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Vectors, SPEC2 2022 VCAA 12 MC

Consider the vectors  \(\underset{\sim}{\text{u}}(x)=-\text{cosec}(x) \underset{\sim}{\text{i}}+\sqrt{3} \underset{\sim}{\text{j}}\)  and  \(\underset{\sim}{\text{v}}(x)=\text{cos}(x) \underset{\sim}{\text{i}}+\underset{\sim}{\text{j}}\).

If \(\underset{\sim}{\text{u}}(x)\) is perpendicular to \(\underset{\sim}{\text{v}}(x)\), then possible values for \(x\) are

  1. \(\dfrac{\pi}{6}\) and \(\dfrac{7 \pi}{6}\)
  2. \(\dfrac{\pi}{3}\) and \(\dfrac{4 \pi}{3}\)
  3. \(\dfrac{5 \pi}{6}\) and \(\dfrac{11 \pi}{6}\)
  4. \(\dfrac{2 \pi}{3}\) and \(\dfrac{5 \pi}{3}\)
  5. \(\dfrac{\pi}{6}\) and \(\dfrac{5 \pi}{6}\)
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\(A\)

Show Worked Solution

\(\underset{\sim}{u} \perp \underset{\sim}{v} \ \Rightarrow \ \underset{\sim}{u} \cdot \underset{\sim}{v}=0\)

\(-\text{cosec}(x) \cos (x)+\sqrt{3}=0\)

\begin{aligned}
-\dfrac{\cos (x)}{\sin (x)}+\sqrt{3} & =0 \\
\tan (x) & =\dfrac{1}{\sqrt{3}}
\end{aligned}

\(\text {Solve by (CAS)}: \ x=\dfrac{\pi}{6}, \dfrac{7 \pi}{6}\)

\(\Rightarrow A\)

Vectors, SPEC2 2023 VCAA 18 MC

What value of \(k\), where \(k \in R\), will make the following planes perpendicular?

\(\Pi_1: \ 2 x-k y+3 z=1\)

\(\Pi_2: \ 2 k x+3 y-2 z=4\)

  1. 2
  2. 4
  3. 6
  4. 8
  5. 10
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\(C\)

Show Worked Solution

\(\underset{\sim}{p_1} = 2\underset{\sim}{i}-k\underset{\sim}{j}+3\underset{\sim}{k}\)

\(\underset{\sim}{p_2} = 2k\underset{\sim}{i}+3\underset{\sim}{j}-2\underset{\sim}{k}\)

\(\text{Solve}\ \ \underset{\sim}{p_1} \cdot \underset{\sim}{p_2}=0\ \ \text{for}\ k\ \text{(by calc)}: \)

\(4k-3k-6=0\ \ \Rightarrow \ k=6\)

\(\Rightarrow C\)

Vectors, SPEC2 2023 VCAA 14 MC

Let  \(\underset{\sim}{\text{a}}=\underset{\sim}{\text{i}}+\underset{\sim}{\text{j}}, \underset{\sim}{\text{b}}=\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}\)  and  \(\text{c}=\underset{\sim}{\text{i}}+2 \underset{\sim}{\text{j}}+3 \underset{\sim}{\text{k}}\).

If \(\underset{\sim}{\text{n}}\) is a unit vector such that  \( \underset{\sim} {\text{a}} \cdot \underset{\sim} {\text{n}}=0\)  and  \( \underset{\sim} {\text{b}} \cdot \underset{\sim} {\text{n}}=0\), then  \(\big{|} \underset{\sim} {\text{c}} \cdot \underset{\sim} {\text{n}}\big{|} \)  is equal to

  1. 2
  2. 3
  3. 4
  4. 5
  5. 6
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\(B\)

Show Worked Solution

\(\underset{\sim}{\text{n}}\ \text{is perpendicular to}\ \underset{\sim}{\text{a}}\ \text{and}\ \underset{\sim}{\text{b}} .\)

\(\underset{\sim}{\text{a}}\ \text{and}\ \underset{\sim}{\text{b}}\ \text{both lie on the}\ \underset{\sim}{\text{i}}\ –\ \underset{\sim}{\text{j}}\ \text{plane.}\)

\(\text{Possible (unit vector)}\ \underset{\sim}{\text{n}} = (0,0,1) \)

\(\therefore \big{|} \underset{\sim} {\text{c}} \cdot \underset{\sim} {\text{n}}\big{|} = 3\)

\(\Rightarrow B\)

Vectors, SPEC1 2020 VCAA 5

Let  `underset ~ a = 2 underset ~i-3 underset ~j + underset ~k`  and  `underset ~b = underset ~i + m underset ~j-underset ~k`, where `m` is an integer.

The vector resolute of  `underset ~a`  in the direction of  `underset ~b`  is  `-11/18 (underset ~i + m underset ~j-underset ~k)`.

  1. Find the value of `m`.   (3 marks)

  2. Find the component of  `underset ~a`  that is perpendicular to  `underset ~b`.   (1 mark)

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  1. `m = 4`
  2. `47/18 underset ~i-5/9 underset ~j + 7/18 underset ~k`

Show Worked Solution

a.    `underset ~b ⋅ underset ~a` `= ((1), (m), (-1))((2), (-3), (1)) = 2-3m-1 = 1-3m`
  `|underset ~b|^2` `= 1^2 + m^2 + (-1)^2 = m^2 + 2`
`(underset ~b ⋅ underset ~a)/|underset ~b|^2 ⋅ underset ~b` `= -11/18 underset ~b`
`(1 -3m)/(m^2 + 2)` `= -11/18`
`18-54m` `= -11m^2-22`
`11m^2-54m + 40` `=0`  
`(11m-10)(m-4)` `=0`  

 
`:. m= 4\ \ (m != 10/11, \ m ∈ Z)`

♦♦ Mean mark part (b) 28%.

 

b.    `underset ~a_(⊥ underset ~b)` `= underset ~a + 11/18 (underset ~i + 4 underset ~j-underset ~k)`
    `= 2 underset ~i + 11/18 underset ~i-3 underset ~j + 44/18 underset ~j + underset ~k-11/18 underset ~k`
    `= 47/18 underset ~i-5/9 underset ~j + 7/18 underset ~k`

Vectors, SPEC2 2012 VCAA 15 MC

The vectors  `underset~a = 2underset~i + m underset~j - 3underset~k`  and  `underset~b = m^2underset~i - underset~j + underset~k`  are perpendicular for

  1. `m = −2/3`  and  `m = 1`
  2. `m = −3/2`  and  `m = 1`
  3. `m = 2/3`  and  `m = −1`
  4. `m = 3/2`  and  `m = −1`
  5. `m = 3`  and  `m = −1`
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`D`

Show Worked Solution

`underset ~a ⊥ underset ~b\ \ =>\ \ underset ~a ⋅ underset ~b=0`

`underset ~a ⋅ underset ~b` `= 2m^2 + m(-1) + (-3)(1)`
`0` `= 2m^2 – m – 3`
`0` `= (2m – 3)(m + 1)`

 
`:. m = 3/2, quad m = -1`

`=> D`

Vectors, SPEC1 2011 VCAA 9

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)

  3. i.  Calculate  `3 underset ~c - underset ~a.`  (1 mark)

  4. ii. Hence find a value of `m` such that  `underset ~a, underset ~b`  and  `underset ~c`  are linearly dependent.  (1 mark)

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  1. `+- sqrt 7`
  2. `1/2`
  3. i.  `2 tilde i + 4 tilde j – 5 tilde k`
  4. ii. `-5/2`

Show Worked Solution

a.    `|underset~b|` `= sqrt(1^2 + 2^2 + m^2)` `= 2sqrt3`
    `1 + 4 + m^2` `= 4 xx 3`
    `m^2` `= 7`
    `m` `= ±sqrt7`

 

b.    `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`  

 

c.i.   `3 underset ~c – underset ~a`

`=3underset~i – underset~i + 3underset~j + underset~j -3underset~k – 2underset~k`

`=2underset~i + 4underset~j-5underset~k`

 

c.ii.   `text(Linear dependence)\ \ =>\ \ 3underset~c – underset~a = t underset~b,\ \ t in RR`

`2 tilde i + 4 tilde j – 5 tilde k= t (tilde i + 2 tilde j + m tilde k)`

`=> t = 2 and tm = -5,`

`:. m=-5/2`

Vectors, SPEC2 2014 VCAA 16 MC

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 `m` and `n` respectively are

  1. `5 sqrt 3, sqrt 3/3`
  2. `5 sqrt 3, sqrt 3`
  3. `−5 sqrt 3, sqrt 3`
  4. `sqrt 93, (5 sqrt 93)/93`
  5. `5, 1`
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`A`

Show Worked Solution

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

`10` `= sqrt(4^2 + m^2 + (-3)^2)`
`m` `= 5sqrt3\ \ \ (text{by CAS,}\ \ m in R^+)`

 

`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 = sqrt 3/3\ \ \ (text{by CAS,}\ \ n in R^+)`

`=> A`

Vectors, SPEC2 2016 VCAA 12 MC

If  `underset ~a = -2 underset ~i - underset ~j + 3 underset ~k`  and  `underset ~b = -m underset ~i + underset ~j + 2 underset ~k`, where  `m`  is a real constant, the vector  `underset ~a - underset ~b`  will be perpendicular to vector  `underset ~b`  where  `m`  equals

A.   0 only

B.   2 only

C.   0 or 2

D.   4.5

E.   0 or −2

Show Answers Only

`C`

Show Worked Solution
`underset ~a – underset ~b` `= (m – 2) underset ~i – 2 underset ~j + underset ~k `
`(underset ~a – underset ~b) ⋅ underset ~b` `= -m(m – 2) – 2(1) + 1(2) = 0`
`0` `= -m^2 + 2m – 2 + 2`
`0` `= 2m – m^2`
`0` `= m(2 – m) \ => \ m = 0 \ or \ 2`

`=>  C`

Vectors, SPEC1 2014 VCAA 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`.
  4. 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

a.    `|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%.

b.    `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^@`

 

c.   `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`