The sum of two unit vectors is a unit vector.
Determine the magnitude of the difference of the two vectors. (3 marks)
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The sum of two unit vectors is a unit vector.
Determine the magnitude of the difference of the two vectors. (3 marks)
\(D\)
\(\text{Vectors can be drawn as two sides of an equilateral triangle.}\)
\(\text{Using the cosine rule for the difference between the two vectors:}\)
\(c^2\) | \(=a^2+b^2-2ab\, \cos C\) | |
\(=1+1-2\times -\dfrac{1}{2} \) | ||
\(=3\) | ||
\(c\) | \(=\sqrt{3}\) |
\(\abs{v_1-v_2} = \sqrt{3}\)
--- 8 WORK AREA LINES (style=lined) --- The point \(P\) lies on the circle centred at \(I(r, 0)\) with radius \(r>0\), such that \(\overrightarrow{I P}\) makes an angle of \(t\) to the horizontal. The point \(Q\) lies on the circle centred at \(J(-R, 0)\) with radius \(R>0\), such that \(\overrightarrow{J Q}\) makes an angle of \(2 t\) to the horizontal. --- 10 WORK AREA LINES (style=lined) --- \(=\dfrac{1}{2} \big{|} r(1+\cos\ t) \cdot R \sin\ 2t-r\ \sin\ t\ \cdot R(\cos\ 2t-1)\ \big{|} \) \(f(\pi) = f(- \pi) = 0\) \(\therefore A_{\triangle AOB}\ \text{is maximum when}\ \ t=\dfrac{\pi}{3}, \ – \dfrac{\pi}{3} \)
\(\Big{|}\text{proj}_{\overrightarrow{OA^{′}}} \overrightarrow{OB} \Big{|}\)
\(=\ \text{⊥ height of}\ \triangle OAB\ \text{from side}\ \overrightarrow{OA} \)
\(= \Bigg{|} \dfrac{\overrightarrow{OB} \cdot \overrightarrow{OA^{′}}}{|\overrightarrow{OA^{′}}|} \Bigg{|} \)
\(= \Bigg{|} \dfrac{b_1a_2-b_2a_1}{|\overrightarrow{OA^{′}}|} \Bigg{|} \)
\(\text{Area}\ \triangle AOB\)
\(=\dfrac{1}{2} \Bigg{|} \dfrac{b_1a_2-b_2a_1}{|\overrightarrow{OA^{′}}|} \Bigg{|} \cdot |\overrightarrow{OA} | \)
\(=\dfrac{1}{2}\left|a_1 b_2-a_2 b_1\right|\ \ \ (\text{noting}\ \ |\overrightarrow{OA} |=|\overrightarrow{OA^{′}} |) \)
ii.
\(\overrightarrow{OP}\)
\(=\overrightarrow{OI}+\overrightarrow{IP}\)
\(= \left(\begin{array}{l}r \\ 0\end{array}\right) + \left(\begin{array}{c} r\ \cos \ t \\ r\ \sin \ t\end{array}\right) \)
\(= \left(\begin{array}{c} r(1+ \cos \ t) \\ r\ \sin \ t\end{array}\right) \)
\(\overrightarrow{OQ}\)
\(=\overrightarrow{OJ}+\overrightarrow{JQ}\)
\(= \left(\begin{array}{c} -R \\ 0 \end{array}\right) + \left(\begin{array}{l} R\ \cos\ 2t \\ R\ \sin\ 2t\end{array}\right) \)
\(= \left(\begin{array}{c} R(\cos\ 2t-1) \\ R\ \sin\ 2t\end{array}\right) \)
\(\text{Using part (i):}\)
\(A_{\triangle OPQ}\)
\(=\dfrac{rR}{2} \big{|} \sin\ 2t+\cos\ t\ \sin\ 2t-\sin\ t\ \cos\ 2t+\sin\ t\ \big{|} \)
\(=\dfrac{rR}{2} \big{|} \sin\ 2t+\sin\ t\ + \sin(2t-t) \big{|} \)
\(=\dfrac{rR}{2} \big{|} 2\sin\ t\ \cos\ t +2\sin\ t \big{|} \)
\(= rR \big{|} \sin\ t(\cos\ t+1) \big{|} \)
\(\text{Let}\ \ f(t)=\sin\ t(\cos\ t+1) \)
\(f^{′}(t) \)
\(=\cos\ t(\cos\ t+1)+\sin\ t(-\sin\ t) \)
\(=\cos^{2}t+\cos\ t-\sin^{2}t\)
\(=\cos^{2}t+\cos\ t-(1-\cos^{2}t) \)
\(=2\cos^{2}t+\cos\ t-1 \)
\(=(2\cos^{2}t-1)(\cos\ t+1) \)
\(\text{SP’s when}\ \ f^{′}(t)=0:\)
\(\cos\ t\)
\(=\dfrac{1}{2} \)
\(\cos\ t\)
\(=-1\)
\(t\)
\(=\dfrac{\pi}{3}, \ – \dfrac{\pi}{3} \)
\(t\)
\(=\pi, \ -\pi \)
\(\text{Testing SP’s:}\)
Point `C` lies on `AB` such that `overset(->)(AC) = lambdaoverset(->)(AB)`.
--- 8 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
i. |
`underset~c = underset~a + overset(->)(AC)`
`overset(->)(AB)` | `= underset~b – underset~a` |
`overset(->)(AC)` | `= lambdaoverset(->)(AB)` |
`= lamda(underset~b – underset~a)` |
`:.underset~c` | `= underset~a + lambda(underset~b – underset~a)` |
`= lambdaunderset~b + (1 – lambda)underset~a` |
ii. | `overset(->)(BC)` | `= underset~c – underset~b` |
`= lambdaunderset~b + (1 – lambda)underset~a – underset~b` | ||
`=(1 – lambda)underset~a +(lambda – 1)underset~b` | ||
`=(1 – lambda)underset~a -(1-lambda)underset~b` | ||
`= (1 – lambda)(underset~a – underset~b)` |
Vectors `underset ~a, underset ~b` and `underset ~c` are shown below.
From the diagram it follows that
A. `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2`
B. `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 - |\ underset ~a\ | |\ underset ~b\ |`
C. `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 + |\ underset ~a * underset ~b\ |`
D. `|\ underset ~c\ |^2 = |\ underset ~a\ |^2 + |\ underset ~b\ |^2 + |\ underset ~a\ | |\ underset ~b\ |`
`D`
`underset~ c + underset~ b = underset~ a\ \ => \ underset~ c = underset~ a – underset~ b`
`:. underset~ c * underset~ c` | `= (underset~ a – underset~ b) * (underset~ a – underset~ b)` |
`= underset~ a * underset~ a – underset~ a * underset~ b -underset~b*underset~a + underset~ b * underset~ b` | |
`= underset~ a * underset~ a + underset~ b * underset~ b-2underset~b*underset~a ` |
`:. |\ underset~ c\ |^2 = |\ underset~ a\ |^2 + |\ underset~ b\ |^2 – 2 |\ underset~ a\ | |\ underset~ b\ | cos 120^@`
`:. |\ underset~ c\ |^2 = |\ underset~ a\ |^2 + |\underset~ b\ |^2 + |\ underset~ a\ | |\ underset~ b\ |`
`=> D`
Points `A` and `B` are on the number plane. The vector `overset(->)(AB)` is `((4),(1))`.
Point `C` is chosen so that the area of `DeltaABC` is `17/2` square units and `|overset(->)(AC)| = sqrt34`.
Find all possible vectors `overset(->)(AC)`. (4 marks)
--- 10 WORK AREA LINES (style=lined) ---
`((3),(5)),((5),(−3)),((−3),(−5))\ text(and)\ ((−5),(3))`
`|overset(->)(AB)| = sqrt(4^2 + 1^2) = sqrt17`
`text(Let)\ \ angleCAB = theta`
`text(Area)\ DeltaABC` | `= 1/2 · |overset(->)(AB)| · |overset(->)(AC)| · sintheta` |
`17/2` | `= 1/2 · sqrt17 · sqrt34 sintheta` |
`17` | `= 17sqrt2 sintheta` |
`sintheta` | `= 1/sqrt2` |
`theta` | `= pi/4\ text(or)\ (3pi)/4\ \ (theta>0)` |
`=> costheta = ±1/sqrt2`
`overset(->)(AB) · overset(->)(AC)` | `= |overset(->)(AB)| · |overset(->)(AC)| · costheta` |
`= ±sqrt17 · sqrt34 · 1/sqrt2` | |
`= ±17` |
`text(Let)\ \ overset(->)(AC) = ((x),(y))`
`overset(->)(AB) · overset(->)(AC)` | `= ((4),(1))((x),(y))` |
`= 4x + y` |
`4x + y = ±17…\ \ (1)`
`|overset(->)(AC)|` | `= sqrt(x^2 + y^2) = sqrt34` |
`x^2 + y^2` | `= 34\ …\ \ (2)` |
`text(Solving simultaneously:)`
`4x + y = 17 \ => \ y = 17 – 4x`
`text{Substitute into (2)}`
`x^2 + (17 – 4x)^2` | `= 34` |
`x^2 + 289 – 136x + 16x^2` | `= 34` |
`17x^2 – 136x + 255` | `= 0` |
`x^2 – 8x + 15` | `= 0` |
`(x – 3)(x – 5)` | `= 0` |
`:. x = 3, y = 5\ \ text(or)\ \ x = 5, y = −3`
`text(Similarly for)\ \ 4x + y = −17 \ => \ y = −4x – 17`
`x^2 + (−4x – 17)^2` | `= 34` |
`x^2 + 8x + 15` | `= 0` |
`(x + 3)(x + 5)` | `= 0` |
`:. x = −3, y = −5\ \ text(or)\ \ x = −5, y = 3`
`:. text(Possible vectors)\ overset(->)(AC)\ text(are)`
`((3),(5)),((5),(−3)),((−3),(−5))\ text(and)\ ((−5),(3))`