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Vectors, EXT2 V1 2025 HSC 1 MC

Points \(A\) and \(B\) are \((-3,1)\) and \((1,4)\) respectively.

Which of the following is a vector equation of the line \(A B\) with parameter \(\lambda\) ?

  1. \(\displaystyle \binom{x}{y}=\binom{1}{4}+\lambda\binom{3}{4}\)
  2. \(\displaystyle\binom{x}{y}=\binom{3}{4}+\lambda\binom{1}{4}\)
  3. \(\displaystyle\binom{x}{y}=\binom{4}{3}+\lambda\binom{-3}{1}\)
  4. \(\displaystyle\binom{x}{y}=\binom{-3}{1}+\lambda\binom{4}{3}\)
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\(D\)

Show Worked Solution

\(\overrightarrow{AB}=\displaystyle \binom{1}{4}-\binom{-3}{1}=\binom{4}{3}\)

\(\text{Line} \ \ AB:\)

\(\displaystyle \binom{x}{y}=\binom{-3}{1}+\lambda\binom{4}{3}\)

\(\Rightarrow D\)

Vectors, EXT2 V1 2024 HSC 12e

The line \(\ell\) passes through the points \(A(3,5,-4)\) and \(B(7,0,2)\).

  1. Find a vector equation of the line \(\ell\).   (1 mark)

  2. Determine, giving reasons, whether the point \(C(10,5,-2)\) lies on the line \(\ell\).   (2 marks)

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i.    \(\text{ Equation of line:} \ \left(\begin{array}{c}3 \\ 5 \\ -4\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right),\ (\lambda \in \mathbb{R} )\)

ii.    \(\text{If \(C\) lies on the line}, \ \exists \lambda \in \mathbb{R}\  \ \text{such that:}\)

\(\left(\begin{array}{c}10 \\ 5 \\ -2\end{array}\right)=\left(\begin{array}{c}3 \\ -5 \\ 6\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right)\)

 
\(\text{Find \(\lambda\) for \(x\)-component: }\)

\(10=3+4 \lambda \ \Rightarrow \ \lambda=\dfrac{7}{4}\)

\(\text{Check \(\lambda=\dfrac{7}{4}\) for \(y\)-component:}\)

\(5=-5+\dfrac{7}{4}(-5) \ \ \Rightarrow\ \ \text{not correct}\)

\(\therefore C\  \text{does not lie on the line.}\)

Show Worked Solution

i.     \(A(3,5,-4), \quad B(7,0,2)\)

\(\overrightarrow{A B}=\left(\begin{array}{l}7 \\ 0 \\ 2\end{array}\right)-\left(\begin{array}{c}3 \\ 5 \\ -4\end{array}\right)=\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right)\)

\(\therefore \text{ Equation of line:} \ \left(\begin{array}{c}3 \\ 5 \\ -4\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right),\ (\lambda \in \mathbb{R} )\)
 

ii.    \(\text{If \(C\) lies on the line}, \ \exists \lambda \in \mathbb{R}\  \ \text{such that:}\)

\(\left(\begin{array}{c}10 \\ 5 \\ -2\end{array}\right)=\left(\begin{array}{c}3 \\ -5 \\ 6\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right)\)

 
\(\text{Find \(\lambda\) for \(x\)-component: }\)

\(10=3+4 \lambda \ \Rightarrow \ \lambda=\dfrac{7}{4}\)

\(\text{Check \(\lambda=\dfrac{7}{4}\) for \(y\)-component:}\)

\(5=-5+\dfrac{7}{4}(-5) \ \ \Rightarrow\ \ \text{not correct}\)

\(\therefore C\  \text{does not lie on the line.}\)

Vectors, EXT2 V1 2023 HSC 11c

Find a vector equation of the line through the points  \(A(-3,1,5)\)  and  \(B(0,2,3)\).  (2 marks)

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\[\underset{\sim}{v}=\left(\begin{array}{c} -3 \\1 \\ 5 \end{array}\right) + \lambda \left(\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right),\ \ \ \text{for some}\ \ \lambda \in \mathbb{R} \]

Show Worked Solution

\[\overrightarrow{AB}=\left(\begin{array}{c} 0-(-3) \\2-1 \\ 3-5 \end{array}\right) = \left(\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right)\]

\[\therefore \underset{\sim}{v}=\left(\begin{array}{c} -3 \\1 \\ 5 \end{array}\right) + \lambda \left(\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right),\ \ \ \text{for some}\ \ \lambda \in \mathbb{R} \]

Vectors, EXT2 V1 EQ-Bank 8

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)}`

Show Worked Solution

`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°)`
 

`text{Consider scalar product of direction vectors:}`

`vec(AB)=((3),(1),(0))+lambda((-2-3),(4-1),(3-0))=((3),(1),(0))+lambda((-5),(3),(3))`
 

`vec(AC)=((3),(1),(0))+mu((3-3),(3-1),(-2-0))=((3),(1),(0))+mu((0),(2),(-2))`
 

`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.}`

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))`
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`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 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 2

  1. Find values of  `a`, `b`, `c`  and  `d`  such that  `underset~v = ((a),(b)) + 2((c),(d))`  is a vector equation of a line that passes through  `((3),(1))`  and  `((−3),(−3))`.  (2 marks)

  2. Determine whether  `underset~u = ((4),(6)) + lambda((−2),(3))`  is perpendicular to  `underset~v`.  (1 mark)

  3. Express  `underset~u`  in Cartessian form.  (1 mark)

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  1. `a = 3, b = 1, c = −3, d = −2`, or

     

    `a = −3, b = −3, c = 3, d = 2`

  2. `text(See worked solutions)`
  3. `y = −3/2x + 12`
Show Worked Solution

i.   `text(Method 1)`

`overset(->)(OA) = underset~a = ((3),(1)),\ \ overset(->)(OB) = underset~b = ((−3),(−3))`

`overset(->)(AB)` `= overset(->)(OB) – overset(->)(OA)`
  `= ((−3),(−3)) – ((3),(1))`
  `= ((−6),(−4))`

 

`underset~v` `= underset~a + lambdaunderset~b`
  `= ((3),(1)) + lambda((−6),(−4))`
  `= ((3),(1)) + 2((−3),(−2))`

 
`:. a = 3, b = 1, c = −3, d = −2`

 

`text(Method 2)`

`overset(->)(BA)` `= overset(->)(OA) – overset(->)(OB)`
  `= ((3),(1)) – ((−3),(−3))`
  `= ((6),(4))`

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

`:. a = −3, b = −3, c = 3, d = 2`
 

ii.   `underset~u = ((4),(6)) + lambda((−2),(3))`

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

`((−2),(3)) · ((−3),(−2)) = 6 – 6 = 0`

`:. underset~u ⊥ underset~v`
 

iii.   `((x),(y))= ((4),(6)) + lambda((−2),(3))`

`x = 4 – 2lambda\ \ \ …\ (1)`

`y = 6 + 3lambda\ \ \ …\ (2)`

`text(Substitute)\ \ lambda = (4 – x)/2\ \ text{from (1) into (2):}`

`y` `= 6 + 3((4 – x)/2)`
`y` `= 6 + 6 – (3x)/2`
`y` `= −3/2x + 12`

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.)`