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Vectors, EXT2 V1 2025 HSC 11d

  1. Force \({\underset{\sim}{F}}_1\) has magnitude 12 newtons in the direction of vector  \(2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}\).   
  2. Show that  \({\underset{\sim}{F}}_1=8 \underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}\).   (1 mark)

  3. Force \({\underset{\sim}{F}}_1\) from part (i) and a second force,  \({\underset{\sim}{F}}_2=-6 \underset{\sim}{i}+12 \underset{\sim}{j}+4 \underset{\sim}{k}\), both act upon a particle.
  4. Show that the resultant force acting on the particle is given by:
  5.      \({\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}.\)   (1 mark)

  6. Calculate  \({\underset{\sim}{F}}_3 \cdot \underset{\sim}{d}\), where \({\underset{\sim}{F}}_3\) is the resultant force from part (ii) and  \(\underset{\sim}{d}=\underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k}\).   (1 mark)

Show Answers Only

i.    \(\abs{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}=\sqrt{2^2+(-2)^2+1^2}=3\)

\({\underset{\sim}{F}}_1=\dfrac{12}{3}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)\)

\({\underset{\sim}{F}}_1=8\underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}\)
 

ii.    \({\underset{\sim}{F}}_3={\underset{\sim}{F}}_1+{\underset{\sim}{F}}_2=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)+\left(\begin{array}{c}-6 \\ 12 \\ 4\end{array}\right)=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\)

\({\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}\)
 

iii.  \({\underset{\sim}{F}}_3 \cdot d=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=2+4+16=22\)

Show Worked Solution

i.    \(\abs{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}=\sqrt{2^2+(-2)^2+1^2}=3\)

\({\underset{\sim}{F}}_1=\dfrac{12}{3}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)\)

\({\underset{\sim}{F}}_1=8\underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}\)
 

ii.    \({\underset{\sim}{F}}_3={\underset{\sim}{F}}_1+{\underset{\sim}{F}}_2=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)+\left(\begin{array}{c}-6 \\ 12 \\ 4\end{array}\right)=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\)

\({\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}\)
 

iii.  \({\underset{\sim}{F}}_3 \cdot d=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=2+4+16=22\)

Vectors, EXT2 V1 EQ-Bank 4

Find all possible values of  `b`  given  `absvec(PQ)=sqrt76`  where  `P(-5,b,-2)`  and  `Q(-3,-2,4)`.   (3 marks)

Show Answers Only

`b=4\ \ text(or)\ \ -8`

Show Worked Solution

`vec(PQ)=((-3),(-2),(4))-((-5),(b),(-2))=((2),(-2-b),(6))`

`absvec(PQ)` `=sqrt(2^2+(-2-b)^2+6^2)`  
`76` `=(-2-b)^2+40`  
`36` `=(-2-b)^2`  
`6` `=+-(-2-b)`  

 
`-2-b=6\ \ =>\ \ b=-8`

`2+b=6\ \ =>\ \ b=4`

`:.b=4\ \ text(or)\ \ -8`

Vectors, EXT2 V1 SM-Bank 11

Given  `lambda_1underset~a + lambda_2underset~b = [(50),(−45),(−8)]`, find  `lambda_1` and `lambda_2`  if 
 

`underset~a = [(2),(−3),(4)]`  and  `underset~b = [(3),(−2),(−3)]`.  (2 marks)

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`lambda_1 = 7, lambda_2 = 12`

Show Worked Solution

`lambda_1underset~a + lambda_2underset~b = [(2lambda_1 + 3lambda_2),(−3lambda_1 – 2lambda_2),(4lambda_1 – 3lambda_2)]`

`2lambda_1 + 3lambda_2 = 50\ \ …\ (1)`  
`4lambda_1 – 3lambda_2 = −8\ \ …\ (2)`  

 
`(1) + (2)`

`6lambda_1` `= 42`
`lambda_1` `= 7`

 
`text(Substitute)\ \ lambda_1=7\ \ text{into (1):}`

`14+3lambda_2` `= 50`
`lambda_2` `= 12`

Vectors, EXT2 V1 2013 SPEC1 3

The coordinates of three points are  `A\ ((– 1), (2), (4)), \ B\ ((1), (0), (5)) and C\ ((3), (5), (2)).`

  1. Find  `vec (AB).`  (1 mark)

  2. The points `A, B` and `C` are the vertices of a triangle. 

     

    Prove that the triangle has a right angle at `A.`  (2 marks)

  3. Find the length of the hypotenuse of the triangle.  (1 mark)

Show Answers Only
  1. `2underset~i – 2underset~j + underset~k`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `sqrt 38`
Show Worked Solution
i.   `vec(AB)` `=(1 – −1)underset~i + (0 – 2)underset~j + (5 – 4)underset~k`
    `= 2underset~i – 2underset~j + underset~k`

 

ii.    `overset(->)(AC)` `= (3 – −1)underset~i + (5 – 2)underset~j + (2 – 4)underset~k`
    `= 4underset~i + 3underset~j – 2underset~k`

 

`overset(->)(AB) · overset(->)(AC)` `= 2 xx 4 + (−2) xx 3 + 1 xx (−2)`
  `= 8 – 6 – 2`
  `= 0`

 
`=>  overset(->)(AB) ⊥ overset(->)(AC)`

`:. DeltaABC\ text(has a right angle at)\ A.`
 

iii.    `overset(->)(BC)` `= (3 – 1)underset~i + (5 – 0)underset~j + (2 – 5)underset~k`
    `= 2underset~i + 5underset~j – 3underset~k`

 

`|overset(->)(BC)|` `= sqrt(2^2 + 5^2 + (−3)^2)`
  `= sqrt(4 + 25 + 9)`
  `= sqrt38`

Vectors, EXT2 V1 SM-Bank 12

Two vectors are given by  `underset ~a = 4 underset ~i + m underset ~j - 3 underset ~k`  and  `underset ~b = −2 underset ~i + n underset ~j - underset ~k`, where  `m`, `n in R^+`.

If  `|\ underset ~a\ | = 10`  and  `underset ~a`  is perpendicular to  `underset ~b`, then find `m` and `n`.   (2 marks)

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`m=5sqrt3, \ n=sqrt3/3`

Show Worked Solution

`text(Using)\ \ |\ underset ~a\ | = 10:`

`10` `= sqrt(4^2 + m^2 + (-3)^2)`
`100` `=m^2 +25`
`m` `= 5sqrt3`

 
`text(S)text(ince)\ \ underset ~a _|_ underset ~b :`

`underset ~a ⋅ underset ~b` `= 0`
`0` `=4 xx (−2) + mn + (−3) xx (−1)`
`0` `= mn-5`
`n` `=5/(5sqrt3)`
  `=sqrt3/3`

 

Vectors, EXT2 V1 SM-Bank 4

Consider the three vectors

`underset ~a = underset ~i - underset ~j + 2underset ~k,\ underset ~b = underset ~i + 2 underset ~j + m underset ~k`  and  `underset ~c = underset ~i + underset ~j - underset ~k`, where  `m in R.`

  1. Find the value(s) of `m` for which  `|\ underset ~b\ | = 2 sqrt 3.`  (2 marks)
  2. Find the value of `m` such that  `underset ~a`  is perpendicular to  `underset ~b.`  (1 mark)
Show Answers Only
  1. `+- sqrt 7`
  2. `1/2`
Show Worked Solution
i.    `|underset~b|` `= sqrt(1^2 + 2^2 + m^2)` `= 2sqrt3`
    `1 + 4 + m^2` `= 4 xx 3`
    `m^2` `= 7`
    `m` `= ±sqrt7`

 

ii.    `underset~a * underset~b` `= 1 xx 1 + (−1) xx 2 + 2 xx m`  
  `0` `= 1 – 2 + 2m\ \ ( underset~a ⊥ underset~b)`  
  `2m` `= 1`  
  `m` `= 1/2`