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Vectors, EXT2 V1 2024 HSC 16a

Consider the function  \(y=\cos (k x)\), where  \(k>0\). The value of \(k\) has been chosen so that a circle can be drawn, centred at the origin, which has exactly two points of intersection with the graph of the function and so that the circle is never above the graph of the function. The point  \(P(a, b)\)  is the point of intersection in the first quadrant, so  \(a>0\)  and  \(b>0\),  as shown in the diagram.

The vector joining the origin to the point \(P(a, b)\) is perpendicular to the tangent to the graph of the function at that point. (Do NOT prove this.)

Show that  \(k>1\).   (4 marks)

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\(y=\cos (k x) \ \Rightarrow \ \dfrac{dy}{dx}=-k \, \sin (k x)\)

\(P(a, b)=P(a, \cos (k a))\)

\(\text{Let}\ \  \underset{\sim}{p}=\overrightarrow{OP}.\)

\(\text{At}\ \ x=a, \ m_{\text{tang}} \perp \underset{\sim}{p} :\)

  \(m_{\text{tang}}\) \(=-k \, \sin (ka)\)
  \(m_{\overrightarrow{OP}}\) \(=\dfrac{\cos(ka)}{a}\)

 
\(-k \, \sin(ka) \times \dfrac{\cos(ka)}{a}=-1\)

  \(k\) \(=\dfrac{a}{\sin(ka) \cos(ka)}\)
    \(=\dfrac{2a}{\sin(2ka)}\)

 

\(\text{For} \ \ \theta>0, \ \sin \theta<\theta\)

\(\Rightarrow k=\dfrac{2a}{\sin (2ka)}>\dfrac{2a}{2ka}>1\).

Show Worked Solution

\(y=\cos (k x) \ \Rightarrow \ \dfrac{dy}{dx}=-k \, \sin (k x)\)

\(P(a, b)=P(a, \cos (k a))\)

\(\text{Let}\ \  \underset{\sim}{p}=\overrightarrow{OP}.\)

\(\text{At}\ \ x=a, \ m_{\text{tang}} \perp \underset{\sim}{p} :\)

  \(m_{\text{tang}}\) \(=-k \, \sin (ka)\)
  \(m_{\overrightarrow{OP}}\) \(=\dfrac{\cos(ka)}{a}\)

 
\(-k \, \sin(ka) \times \dfrac{\cos(ka)}{a}=-1\)

  \(k\) \(=\dfrac{a}{\sin(ka) \cos(ka)}\)
    \(=\dfrac{2a}{\sin(2ka)}\)

 

\(\text{For} \ \ \theta>0, \ \sin \theta<\theta\)

\(\Rightarrow k=\dfrac{2a}{\sin (2ka)}>\dfrac{2a}{2ka}>1\).

Vectors, EXT2 V1 2023 HSC 10 MC

Consider any three-dimensional vectors  \(\underset{\sim}{a}=\overrightarrow{O A}, \underset{\sim}{b}=\overrightarrow{O B}\)  and  \(\underset{\sim}{c}=\overrightarrow{O C}\)  that satisfy these three conditions

\(\underset{\sim}{a} \cdot \underset{\sim}{b}=1\)

\(\underset{\sim}{b} \cdot \underset{\sim}{c}=2\)

\(\underset{\sim}{c} \cdot \underset{\sim}{a}=3\).

Which of the following statements about the vectors is true?

  1. Two of \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\) could be unit vectors.
  2. The points \(A, B\) and \(C\) could lie on a sphere centred at \(O\).
  3. For any three-dimensional vector \(\underset{\sim}{a}\), vectors \(\underset{\sim}{b}\) and \(\underset{\sim}{c}\) can be found so that \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\) satisfy these three conditions.
  4. \(\forall \ \underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\) satisfying the conditions, \(\exists \ r, s\) and \(t\) such that \(r, s\) and \(t\) are positive real numbers and  \(r\underset{\sim}{a}+s \underset{\sim}{b}+t \underset{\sim}{c}=\underset{\sim}{0}\).
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\(B\)

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\(\text{By elimination}\)

\(\text{Option}\ A:\)

\(\text{If}\ \underset{\sim}{a}\ \text{and}\ \underset{\sim}{b}\ \text{are the two unit vectors,}\ \ \underset{\sim}{a} \cdot \underset{\sim}{b} = \abs{\underset{\sim}{a}} \abs{\underset{\sim}{b}} \cos\,\theta = \cos\,\theta\)

\(-1 \leq \cos\,\theta \leq1\ \ \Rightarrow \ \ -1 \leq \underset{\sim}{a} \cdot \underset{\sim}{b} \leq1 \)

\(\text{Given}\ \ \underset{\sim}{a} \cdot \underset{\sim}{b}=1\ \Rightarrow\ \underset{\sim}{a} = \underset{\sim}{b}\ \Rightarrow\ \underset{\sim}{b} \cdot \underset{\sim}{c} = \underset{\sim}{c} \cdot \underset{\sim}{a}\)

\(\text{Contradicts}\ \ \underset{\sim}{b} \cdot \underset{\sim}{c} = 2\ \ \text{and}\ \ \underset{\sim}{c} \cdot \underset{\sim}{a} = 3\)

\(\text{Similar reasoning rules out any pair satisfying all conditions (eliminate}\ A). \)
 

\(\text{Option}\ C:\ \text{If}\ \ \underset{\sim}{a} = \underset{\sim}{0}, \  \underset{\sim}{a} \cdot \underset{\sim}{b} = 0 \neq 1\ \text{(eliminate}\ C). \)

\(\text{Option}\ D:\)

\(\text{Consider the vectors below that satisfy the conditions,}\)

\[\underset{\sim}{a}=\left(\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right),\ \  \underset{\sim}{b}=\left(\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right), \ \  \underset{\sim}{c}=\left(\begin{array}{c} 2 \\ 0 \\ 1 \end{array}\right) \]

\(\text{However,}\ r\underset{\sim}{a}+s \underset{\sim}{b}+t \underset{\sim}{c}=\underset{\sim}{0}\ \text{requires}\ \ r=s=t=0\ \ \text{which are not}\)

\(\text{positive constants (eliminate}\ D).\)

\(\Rightarrow B\)

♦♦♦ Mean mark 15%.

Vectors, EXT2 V1 2023 HSC 15c

A curve \( \mathcal{C}\) spirals 3 times around the sphere centred at the origin and with radius 3, as shown.

A particle is initially at the point \((0,0,-3)\) and moves along the curve \(\mathcal{C}\) on the surface of the sphere, ending at the point \((0,0,3)\).
 

By using the diagram below, which shows the graphs of the functions  \(f(x)=\cos (\pi x)\)  and  \(g(x)=\sqrt{9-x^2}\), and considering the graph  \(y=f(x)g(x)\), give a possible set of parametric equations that describe the curve \( \mathcal{C}\).  (3 marks)
 

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\(x= \cos{(\pi t)}\sqrt{9-t^2} \)

\(y= -\sin{(\pi t)}\sqrt{9-t^2} \)

\( z=t \)

Show Worked Solution

\(\text{Since the curve lies on a sphere with radius 3:}\)

\(x^2+y^2+z^2=3^3 \)

\(\text{Considering the graph}\ \ y=\cos (\pi t)\sqrt{9-t^2}\ \ \text{(as per hint)} \)

\(\Big(\cos (\pi t)\sqrt{9-t^2}\Big)^2+\Big(\sin (\pi t)\sqrt{9-t^2}\Big)^2+t^2=3^2 \ \ …\ (1) \)

\(\text{Since}\ z\ \text{increases and}\ x\ \text{and}\ y\ \text{change signs} \)

\( \Rightarrow z=t \)
 

\(\text{In order to satisfy the equation in (1): } \)

\( x,y\ \text{must be one of }\ \ \pm \cos{(\pi t)}\sqrt{9-t^2}\ \ \text{or}\ \ \pm \sin{(\pi t)}\sqrt{9-t^2} \)
 

\(\text{At}\ \ z=0,\ t=0, \ x=3\ \ \text{(from graph):} \)

\( \Rightarrow x= \cos{(\pi t)}\sqrt{9-t^2} \)
 

\(\text{At}\ \ z=0+\epsilon,\ t=0+\epsilon, \ y \lt 0\ \ \text{(from graph):} \)

\( \Rightarrow y= -\sin{(\pi t)}\sqrt{9-t^2} \)

♦♦♦ Mean mark 22%.

Vectors, EXT2 V1 EQ-Bank 11

Point `O` is the circumcentre of triangle `ABC` which is the centre of the circle that passes through each of its vertices.

Point `P` is the centroid of triangle `ABC` where the bisectors of each angle within the triangle intersect.

Point `Q` is such that `vec(OQ)=3vec(OP)`.
 

Prove that  `vec(CQ) ⊥ vec(AB)`   (5 marks)

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`text{See Worked Solution}` 

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`vec(OP)=(vec(OA)+vec(OB)+vec(OC))/3`

`vec(OQ)` `=3\ vec(OP)`  
  `=3((vec(OA)+vec(OB)+vec(OC))/3)`  
  `=vec(OA)+vec(OB)+vec(OC)`  
  `=((x_1+x_2+x_3),(y_1+y_2+y_3))`  

 

`vec(CQ)` `=vec(OQ)-vec(OC)`  
  `=((x_1+x_2+x_3),(y_1+y_2+y_3))-((x_3),(y_3))`  
  `=((x_1+x_2),(y_1+y_2))`  

 

`vec(AB)` `=vec(OB)-vec(OA)`  
  `=((x_2),(y_2))-((x_1),(y_1))`  
  `=((x_2-x_1),(y_2-y_1))`  

 

`text{Test}\ \ vec(CQ)*vec(AB)=0:`

`text{LHS}` `=((x_1+x_2),(y_1+y_2))((x_2-x_1),(y_2-y_1))`  
  `=(((x_1+x_2)(x_2-x_1)),((y_1+y_2)(y_2-y_1)))`  
  `=(x_2)^2-(x_1)^2+(y_2)^2-(y_1)^2`  
  `=(x_2)^2+(y_2)^2-((x_1)^2+(y_1)^2)`  
  `=abs(vec(OB))^2-abs(vec(OA))^2`  
  `=OB^2-OA^2\ \ \ text{(radii)}`  
  `=0`  
 
`:.vec(CQ) ⊥ vec(AB)\ \ text{… as required}` 

Vectors, EXT2 V1 2021 HSC 16a

  1. The point  `P(x, y, z)`  lies on the sphere of radius 1 centred at the origin `O`.
  2. Using the position vector of `P, overset->{OP} = x underset~i + y underset~j + z underset~k` , and the triangle inequality, or otherwise, show that `| x | + | y | + |z | ≥ 1`.  (2 marks)

  3. Given the vectors  `underset~a = ((a_1),(a_2),(a_3))`  and  `underset~b = ((b_1),(b_2),(b_3))` , show that 
  4.    `|a_1 b_1 + a_2 b_2 + a_3 b_3 | ≤ sqrt{a_1^2 + a_2^2 + a_3^2 }\ sqrt{b_1^2 + b_2^2 + b_3^2}`.  (3 marks)

  5. As in part (i), the point  `P (x, y, z)`  lies on the sphere of radius 1 centred at the origin `O`.
  6. Using part (ii), or otherwise, show that  `| x | + | y | + | z | ≤ sqrt3`.  (2 marks)

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  1. `text{See Worked Solution}`
  2. `text{See Worked Solution}`
  3. `text{See Worked Solution}`
Show Worked Solution
i.      `text{Triangle inequality:} \ |x| + |y| ≥ |x + y|`
♦ Mean mark (i) 50%.
`|x| + |y| + |z|` ` = |x_underset~i| + |y_underset~j| + |z_underset~k|`
  `≥ |x_underset~i + y_underset~j| + |z_underset~k|`
  `≥ |x_underset~i + y_underset~j + z_underset~k|`
  `≥ 1\ \ (|overset->{OP}| = | x_underset~i + y_underset~j + z_underset~k | = 1)`

 

ii.   `text{Using the dot product:}`

♦ Mean mark (ii) 42%.

`underset~a * underset~b = a_1 b_1 + a_2 b_2 + a_3 b_3`

`underset~a * underset~b = |underset~a| |underset~b| cos theta`

 

`a_1 b_1 + a_2 b_2 + a_3 b_3` `= sqrt{a_1^2 + a_2^2 + a_3^2} * sqrt{b_1^2 + b_2^2 + b_3^2} * cos theta`
`|a_1 b_1 + a_2 b_2 + a_3 b_3|` `= sqrt{a_1^2 + a_2^2 + a_3^2} * sqrt{b_1^2 + b_2^2 + b_3^2} * |cos theta|`

 

`text{S} text{ince} \ -1 ≤ cos theta ≤ 1 \ => \ |cos theta| ≤ 1`

`:. \ | a_1 b_1 + a_2 b_2 + a_3 b_3 | ≤ sqrt{a_1^2 + a_2^3 + a_3^2} * sqrt{b_1^2 + b_2^2 + b_3^2}`

 

iii.  `text{Using part (ii) with vectors:}`

♦♦♦ Mean mark (iii) 14%.

`underset~a = ((1),(1),(1)) \ , \ underset~b = (( | x| ),( |y| ),( |z| ))`

`|\ |x| + |y| + |z|\ |` `≤ sqrt{1^2 + 1^2 + 1^2} * sqrt{ x^2 + y^2 + z^2}`
`|x| + |y| + |z|` `≤ sqrt3`