),(3),(-1))` , where `p` and `q` are real numbers. These lines intersect and are perpendicular.

Find the values of `p` and `q`.  (3 marks)

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`p = 2 \ , \ q = 20`

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`text{S} text{ince vectors are perpendicular}`

`((1),(0),(2)) ((p),(3),(-1))` `= 0`
`p – 2` `= 0`
`p` `= 2`
 

`text{S} text{ince lines intersect, equate}\ y text{-coordinates:} `

`1 + lambda 0` `= -2 + 3 mu`
`mu` `= 1`

 

`text{Find} \ lambda \ text{by equating}\ xtext{-coordinates:}`

`-2 + lambda` `= 4 + 1 xx 2`
`lambda` `= 8`

 
`text{Equating}\ ztext{-coordinates:}`

`3 + 8 xx 2` `= q – 1 xx 1`
`q` `= 20`

Vectors, EXT2 V1 2020 HSC 13b

Consider the two lines in three dimensions given by
 

`underset~r = ((3),(-1),(7)) + λ_1 ((1),(2),(1))`  and  `underset~r = ((3),(-6),(2)) + λ_2 ((-2),(1),(3))`.
 

By equating components, find the point of intersection of the two lines.   (3 marks)

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`((1),(-5),(5))`

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`underset~(r_1) = ((3),(-1),(7)) + λ_1 ((1),(2),(1)) = ((3 + λ_1),(-1 + 2λ_1),(7 + λ_1))`

`underset~(r_2) = ((3),(-6),(2)) + λ_2 ((-2),(1),(3)) = ((3 – 2λ_2),(-6 + λ_2),(2 + 3λ_2))`

 
`text{Intersection occurs when:}`

`3 + λ_1 ` `= 3 – 2λ_2 \ … \ (1)`
`-1 + 2λ_1` `= -6 + λ_2 \ … \ (2)`
`7 + λ_1` `= 2 + 3λ_2 \ … \ (3)`

 
`text{Subtract} \ (3) – (1):`

`4` `= -1 + 5 λ_2`
`λ_2` `=1`

 
`text{Substitute} \ \ λ_2 = 1\ \ text{into} \ (1):`

`3 + λ_1` `= 1`
`λ_1` `= -2`

 

`text{Test that}\  \ λ_1 = -2 \ , \  λ_2 = 1\ \ text{satisfies} \ (2):`

`-1 + 2 xx  – 2` `= -6 + 1`
`-5` `= -5`

 
`∃ \ λ_1,  λ_2, \ text{that satisfy all equations}`

`=> \ text{3 lines intersect at a point}`

`:.\ text{Point of intersection}`

`= ((3),(-6),(2)) + 1 ((-2),(1),(3)) = ((1),(-5),(5))`

Vectors, EXT2 V1 SM-Bank 15

Consider the two vector line equations

`underset~(v_1) = ((1),(4),(−2)) + lambda_1((3),(0),(−1)), qquad underset~(v_2) = ((3),(2),(2)) + lambda_2((4),(2),(−6))`

  1. Show that  `underset~(v_1)`  and  `underset~(v_2)`  intersect and determine the point of intersection . (2 marks)

  2. What is the acute angle between the vector lines, to the nearest minute.  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `40°29’\ \ (text(nearest minute))`
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i.   `text(Solve simultaneously:)`

`1 + 3lambda_1` `= 3 + 4lambda_2` `\ \ …\ (1)`
`4 + 0lambda_1` `= 2 + 2lambda_2` `\ \ …\ (2)`
`−2 – lambda_1` `= 2 – 6lambda_2` `\ \ …\ (3)`

 
`=> lambda_2 = 1\ \ \ text{(from (2))}`

`=>lambda_1 = 2\ \ \ text{(from (1) and (3))}`

`:.\ text(vector lines intersect)`
  

`text(P.O.I.) = ((1),(4),(−2)) + 2((3),(0), (−1)) = ((7),(4),(−4))`

 

ii.   `underset~(v_1) = underset~(a_1) + lambda_1*underset~(b_1)`

`underset~(v_2) = underset~(a_2) + lambda_2*underset~(b_2)`

`costheta` `= (underset~(b_1) · underset~(b_2))/(|underset~b_1||underset~b_2|)`
  `= (12 + 0 + 6)/(sqrt10 sqrt56)`
  `= 0.7606…`

 

`theta` `= 40.479…`
  `= 40°29’\ \ (text(nearest minute))`

Vectors, EXT1 V1 SM-Bank 11 MC

Which pair of line segments intersect at exactly one point

A.    `{(underset ~u = ((3), (2)) + lambda ((text{−1}),(2)) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((2), (1)) + lambda ((2), (text{−4})) text{,} quad qquad 0 <= lambda <= 1):}`
   
B.    `{(underset ~u = ((4), (1)) + lambda ((3), (text{−1})) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((3), (2)) + lambda ((2), (2)) text{,} quad qquad 0 <= lambda <= 1):}`
   
C.    `{(underset ~u = ((4), (0)) + lambda ((text{−3}),(6)) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((0), (1)) + lambda ((1), (text{−2})) text{,} quad qquad 0 <= lambda <= 1):}`
   
D.    `{(underset ~u = ((0), (2)) + lambda ((3), (text{−2})) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((0), (1)) + lambda ((1), (1)) text{,} quad qquad 0 <= lambda <= 1):}`
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`D`

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`text(S) text(ince)\ ((2), (text{−4})) = -2((text{−1}), (2)) and ((text{−3}), (6)) = -3((text{−1}), (2))`

`=> A and C\ text(are parallel lines.)`
 

`text(Consider)\ D:`

`3 lambda_1` `= lambda_2` `\ text{… (1)}`
`2 – 2 lambda_1` `= 1 + lambda_2` `\ text{… (2)}`

 
`text(Substitute)\ text{(1) into (2)}`

`2 – 2 lambda_1` `= 1 + 3 lambda_1`
`lambda_1` `= 1/5`
`lambda_2` `= 3/5`

 
`text(Similarly,)\ lambda_1, lambda_2\ text(in)\ B\ \ text(can be calculated)`

`text(and found to be outside)\ \ 0 <= lambda <= 1.`

`=> D`

Vectors, EXT2 V1 SM-Bank 10

  1. Determine the point of intersection of  `underset ~a`  and  `underset ~b`  given.

`qquad underset ~a = ((3), (5), (1)) + lambda ((1), (3), (text{−2})),`  and
 

`qquad underset ~b = ((text{−2}), (2), (text{−1})) + mu ((1), (text{−1}), (2))`  (2 marks)

 
  1. Determine if the point  `(2, text{−2}, 5)`  lies on  `underset ~b`.  (1 mark)

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  1. `((1), (text{−1}), (5))`
  2. `text(See Worked Solutions)`
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i.     `text(At point of intersection:)`

`3 + lambda` `= -2 + mu\ \ text{… (1)}`
`5 + 3 lambda` `= 2 – mu\ \ text{… (2)}`
`1 – 2 lambda` `= -1 + 2 mu\ \ text{… (3)}`

 
`(1) + (2)`

`8 + 4 lambda` `= 0`
`lambda` `= -2,\ \ mu = 3`

 
`text{Intersection (using}\ lambda = –2 text{)}:`
 

`((x), (y), (z)) = ((3 – 2 xx 1), (5 – 2 xx 3), (1 – 2 xx text{−2})) = ((1), (text{−1}), (5))`

 

ii.   `text(If)\ \ (2, text{−2}, text{−10})\ \ text(lies on)\ underset ~b, ∃ mu\ \ text(that satisfies:)`

`-2 + mu` `= 2\ \ text{… (1)}\ => \ mu = 4`
`2 – mu` `= ­text{−2}\ \ text{… (2)}\ => \ mu = 4`
`-1 + 2 mu` `= 5\ \ text{… (3)}\ => \ mu = 3`

 
`=>\  text(No solution)`

`:. (2, text{−2}, 5)\ \ text(does not lie on)\ underset ~b.`