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Vectors, EXT2 V1 2024 HSC 10 MC

Three unit vectors \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\), in 3 dimensions, are to be chosen so that  \(\underset{\sim}{a} \perp \underset{\sim}{b}, \ \underset{\sim}{b} \perp \underset{\sim}{c}\)  and the angle \(\theta\) between \(\underset{\sim}{a}\) and  \(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}\)  is as small as possible.

What is the value of \(\cos \theta\) ?

  1. \(0\)
  2. \(\dfrac{1}{\sqrt{3}}\)
  3. \(\dfrac{1}{\sqrt{2}}\)
  4. \(\dfrac{2}{\sqrt{5}}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\cos \theta=\dfrac{\underset{\sim}{a} \cdot(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})}{\abs{\underset{\sim}{a}}\abs{\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}}}\)

 
\(\theta_{\text{min}} \ \Rightarrow \ \cos\,\theta_{\text{max}}:\)

\(\underset{\sim}{a} \cdot\left(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}\right)=\abs{\underset{\sim}{a}}^2+\underset{\sim}{a} \cdot \underset{\sim}{b}+\underset{\sim}{a} \cdot \underset{\sim}{c}=1+\underset{\sim}{a} \cdot \underset{\sim}{c}\)

\(\text{Find}\ \ \abs{\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}}:\)

  \(\abs{\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}}^2\) \(=\abs{\underset{\sim}{a}}^2+\abs{\underset{\sim}{b}}^2+\abs{\underset{\sim}{c}}^2+2\left(\underset{\sim}{a} \cdot \underset{\sim}{b}+\underset{\sim}{b} \cdot \underset{\sim}{c}+\underset{\sim}{a} \cdot \underset{\sim}{c}\right)\)
    \(=3+2 \underset{\sim}{a} \cdot \underset{\sim}{c}\)

 
\(\cos \theta=\dfrac{1+a \cdot c}{\sqrt{3+2 a \cdot c}}\)

\(\underset{\sim}{a} \cdot\underset{\sim}{c}=\abs{\underset{\sim}{a}}\abs{\underset{\sim}{c}} \cos\, \alpha = \cos\,\alpha \)

\(0 \leqslant \cos\, \alpha \, \leqslant 1\)

\(\therefore \cos \theta_{\text{max}}=\dfrac{2}{\sqrt{5}}\)

\(\Rightarrow D\)

Vectors, EXT2 V1 2024 HSC 11c

Find the angle between the two vectors  \(\underset{\sim}{u}=\left(\begin{array}{c}1 \\ 2 \\ -2\end{array}\right)\) and  \(\underset{\sim}{v}=\left(\begin{array}{c}4 \\ -4 \\ 7\end{array}\right)\), giving your answer in radians, correct to 1 decimal place.   (2 marks)

Show Answers Only

 

Show Worked Solution

\(\underset{\sim}{u}=\left(\begin{array}{c}1 \\ 2 \\ -2\end{array}\right),\abs{\underset{\sim}{u}}=\sqrt{1+4+4}=3\)

\(\underset{\sim}{v}=\left(\begin{array}{c}4 \\ -4 \\ 7\end{array}\right),\abs{\underset{\sim}{v}}=\sqrt{16+16+49}=9\)

\(\cos \theta=\dfrac{\underset{\sim}{u} \cdot \underset{\sim}{v}}{|\underset{\sim}{u}||\underset{\sim}{v}|}=\dfrac{1 \times 4-2 \times 4-2 \times 7}{3 \times 9}=-\dfrac{2}{3}\)

\(\theta=\cos ^{-1}\left(-\dfrac{2}{3}\right)=2.30 \ldots=2.3^c \ \ \text{(1 d.p.)}\)

Vectors, EXT2 V1 2022 SPEC1 6

Find the cosine of the acute angle between the vectors  `underset~a=2underset~i-3underset~j+6underset~k`  and  `underset~b=underset~i+2underset~j+2underset~k`.   (2 marks)

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`8/21`

Show Worked Solution
`cos \ theta` `=(underset~a*underset~b)/(|underset~a||underset~b|)`  
  `=(2-6+12)/(sqrt(4+9+36)\ sqrt(1+4+4))`  
  `=8/(7 xx 3)`  
  `=8/21`  

Complex Numbers, EXT2 N2 2023 14a

Let \(z\) be the complex number  \(z=e^{\small{\dfrac{i \pi}{6}}} \)  and \(w\) be the complex number  \(w=e^{\small{\dfrac{3 i \pi}{4}}} \).

  1. By first writing \(z\) and \(w\) in Cartesian form, or otherwise, show that
  2.    \(|z+w|^2=\dfrac{4-\sqrt{6}+\sqrt{2}}{2}\).  (3 marks)

  3. The complex numbers \(z, w\) and \(z+w\) are represented in the complex plane by the vectors \(\overrightarrow{O A},\overrightarrow{O B}\) and \(\overrightarrow{O C}\) respectively, where \(O\) is the origin.
  4. Show that  \(\angle A O C=\dfrac{7 \pi}{24}\).  (2 marks)

  5. Deduce that  \(\cos \dfrac{7 \pi}{24}=\dfrac{\sqrt{8-2 \sqrt{6}+2 \sqrt{2}}}{4}\).  (1 mark)

Show Answers Only
  1. \(\text{See Worked Solutions}\)
  2. \(\text{See Worked Solutions}\)
  3. \(\text{See Worked Solutions}\)
Show Worked Solution

i.    \(z=e^{\small\dfrac{i \pi}{6}} = \cos\,\dfrac{\pi}{6} + i \,\sin\,\dfrac{\pi}{6} = \dfrac{\sqrt3}{2} + \dfrac{1}{2}i \)

\(w=e^{\small\dfrac{3i \pi}{4}} = \cos\,\dfrac{3\pi}{4} + i \,\sin\,\dfrac{3\pi}{4} = -\dfrac{1}{\sqrt2} + \dfrac{i}{\sqrt2} \)

\(|z+w|^2\) \(=\Bigg{|} \dfrac{\sqrt3}{2}+\dfrac{1}{2}i-\dfrac{1}{\sqrt2}+\dfrac{i}{\sqrt2} \Bigg{|}\)  
  \(=\Bigg{|} \Bigg{(}\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2} \Bigg{)} +\Bigg{(}\dfrac{1}{2}+\dfrac{1}{\sqrt2}\Bigg{)}\,i \Bigg{|}\)  
  \(=\Bigg{|} \dfrac{\sqrt6-2}{2\sqrt2}+\dfrac{\sqrt2+2}{2\sqrt2}\,i \Bigg{|}\)  
  \(= \dfrac{(\sqrt6-2)^2+(\sqrt2+2)^2}{(2\sqrt2)^2}\)  
  \(= \dfrac{6-4\sqrt6+4+2+4\sqrt2+4}{8}\)  
  \(=\dfrac{16-4\sqrt6+4\sqrt2}{8} \)  
  \(=\dfrac{4-\sqrt6+\sqrt2}{2} \)  

 
ii.   

\(\angle AOB= \arg(w)-\arg(z)=\dfrac{3\pi}{4}-\dfrac{\pi}{6}=\dfrac{7\pi}{12} \)

\( |z|=|w|=1\ \Rightarrow AOBC\ \text{is a rhombus.} \)

\(\overrightarrow{OC}\ \text{is a diagonal of rhombus}\ AOBC \)

\(\Rightarrow \overrightarrow{OC}\ \text{bisects}\ \angle AOB \)

\(\therefore \angle AOC= \dfrac{1}{2} \times \dfrac{7\pi}{12}=\dfrac{7\pi}{24} \)
  

iii.   \(\text{In}\ \triangle AOC: \)

\( \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA} = \overrightarrow{OB} \)

\(\Rightarrow \overrightarrow{OB}\ \text{is represented by}\ w. \)
 

\(\text{Using the cos rule in}\ \triangle AOC: \)

\(\cos\,\dfrac{7\pi}{24}\) \(=\dfrac{|z|^2+|z+w|^2-|w|^2}{2|z||z+w|}\)  
  \(=\dfrac{ 1+\frac{4-\sqrt6+\sqrt2}{2}-1}{2 \times 1  \sqrt{\frac{4-\sqrt6+\sqrt2}{2}}} \)  
  \(=\dfrac{\sqrt{\frac{4-\sqrt6+\sqrt2}{2}} \times 2} {2 \times 2} \)  
  \(=\dfrac{\sqrt{4( \frac{4-\sqrt6+\sqrt2}{2})}} {4} \)  
  \(=\dfrac{8-2\sqrt6+2\sqrt2}{4} \)  
♦♦ Mean mark (iii) 26%.

Vectors, EXT2 V1 2023 HSC 11b

Find the angle between the vectors

\(\underset{\sim}{a}=\underset{\sim}{i}+2 \underset{\sim}{j}-3 \underset{\sim}{k}\)

\(\underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}+2 \underset{\sim}{k}\),

giving your answer to the nearest degree.  (3 marks)

Show Answers Only

\(87^{\circ} \)

Show Worked Solution

\[\underset{\sim}{a}=\left(\begin{array}{c} 1 \\ 2 \\ -3 \end{array}\right),\ \  \underset{\sim}{b}=\left(\begin{array}{c} -1 \\ 4 \\ 2 \end{array}\right) \]

\(\Big{|} \underset{\sim}{a} \Big{|} = \sqrt{1+4+9} = \sqrt{14} \)

\(\Big{|} \underset{\sim}{b} \Big{|} = \sqrt{1+16+4} = \sqrt{21} \)

\( \underset{\sim}{a} \cdot \underset{\sim}{b} = -1 + 8-6=1 \)

\(\cos\ \theta \) \(=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\Big{|}\underset{\sim}{a}\Big{|} \cdot \Big{|}\underset{\sim}{b}\Big{|}} \)  
  \(=\dfrac{1}{\sqrt{294}} \)  
\( \theta\) \(=\cos ^{-1} \Big{(}\dfrac{1}{\sqrt{294}}\Big{)} \)  
  \(=86.65…\)  
  \(=87^{\circ} \)  

Vectors, EXT2 V1 EQ-Bank 12

Two planes are flying in directions given by the vectors  `-200underset~i+400underset~j-3underset~k`  and  `300underset~i+250underset~j+underset~k`.

A person in the flight control centre is plotting their paths on a map.

Calculate the acute angle between their projected flight paths, giving your answer correct to one decimal place.  (2 marks)

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`76.8°`

Show Worked Solution

`text{Let}\ \ underset~a=-200underset~i+400underset~j-3underset~k,\ \ underset~b=300underset~i+250underset~j+underset~k`

`underset~a*underset~b=((-200),(400),(-3))((300),(250),(1))=-60\ 000+100\ 000-3=39\ 997`

`abs(underset~a)=sqrt(200^2+400^2+9)=447.2236…`

`abs(underset~b)=sqrt(300^2+250^2+1)=390.5137…`

`costheta` `=(39\ 997)/(447.2236xx390.5137)`  
  `=0.2291`  
`:.theta` `=76.8°\ \ text{(to 1 d.p.)}`  

Vectors, EXT2 V1 EQ-Bank 2

Find the angle between the vectors  `underset~r = ((3),(-2),(-1))`  and  `underset~s = ((2),(1),(1))`, giving the angle in degrees correct to 1 decimal place. (3 marks)

Show Answers Only

`70.9^@`

Show Worked Solution

`underset~r = ((3),(- 2),(-1)) \ , \ |underset~r| \ = sqrt{3^2 + (-2)^2+(-1)^2} = sqrt14`

`underset~s = ((2),(1),(1)) \ , \ |underset~s| \ = sqrt{2^2 + 1^2 + 1^2} = sqrt6`

`underset~r * underset~s` `= ((3),(-2),(-1)) ((2),(1),(1)) = 6-2-1 = 3`
`underset~r * underset~s` `= |underset~a| |underset~b| \ cos theta`
`3` `= sqrt14 sqrt6 \ cos theta`
`cos theta` `= 3/sqrt84`
`theta` `= cos^(-1) (3/sqrt84)`
  `= 70.9^@ \ text{(1 d.p.)}`

Vectors, EXT2 V1 2022 HSC 11d

A triangle is formed in three-dimensional space with vertices  `A(1,-1,2)`, `B(0,2,-1)`  and  `C(2,1,1)`.

Find the size of `/_ABC`, giving your answer to the nearest degree.  (3 marks)

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`33°`

Show Worked Solution

`vec(BA)=((1),(-1),(2))-((0),(2),(-1))=((1),(-3),(3))`

`abs(vec(BA))=sqrt(1^2+3^2+3^2)=sqrt19`
 

`vec(BC)=((2),(1),(1))-((0),(2),(-1))=((2),(-1),(2))`

`abs(vec(BC))=sqrt(2^2+1^2+2^2)=sqrt9=3`
 

`vec(BA)*vec(BC)=1xx2+ -3xx-1+3xx2=11`

`cos/_ABC` `=(vec(BA)*vec(BC))/(abs(vec(BA)abs(vec(BC))`  
  `=11/(3sqrt19)`  
`:./_ABC` `=cos^(-1)(11/(3sqrt19))`  
  `=32.733…`  
  `=33°\ \ text{(nearest degree)}`  

Vectors, EXT2 V1 2021 HSC 11c

Find the angle between the vectors  `underset~a = ((2),(0),(4))`  and  `underset~b = ((-3),(1),(2))`, giving the angle in degrees correct to 1 decimal place. (3 marks)

Show Answers Only

`83.1^@`

Show Worked Solution

`underset~a = ((2),(0),(4)) \ , \ |underset~a| \ = sqrt{2^2 + 4^2} = sqrt20`

`underset~b = ((-3),(1),(2)) \ , \ |underset~b| \ = sqrt{(-3)^2 + 1^2 + 2^2} = sqrt14`

`underset~a * underset~b` `= ((2),(0),(4)) ((-3),(1),(2)) = – 6 + 0 + 8 = 2`
`underset~a * underset~b` `= |underset~a| |underset~b| \ cos theta`
`2` `= sqrt20 sqrt14 \ cos theta`
`cos theta` `= 2/sqrt280`
`theta` `= cos^(-1) (1/sqrt70)`
  `= 83.1^@ \ text{(1 d.p,)}`

Vectors, EXT2 V1 2020 SPEC2 16 MC

Let  `underset~a = underset~i + 2underset~j + 2underset~k`  and  `underset~b = 2underset~i - 4underset~j + 4underset~k`, where the acute angle between these vectors is  `theta`.

The value of  `sin(2theta)`  is

  1. `1/9`
  2. `(4sqrt5)/9`
  3. `(4sqrt5)/81`
  4. `(8sqrt5)/81`
Show Answers Only

`D`

Show Worked Solution

`underset~a = ((1),(2),(2)), |underset~a| = sqrt(1 + 4 + 4) = 3`

`underset~b = ((2),(−4),(4)) , |underset~b| = sqrt(4 + 16 + 16) = 6`

`underset~a · underset~b = ((1),(2),(2))((1),(−4),(4)) = 2 – 8 + 8 = 2`

`costheta = (underset~a · underset~b)/(|underset~a||underset~b|) = 2/(3 xx 6) = 1/9`
 

`sintheta = sqrt(81 – 1)/9 = (4sqrt5)/9`

`:. sin(2theta)` `= 2 xx 1/9 xx (4sqrt5)/9`
  `= (8sqrt5)/81`

 
`=>D`

Vectors, EXT2 V1 SM-Bank 25

The acute angle  `theta`  is the angle between the vectors  `underset~a = −2underset~i + 2underset~j - underset~k`  and  `underset~b = −4underset~i + 4underset~j + 7underset~k`.

Find the exact value of  `sin(2theta)`.   (2 marks)

Show Answers Only

`(4sqrt2)/9`

Show Worked Solution

`underset~a = ((−2),(2),(−1)), \ |underset~a| = sqrt(4 + 4 + 1) = 3`

`underset~b = ((−4),(4),(7)), \ |underset~b| = sqrt(16 + 16 + 49) = 9`

`underset~a · underset~b = ((−2),(2),(−1))((−4),(4),(7)) = 8 + 8 – 7 = 9`

`costheta = (underset~a · underset~b)/(|underset~a||underset~b|) = 9/(3 xx 9) = 1/3`
 

`sintheta` `= sqrt8/3`
`sin2theta` `= 2sinthetacostheta`
  `= 2 · sqrt8/3 · 1/3`
  `= (4sqrt2)/9`

Vectors, EXT2 V1 2017 SPEC1 5

Relative to a fixed origin, the points `B`, `C` and `D` are defined respectively by the position vectors  `underset~b = underset~i - underset~j + 2underset~k, \ underset~c = 2underset~i - underset~j + underset~k`  and  `underset~d = aunderset~i - 2underset~j`  where `a` is a real constant.

Given that the magnitude of angle `BCD` is  `pi/3`, find `a`.  (4 marks)

Show Answers Only

`a = −2`

Show Worked Solution

`text(Angle between)\ overset(->)(CB)\ text(and)\ overset(->)(CD) = pi/3`

♦ Mean mark 45%.

`overset(->)(CB)` `= (1 – 2)underset~i + (−1 – −1)underset~j + (2 – 1)underset~k`
  `= −underset~i + underset~k`
`overset(->)(CD)` `= (a – 2)underset~i + (−2 – −1)underset~j + (−1 + 0)underset~k`
  `= (a – 2)underset~i – underset~j – underset~k`

 

`overset(->)(CD) · overset(->)(CB)` `= −(a – 2) + 0 – 1`
  `= 1 – a`
  `= |overset(->)(CD)||overset(->)(CB)|cos(pi/3)`

 

`1 – a` `= sqrt((a – 2)^2 + (−1)^2 + (−1)^2)sqrt((−1)^2 + (1)^2)cos(pi/3)`
`1 – a` `= sqrt(a^2 – 4a + 4 + 1 + 1) xx sqrt2 xx 1/2`
`2(1 – a)` `= sqrt(2a^2 – 8a + 12), \ \ a < 1`
`4(1 – a)^2` `= 2a^2 – 8a + 12, \ \ a < 1`
`4(1 – 2a + a^2)` `= 2a^2 – 8a + 12`
`4 – 8a + 4a^2` `= 2a^2 – 8a + 12`
`2a^2` `= 8`
`a^2` `= 4`
`:. a` `= −2\ \ \ (a<1)`

Vectors, EXT2 V1 2017 SPEC1 10

Consider the vectors  `underset ~a = - underset ~i - 2 underset ~j + 3 underset ~k`  and  `underset ~b = 2 underset ~i + c underset ~j + underset ~k`.

Find the value of  `c, \ c in R`, if the angle between  `underset ~a`  and  `underset ~b`  is  `pi/3`.  (4 marks)

Show Answers Only

`c = -3`

Show Worked Solution
`underset ~a ⋅ underset ~b` `= -1 xx 2 + (-2) xx c + 3 xx 1`
  `= -2 – 2c + 3`
  `= 1 – 2c`

 

`1-2c` `= sqrt((-1)^2 + (-2)^2 + 3^3) *sqrt(2^2 + c^2 + 1^2) xx cos (pi/3)`
`1 – 2c` `= 1/2(sqrt 14 ⋅ sqrt(5 + c^2))`
`2 – 4c` `= sqrt(14(5 + c^2))`
`(2 – 4c)^2` `= 14(5 + c^2)`
`4 – 16c + 16c^2` `= 70 + 14c^2`
`2c^2 – 16c – 66` `= 0`
`c^2 – 8c – 33` `= 0`
`(c – 11)(c + 3)` `= 0`

 
`c = 11 or c = -3`

`text(S)text(ince)\ \ 2 – 4c = sqrt(15(5 + c^2))`

`2 – 4c > 0\ \ =>\ \ c<2`

`:. c = -3`

Vectors, EXT2 V1 2014 SPEC1 1

Consider the vector  `underset ~a = sqrt 3 underset ~i - underset ~j - sqrt 2 underset ~k`, where  `underset ~i, underset ~j`  and  `underset ~k`  are unit vectors in the positive directions of the `x, y` and `z` axes respectively.

  1.  Find the unit vector in the direction of  `underset ~a`.  (1 mark)

  2.  Find the acute angle that  `underset ~a`  makes with the positive direction of the `x`-axis.  (2 marks)

  3.  The vector  `underset ~b = 2 sqrt 3 underset ~i + m underset ~j - 5 underset ~k`.

     

     Given that  `underset ~b`  is perpendicular to  `underset ~a,` find the value of `m`.  (2 marks)

Show Answers Only
  1. `1/sqrt 6 (sqrt 3 underset ~i – underset ~j – sqrt 2 underset ~k)`
  2. `theta = 45^@`
  3. `m = 6 + 5 sqrt 2`
Show Worked Solution
i.    `|underset ~a|` `= sqrt((sqrt 3)^2 + (-1)^2 + (-sqrt 2)^2)`
    `= sqrt 6`
`:. hat underset ~a` `= underset ~a/|underset ~a|`
  `= 1/sqrt 6 (sqrt 3 underset ~i – underset ~j – sqrt 2 underset ~k)`

 

Mean mark part (b) 51%.

ii.    `underset ~a ⋅ underset ~i` `= sqrt 3 xx 1 = sqrt 3`
  `underset ~a ⋅ underset ~i` `= |underset ~a||underset ~i| cos theta`
    `= sqrt 6 cos theta`
  `sqrt 3` `= sqrt 6 cos theta`
`cos theta` `= 1/sqrt 2`
`:. theta` `= pi/4 = 45^@`

 

iii.   `underset ~a ⋅ underset ~b = sqrt 3 (2 sqrt 3) + (-1)(m) + (-sqrt 2)(-5) = 0`

`6 – m + 5 sqrt 2` `=0`  
`:. m` `=6 + 5 sqrt 2`  

Vectors, EXT2 V1 SM-Bank 9

Points  `A`, `B` and `C` have position vectors  `underset~a = 2underset~i + underset~j`,  `underset~b = 3underset~i - underset~j + underset~k`  and  `underset~c = -3underset~j + underset~k`  respectively.

Find the cosine of angle `ABC`.   (2 marks)

Show Answers Only

`(-1)/(sqrt6sqrt13)`

Show Worked Solution

`vec(BA) = vec(OA) – vec(OB) = – underset~ i + 2underset~j – underset~k`

♦ Mean mark 48%.

`=>\ |\ vec(BA)\ | = sqrt6`

`vec(BC) = vec(OC) – vec(OB) = -3underset~i – 2underset~j`

`=>\ |\ vec(BC)\ | = sqrt13`
 

`vec(BA).vec(BC)` `= |\ vec(BA)\ ||\ vec(BC)\ |cos angleABC`
`cos angleABC` `= (vec(BA).vec(BC))/(|\ vec(BA)\ ||\ vec(BC)\ |)`
  `= (-3xx-1 + 2 xx – 2)/(sqrt6sqrt13)`
  `= (-1)/(sqrt6sqrt13)`

Vectors, EXT2 V1 SM-Bank 7

If  `theta`  is the angle between  `underset ~a = sqrt 3 underset ~i + 4 underset ~j - underset ~k`  and  `underset ~b = underset ~i - 4 underset ~j + sqrt 3 underset ~k`, then find  `cos(2 theta)`.   (2 marks)

Show Answers Only

`7/25`

Show Worked Solution
`underset ~a ⋅ underset ~b` `= sqrt 3 xx 1 + 4 xx (-4) + (-1) xx sqrt 3`
  `= -16`

 
`|\ underset~a\ | = sqrt20,\ \ |\ underset~b\ | = sqrt20,`

`cos(theta)` `=(underset ~a ⋅ underset ~b)/(|\ underset~a\ |\ |\ underset~b\ | `
  `= (-16)/(sqrt 20 xx sqrt 20)`
  `= -4/5`

 

`cos (2 theta)` `= 2cos^2theta – 1`
  `= 2(-4/5)^2 – 1`
  `= 7/25`

Vectors, EXT2 V1 2011 SPEC2 12 MC

The angle between the vectors  `3underset~i + 6underset~j - 2underset~k`  and  `2underset~i - 2underset~j + underset~k`, correct to the nearest tenth of a degree, is

A.       2.0°

B.     91.0°

C.   112.4°

D.   121.3°

Show Answers Only

`C`

Show Worked Solution

`|3underset~i + 6underset~j – 2underset~k| = sqrt(9 + 36 + 4) = sqrt49 = 7`

`|2underset~i – 2underset~j + underset~k| = sqrt(4 + 4 + 1) = sqrt9 = 3`

`(3underset~i + 6underset~j – 2underset~k) * (2underset~i – 2underset~j + underset~k)`

`= 3 xx 2 + 6 xx (−2) + (−2) xx 1`

`= 6 – 12 – 2`

`= -8`
  

`costheta` `= ((3tildei + 6tildej – 2tildek).(2tildei – 2tildej + tildek))/(|\ 3tildei + 6tildej – 2tildek\ ||\ 2tildei – 2tildej + tildek\ |)`
  `= −8/21`
`:. theta` `= cos^(−1)(−8/12)`
  `~~ 112.4^@`

 
`=> C`