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Vectors, EXT2 V1 2024 HSC 13a

The point \(A\) has position vector  \(8 \underset{\sim}{i}-6 \underset{\sim}{j}+5 \underset{\sim}{k}\). The line \(\ell\) has vector equation

\(x \underset{\sim}{i}+y \underset{\sim}{j}+z \underset{\sim}{k}=t(\underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k})\).

The point \(B\) lies on \(\ell\) and has position vector  \(p \underset{\sim}{i}+p \underset{\sim}{j}+2 p \underset{\sim}{k}\).

  1. Show that  \(\abs{A B}^2=6 p^2-24 p+125\).   (1 mark)

  2. Hence, or otherwise, determine the shortest distance between the point \(A\) and the line \(\ell\).  (2 marks)

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i.     \(A\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right), \quad \ell: \left(\begin{array}{l}x \\ y \\ z\end{array}\right)+t\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right), \quad B\left(\begin{array}{c}p \\ p \\ 2 p\end{array}\right)\)
 

\(\overrightarrow{A B}=\left(\begin{array}{l}p \\ p \\ 2 p\end{array}\right)-\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right)=\left(\begin{array}{c}p-8 \\ p+6 \\ 2 p-5\end{array}\right)\)
 

  \(\abs{AB}^2\) \(=(p-8)^2+(p+6)^2+(2 p-5)^2\)
    \(=p^2-16 p+64+p^2+12 p+36+4 p^2-20 p+25\)
    \(=6 p^2-24 p+125\)

 

ii.    \(\abs{AB}_{\text {min}}=\sqrt{101} \text { units}\)

Show Worked Solution

i.     \(A\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right), \quad \ell: \left(\begin{array}{l}x \\ y \\ z\end{array}\right)+t\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right), \quad B\left(\begin{array}{c}p \\ p \\ 2 p\end{array}\right)\)
 

\(\overrightarrow{A B}=\left(\begin{array}{l}p \\ p \\ 2 p\end{array}\right)-\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right)=\left(\begin{array}{c}p-8 \\ p+6 \\ 2 p-5\end{array}\right)\)
 

  \(\abs{AB}^2\) \(=(p-8)^2+(p+6)^2+(2 p-5)^2\)
    \(=p^2-16 p+64+p^2+12 p+36+4 p^2-20 p+25\)
    \(=6 p^2-24 p+125\)

  

ii.    \(\text{Find shortest distance between \(A\) and \(\ell\).}\)

\(\Rightarrow \text { Find \(p\) when \(\abs{A B}\) is a minimum:}\)

\(\text{Minimum occurs when}\ \ p=\dfrac{-b}{2 a}=\dfrac{24}{2 \times 6}=2\)

\(\therefore \abs{AB}_{\text {min}}=\sqrt{6(2)^2-24(2)+125}=\sqrt{101} \text { units}\)

♦ Mean mark (ii) 45%.

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 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 8

Use the vector form of the linear equations

`3x - 2y = 4`  and  `3y + 2x - 6 = 0`

to show they are perpendicular.  (3 marks)

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

Show Worked Solution
`3x-2y` `= 4`
`3x` `= 2y + 4`
`3/2 x` `=y+2`
`x/(2/3)` `= y + 2`

 
`underset ~(v_1) = ((0), (-2)) + lambda ((2/3), (1))`

 

`3y + 2x-6` `= 0`
`2x` `= -3y + 6`
`-2/3 x` `= y-2`
`x/(-3/2)` `= y-2`

 
`underset ~(v_2) = ((0), (2)) + lambda ((-3/2), (1))`

`((2/3), (1)) ((-3/2), (1)) = -1 + 1 = 0`

`:. underset ~(v_1) _|_ underset ~(v_2)`

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`