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Vectors, EXT2 V1 2025 HSC 16c

Consider the point \(B\) with three-dimensional position vector \(\underset{\sim}{b}\) and the line  \(\ell: \underset{\sim}{a}+\lambda \underset{\sim}{d}\), where \(\underset{\sim}{a}\) and \(\underset{\sim}{d}\) are three-dimensional vectors, \(\abs{\underset{\sim}{d}}=1\) and \(\lambda\) is a parameter.

Let \(f(\lambda)\) be the distance between a point on the line \(\ell\) and the point \(B\).

  1. Find \(\lambda_0\), the value of \(\lambda\) that minimises \(f\), in terms of \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{d}\).   (2 marks)

  2. Let \(P\) be the point with position vector  \(\underset{\sim}{a}+\lambda_0 \underset{\sim}{d}\).
  3. Show that \(PB\) is perpendicular to the direction of the line \(\ell\).   (1 mark)

  4. Hence, or otherwise, find the shortest distance between the line \(\ell\) and the sphere of radius 1 unit, centred at the origin \(O\), in terms of \(\underset{\sim}{d}\) and \(\underset{\sim}{a}\).
  5. You may assume that if \(B\) is the point on the sphere closest to \(\ell\), then \(O B P\) is a straight line.   (3 marks)

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i.    \(\lambda_0=\underset{\sim}{d}(\underset{\sim}{b}-\underset{\sim}{a})\)

ii.   \(\text{See Worked Solutions.}\)

iii.  \(d_{\min }= \begin{cases}\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}-1, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}>1 \\ 0, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2} \leqslant 1 \ \ \text{(i.e. it touches sphere) }\end{cases}\)

Show Worked Solution

i.    \(\ell=\underset{\sim}{a}+\lambda \underset{\sim}{d}, \quad\abs{\underset{\sim}{d}}=1\)

\(\text{Vector from point \(B\) to a point on \(\ell\)}:\ \underset{\sim}{a}+\lambda \underset{\sim}{d}-\underset{\sim}{b}\)

\(f(\lambda)=\text{distance between \(\ell\) and \(B\)}\)

\(f(\lambda)=\abs{\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d}}\)

\(\text{At} \ \ \lambda_0, f(\lambda) \ \ \text{is a min}\ \Rightarrow \ f(\lambda)^2 \ \ \text {is also a min}\)

\(f(\lambda)^2\) \(=\abs{\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d}}^2\)
  \(=(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})\)
  \(=(\underset{\sim}{a}-\underset{\sim}{b})\cdot (\underset{\sim}{a}-\underset{\sim}{b})+2\lambda (\underset{\sim}{a}-\underset{\sim}{b}) \cdot \underset{\sim}{d}+\lambda^2 \underset{\sim}{d} \cdot  \underset{\sim}{d}\)
  \(=\lambda^2|\underset{\sim}{d}|^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda +\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2\)
  \(=\lambda^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda+\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2\)

 

\(f(\lambda)^2 \ \ \text{is a concave up quadratic.}\)

\(f(\lambda)_{\text {min}}^2 \ \ \text{occurs at the vertex.}\)

\(\lambda_0=-\dfrac{b}{2 a}=-\dfrac{2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b})}{2}=\underset{\sim}{d} \cdot (\underset{\sim}{b}-\underset{\sim}{a})\)
 

ii.    \(P \ \text{has position vector} \ \ \underset{\sim}{a}+\lambda_0 \underset{\sim}{d}\)

\(\text{Show} \ \ \overrightarrow{PB} \perp \ell:\)

\(\overrightarrow{PB}=\underset{\sim}{b}-\underset{\sim}{p}=\underset{\sim}{b}-\underset{\sim}{a}-\lambda_0 \underset{\sim}{d}\)

\(\overrightarrow{P B} \cdot \underset{\sim}{d}\) \(=\left(\underset{\sim}{b}-\underset{\sim}{a}-\lambda_0 \underset{\sim}{d}\right) \cdot \underset{\sim}{d}\)
  \(=(\underset{\sim}{b}-\underset{\sim}{a}) \cdot \underset{\sim}{d}-\lambda_0 \underset{\sim}{d} \cdot \underset{\sim}{d}\)
  \(=\lambda_0-\lambda_0\abs{\underset{\sim}{d}}^2\)
  \(=0\)

  

\(\therefore \overrightarrow{PB}\ \text{is perpendicular to the direction of the line}\ \ell. \)
 

iii.   \(\text{Shortest distance between} \ \ell \ \text{and sphere (radius\(=1\))}\)

\(=\ \text{(shortest distance \(\ell\) to \(O\))}-1\)
 

\(f\left(\lambda_0\right)=\text{shortest distance \(\ell\) to point \(B\)}\)

\(\text{Set} \ \ \underset{\sim}{b}=0 \ \Rightarrow \ f\left(\lambda_0\right)=\text{shortest distance \(\ell\) to \(0\)}\)

\(\Rightarrow \lambda_0=\underset{\sim}{d} \cdot (\underset{\sim}{b}-\underset{\sim}{a})=-\underset{\sim}{d} \cdot \underset{\sim}{a}\)

\(f\left(\lambda_0\right)\) \(=\abs{\underset{\sim}{a}-\underset{\sim}{b}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}=\abs{\underset{\sim}{a}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}\)
\(f\left(\lambda_0\right)^2\) \(=\abs{\underset{\sim}{a}}^2-2( \underset{\sim}{a}\cdot \underset{\sim}{d})^2+(\underset{\sim}{d} \cdot \underset{\sim}{a})^2\abs{\underset{\sim}{d}}^2\)
  \(=\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2\)
\(f\left(\lambda_0\right)\) \(=\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}\)

 

\(\text {Shortest distance of \(\ell\) to sphere \(\left(d_{\min }\right)\):}\)

\(d_{\min }= \begin{cases}\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}-1, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}>1 \\ 0, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2} \leqslant 1 \ \ \text{(i.e. it touches sphere) }\end{cases}\)

Vectors, EXT2 V1 2022 HSC 14a

  1. The two non-parallel vectors `\vec{u}` and `\vec{v}` satisfy  `lambda vec(u)+mu vec(v)= vec(0)`  for some real numbers `\lambda` and `\mu`.
  2. Show that  `\lambda=\mu=0`.  (2 marks)

  3. The two non-parallel vectors `\vec{u}` and `\vec{v}` satisfy  `\lambda_1 \vec{u}+\mu_1 \vec{v}=\lambda_2 \vec{u}+\mu_2 \vec{v}`  for some real numbers `\lambda_1, \lambda_2, \mu_1` and `\mu_2`.
  4. Using part (i), or otherwise, show that  `\lambda_1=\lambda_2`  and  `\mu_1=\mu_2`. (1  mark)

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`.
 
                       
 

  1. Using part (ii), or otherwise, determine the position of `L` by showing that  `vec(BL)=(4)/(7) vec(BC)`.  (2 marks)

  2. The point `P` is defined by  `vec(AP)=-6 vec(AB)-8 vec(AC)`.
  3. Does `P` lie on the line `A L`? Justify your answer.  (2 marks)

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  1. `text{Proof (See Worked Solution)}`
  2. `text{Proof (See Worked Solution)}`
  3. `text{Proof (See Worked Solution)}`
  4. `P\ text{lies on}\ vec(AL)\ \ text{(See Worked Solution)}`
Show Worked Solution
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}`


Mean mark (ii) 56%.
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)`  


♦♦♦ Mean mark (iii) 25%.

`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)`


♦♦♦ Mean mark (iv) 23%.
`-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.`

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`

Show Worked Solution

`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 2021 HSC 7 MC

Which diagram best shows the curve described by the position vector

`underset~r (t) = -5 text{cos}(t) underset~i + 5 text{sin}(t) underset~j + t underset~k`  for  `0 ≤ t ≤ 4 pi` ?

 

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`D`

Show Worked Solution

`text{By elimination}`

`text{Check graph coordinates for specific values of}\ t:`

`text{When} \ \ t = 4 pi \ , \ underset~r = -5 underset~i + 0 underset~j + 4 pi underset~k`

`-> \ text{Eliminate A and B}`

`text{When} \ t = pi/4 \ , \ underset~r = (-5)/sqrt(2) underset~i + 5/sqrt(2) underset~j + pi/4 underset~k`

`-> \ text{Eliminate C}`
 

`=>\ D`

Vectors, EXT2 V1 2021 HSC 3 MC

Which of the following is a vector equation of the line joining the points  `A (4, 2, 5)`  and  `B (–2, 2, 1)`?

  1. `underset~r = ((4), (2), (5)) + λ ((1),(2),(3))`
  2. `underset~r = ((4), (2), (5)) + λ ((3),(0),(2))`
  3. `underset~r = ((1), (2), (3)) + λ ((4),(2),(5))`
  4. `underset~r = ((3), (0), (2)) + λ ((4),(2),(5))`
Show Answers Only

`B`

Show Worked Solution
`overset->{AB}` `= ((-2),(2),(1)) – ((4),(2),(5)) = ((-6),(0),(-4))`  
`underset~r` `= ((4),(2),(5)) + λ_1 ((-6),(0),(-4))`  
  `= ((4), (2), (5)) + λ_2 ((3),(0),(2))`  

 
`=>\ B`

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))`

Show Worked Solution

`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 24

Show that the points  `A(2, 1, text{−1}), \ B(4, 2, text{−3})`  and  `C(text{−4}, text{−2}, 5)`  are collinear.  (2 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`vec(AB) = ((4), (2), (text{−3})) – ((2), (1), (text{−1})) = ((2), (1), (text{−2}))`

`vec(AC) = ((text{−4}), (text{−2}), (5)) – ((2), (1), (text{−1})) = ((text{−6}), (text{−3}), (6)) = -3((2), (1), (text{−2}))`

 

`text(S) text(ince)\ vec(AB)\ text(||)\ vec(AC) and A\ text(lies on both lines,)`

`A, B, C\ \ text(are collinear).`

Vectors, EXT2 V1 SM-Bank 21

  1. Find the equation of the vector line  `underset~v`  that passes through  `Atext{(5, 2, 3)}`  and  `B(7, 6, 1)`.  (1 mark)

  2. A sphere has centre  `underset~c`  at  `text{(2, 3, 5)}`  and a radius of  `5sqrt2`  units.
    Find the points where the vector line  `underset~v`  meets the sphere.  (3 marks)

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  1. `underset~v = ((5),(2),(3)) + lambda((1),(2),(−1))`
  2. `((2),(–4),(6)), \ ((7),(6),(1))`
Show Worked Solution

i.   `overset(->)(BA) = ((7-5),(6-2),(1-3)) = ((2),(4),(−2)) = 2((1),(2),(−1))`

`underset~v = ((5),(2),(3)) + lambda((1),(2),(−1))`
 

ii.   `text(General point)\ underset~v:`

`x = 5 + lambda`

`y = 2 + 2lambda`

`z = 3-lambda`
 

`text(Equation of sphere,)\ underset~c = (2, 3, 5),\ text(radius)\ 5sqrt2:`

`(x-2)^2 + (y-3)^2 + (z -5)^2` `= (5sqrt2)^2`
`(lambda + 3)^2 + (2lambda-1)^2 + (−lambda-2)^2` `= 50`
`lambda^2 + 6lambda + 9 + 4lambda^2-4lambda + 1 + lambda^2 + 4lambda + 4` `= 50`
`6lambda^2 + 6lambda + 14` `= 50`
`6lambda^2 + 6lambda-36` `= 0`
`6(lambda + 3)(lambda-2)` `= 0`
`lambda` `= –3\ text(or)\ 2`

 
`text(When)\ \ lambda = –3,`

`text(Intersection) = ((5),(2),(3))-3((1),(2),(−1)) = ((2),(–4),(6))`

`text(When)\ \ lambda = 2,`

`text(Intersection) = ((5),(2),(3)) + 2((1),(2),(−1)) = ((7),(6),(1))`

Vectors, EXT2 V1 SM-Bank 18

A sphere is represented by the equation

`x^2-4x + y^2 + 8y + z^2-3z + 2 = 0`

  1. Determine the centre  `underset~c`  and radius of the sphere.  (2 marks)

  2. Find the vector equation of the sphere.  (1 mark)

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  1.  `underset~c = ((2),(-4),({3}/{2})) \ , \ text(radius) = (9)/(2)`
  2. `| \ underset~r-((2),(-4),({3}/{2})) | = (9)/(2)`
Show Worked Solution

i.  `x^2-4x + y^2 + 8y + z^2-3z + 2 = 0`

`(x-2)^2 + (y+4)^2 + (z-{3}/{2})^2 + 2-(89)/(4) = 0`

`(x-2)^2 + (y+4)^2 + (z-{3}/{2})^2 = (81)/(4)`
 
`:. \ underset~c = ((2),(-4),({3}/{2})) \ , \ text(radius) = (9)/(2)`
 

ii. `text(Vector equation:)`

`| \ underset~r-((2),(-4),({3}/{2})) | = (9)/(2)`

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))`
Show Worked Solution

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, EXT2 V1 SM-Bank 9

  1. Find the equation of line vector  `underset ~r`, given it passes through  `(1, 3, –2)`  and  `(2, –1, 2)`.   (2 marks)

  2. Determine if  `underset ~r`  passes through  `(4, –9, 10)`.   (1 mark)

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  1. `underset ~r = ((1), (3), (-2)) + lambda ((1), (-4), (4)) or`

       
      

    `underset ~r = ((2), (-1), (2)) + lambda ((-1), (4), (-4))`

  2. `text(See Worked Solutions)`
Show Worked Solution

i.     `text(Method 1)`

`text(Let)\ \ A(1, 3, –2) and B(2, –1, 2)`

`vec (AB)` `= ((2), (-1), (2))-((1), (3), (-2)) = ((1), (-4), (4))`
`underset ~r` `= ((1), (3), (-2)) + lambda ((1), (-4), (4))`

 

`text (Method 2)`

`vec (BA)` `= ((1), (3), (-2))-((2), (-1), (2)) = ((-1), (4), (-4))`
`underset ~r` `= ((2), (-1), (2)) + lambda ((-1), (4), (-4))`

 

ii.   `text(If)\ \ (4, –9, 10)\ \ text(lies on the vector line,)`

`∃ lambda\ \ text(that satisfies:)`

`1 + lambda` `= 4\ \ text{… (1)}`
`3-4 lambda` `= -9\ \ text{… (2)}`
`-2 + 4 lambda` `= 10\ \ text{… (3)}`

 

`lambda = 3\ \ text(satisfies all equations)`

`:. (4, –9, 10)\ \ text(lies on the line.)`

Vectors, EXT2 V1 SM-Bank 7

Find the value of `x` and `y`, given

`underset ~r = ((5), (-1), (2)) + lambda ((x), (y), (-3))`

and  `underset ~r`  is perpendicular to both `underset ~v` and `underset ~w`, where

`underset ~v = ((1), (2), (1)) + mu_1 ((3), (-3), (-1))`  and  `underset ~w = ((-3), (1), (1)) + mu_2 ((-4), (-1), (-2))`.  (2 marks)

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`x = 1, y = 2`

Show Worked Solution

`text(S) text(ince)\ underset ~r\ text(is perpendicular to)\ underset ~v and underset ~w:`
 

`((x), (y), (-3))((3), (-3), (-1)) = 0`

`3x – 3y + 3` `= 0`
`x – y` `= -1\ text{… (1)}`

 
`((x), (y), (-3))((-4), (-1), (-2)) = 0`

`-4x – y + 6` `= 0`
`4x + y` `= 6\ text{… (2)}`

 
`(1) + (2)`

`5x` `= 5`
`:.x` `= 1`
`:. y` `= 2`