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Vectors, EXT2 V1 2024 HSC 15a

Consider the three vectors  \(\underset{\sim}{a}=\overrightarrow{O A}, \underset{\sim}{b}=\overrightarrow{O B}\) and \(\underset{\sim}{c}=\overrightarrow{O C}\), where \(O\) is the origin and the points \(A, B\) and \(C\) are all different from each other and the origin.

The point \(M\) is the point such that  \(\dfrac{1}{2}(\underset{\sim}{a}+\underset{\sim}{b})=\overrightarrow{O M}\).

  1. Show that \(M\) lies on the line passing through \(A\) and \(B\).   (1 mark)

  2. The point \(G\) is the point such that  \(\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})=\overrightarrow{O G}\).
  3. Show that \(G\) lies on the line passing through \(M\) and \(C\), and lies between \(M\) and \(C\).   (2 marks)

  4. The complex numbers \(x, w\) and \(z\) are all different and all have modulus 1.
  5. Using part (ii), or otherwise, show that  \(\dfrac{1}{3}(x+w+z)\) is never a cube root of \(x w z\).   (2 marks)

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i.    \(\text{Equation of line through \(A\) and \(B\)}\)

\(\Rightarrow \ell_1=\overrightarrow{O A}+\lambda \overrightarrow{A B}\)

  \(\overrightarrow{O M}\) \(=\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\underset{\sim}{a}-\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\overrightarrow{O A}+\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{a})\)
    \(=\overrightarrow{O A}+\dfrac{1}{2} \overrightarrow{A B}\)

 
\(\therefore \overrightarrow{OM} \ \text{lies on} \ \ell_1\).
 

ii.    \(\text{Equation of line through \(M\) and \(C\)}\)

\(\Rightarrow \ell_2=\overrightarrow{OC}+\lambda \overrightarrow{CM}\)

  \(\overrightarrow{O G}\) \(=\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})\)
    \(=\underset{\sim}{c}-\dfrac{2}{3} \underset{\sim}{c}+\dfrac{1}{3} \underset{\sim}{a}+\dfrac{1}{3} \underset{\sim}{b}\)
    \(=\overrightarrow{OC}+\dfrac{2}{3}\left(\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}-\underset{\sim}{c}\right)\)
    \(=\overrightarrow{OC}+\dfrac{2}{3} \overrightarrow{CM}\)

 
\(\therefore \overrightarrow{O G} \ \ \text{lies on} \ \ \ell_2\)

\(\ \ \overrightarrow{O G} \neq \overrightarrow{O C}  \ \ \text{and} \ \ \overrightarrow{O G} \neq \overrightarrow{O M}\)

\(\therefore G \ \ \text{lies between} \ \ C \ \text{and} \ M\).
 

iii.  \(\text{Place}\ x, w,\ \text{and}\ z\ \text{on unit circle.}\)
 

\(\abs{w}=\abs{x}=\abs{z}=1\)

\(\text{Using part (ii):}\)

\(G \equiv \dfrac{1}{3}(x+w+z)\)

\(G \ \text{lies on} \ CM \Rightarrow G \ \text{is inside the unit circle.}\)

\(\Rightarrow\left|\dfrac{1}{3}(x+w+z)\right|<1\)

\(\text{Since}\ \ \abs{xwz}=\abs{x}\abs{w}\abs{z}=1\)

\(\Rightarrow \ \text{All cube roots have modulus = 1.}\)

\(\therefore \dfrac{1}{3}(x+w+z) \ \ \text{cannot be a cube root of  \(xwz\).}\)

Show Worked Solution

i.    \(\text{Equation of line through \(A\) and \(B\)}\)

\(\Rightarrow \ell_1=\overrightarrow{O A}+\lambda \overrightarrow{A B}\)

  \(\overrightarrow{O M}\) \(=\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\underset{\sim}{a}-\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\overrightarrow{O A}+\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{a})\)
    \(=\overrightarrow{O A}+\dfrac{1}{2} \overrightarrow{A B}\)

 
\(\therefore \overrightarrow{OM} \ \text{lies on} \ \ell_1\).
 

ii.    \(\text{Equation of line through \(M\) and \(C\)}\)

\(\Rightarrow \ell_2=\overrightarrow{OC}+\lambda \overrightarrow{CM}\)

  \(\overrightarrow{O G}\) \(=\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})\)
    \(=\underset{\sim}{c}-\dfrac{2}{3} \underset{\sim}{c}+\dfrac{1}{3} \underset{\sim}{a}+\dfrac{1}{3} \underset{\sim}{b}\)
    \(=\overrightarrow{OC}+\dfrac{2}{3}\left(\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}-\underset{\sim}{c}\right)\)
    \(=\overrightarrow{OC}+\dfrac{2}{3} \overrightarrow{CM}\)

 
\(\therefore \overrightarrow{O G} \ \ \text{lies on} \ \ \ell_2\)

\(\ \ \overrightarrow{O G} \neq \overrightarrow{O C}  \ \ \text{and} \ \ \overrightarrow{O G} \neq \overrightarrow{O M}\)

\(\therefore G \ \ \text{lies between} \ \ C \ \text{and} \ M\).

♦ Mean mark (ii) 43%.

iii.  \(\text{Place}\ x, w,\ \text{and}\ z\ \text{on unit circle.}\)
 

♦♦♦ Mean mark (iii) 10%.

\(\abs{w}=\abs{x}=\abs{z}=1\)

\(\text{Using part (ii):}\)

\(G \equiv \dfrac{1}{3}(x+w+z)\)

\(G \ \text{lies on} \ CM \Rightarrow G \ \text{is inside the unit circle.}\)

\(\Rightarrow\left|\dfrac{1}{3}(x+w+z)\right|<1\)

\(\text{Since}\ \ \abs{xwz}=\abs{x}\abs{w}\abs{z}=1\)

\(\Rightarrow \ \text{All cube roots have modulus = 1.}\)

\(\therefore \dfrac{1}{3}(x+w+z) \ \ \text{cannot be a cube root of  \(xwz\).}\)

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 5 MC

Which of the following is a true statement about the lines  \(\ell_1={\displaystyle\left(\begin{array}{cc}-1 \\ 2 \\ 5\end{array}\right)+\lambda\left(\begin{array}{c}-1 \\ 3 \\ 1\end{array}\right)}\)  and  \(\ell_2=\left(\begin{array}{c}3 \\ -10 \\ 1\end{array}\right)+\mu\left(\begin{array}{c}1 \\ -3 \\ -1\end{array}\right) ?\)

  1. \(\ell_1\) and \(\ell_2\) are the same line.
  2. \(\ell_1\) and \(\ell_2\) are not parallel and they intersect.
  3. \(\ell_1\) and \(\ell_2\) are parallel and they do not intersect.
  4. \(\ell_1\) and \(\ell_2\) are not parallel and they do not intersect.
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\(A\)

Show Worked Solution

\(\text{Since}\ \ \left(\begin{array}{c}-1 \\ 3 \\ 1\end{array}\right) = -1 \left(\begin{array}{c}1 \\ -3 \\ -1\end{array}\right), \ \ell_1\ \text{is parallel to}\ \ell_2 \)

\(\text{Test if point}\ (3,-10,1)\ \text{lies on}\ \ell_1: \)

\(\text{i.e.}\ \ \exists \lambda\ \ \text{such that} \)

♦ Mean mark 49%.

\( \left(\begin{array}{cc}3 \\ -10 \\ 1\end{array}\right) = \left(\begin{array}{cc}-1 \\ 2 \\ 5\end{array}\right) + \lambda \left(\begin{array}{cc}-1 \\ 3 \\ 1\end{array}\right)\)

\(\lambda = -4\ \ \text{satisfies equation} \)

\(\therefore\ \ell_1\ \text{and}\ \ell_2\ \text{are the same line.}\)

\(\Rightarrow A\)

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 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 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)`
Show Worked Solution

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.`

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.)`