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Vectors, SPEC2 2022 VCAA 11 MC

Consider the vectors  \(\underset{\sim}{\text{a}}=2 \underset{\sim}{\text{i}}-3 \underset{\sim}{\text{j}}+p \underset{\sim}{\text{k}}, \ \underset{\sim}{\text{b}}=\underset{\sim}{\text{i}}+2 \underset{\sim}{\text{j}}-q \underset{\sim}{\text{k}}\)  and  \(\underset{\sim}{\text{c}}=-3 \underset{\sim}{\text{i}}+2 \underset{\sim}{\text{j}}+5 \underset{\sim}{\text{k}}\), where \(p\) and \(q\) are real numbers.

If these vectors are linearly dependent, then

  1. \(8p=5q-35\)
  2. \(5p=8q-35\)
  3. \(8p=-5q-35\)
  4. \(8p=5q+35\)
  5. \(5p=8q+35\)
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\(A\)

Show Worked Solution

\(\text{Linearly dependent when}\ \underset{\sim}{c}=m\underset{\sim}{a}+n\underset{\sim}{b}\)

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

\(m\underset{\sim}{a}+n \underset{\sim}{b}=(2 m+n) \underset{\sim}{i}-(3 m-2 n) \underset{\sim}{j}+(m p-q n) \underset{\sim}{k}\)

\(\text{Equating co-efficients:}\)

  \(2 m+n=-3 \ \ldots\ (1)\) 

  \(-3 m+2 n=2\ \ldots\ (2)\)

  \(p m-q n=5\ \ldots\ (3)\)

\(\text{Solve (1) and (2) by CAS:}\)

  \(m=-\dfrac{8}{7}, n=-\dfrac{5}{7}\)

\(\text { Substitute } m,n \text { into (3): }\)

\begin{aligned}
-\dfrac{8}{7}\,p+\dfrac{5}{7}\,q & =5 \\
8 p & =5 q-35
\end{aligned}

\(\Rightarrow A\)

Vectors, SPEC1 2021 VCAA 6

Consider the three vectors  `underset~a = -underset~i + 6underset~j - 3underset~k, underset~b = 2underset~i - 8underset~j + 5underset~k`  and  `underset~c = 3underset~i + 2underset~j + |1 - p^2|underset~k`,  where  `p`  is a real constant.

Find the values of `p` for which the three vectors are linearly independent.  (4 marks)

Show Answers Only

`p ∈ R\ text(\) {±sqrt5}`

Show Worked Solution

`text(Linearly dependent when)\ \ underset~a = m underset~b + n underset~c`

`text(Equating)\ \ underset~i, underset~j\ \ text(and)\ \ underset~k\ \ text(components:)`

`-1` `= 2m + 3n\ \ …\ \ (1)`  
`6` `= -8m + 2n\ \ …\ \ (2)`  
`-3` `= 5m + |1 – p^2|n\ \ …\ \ (3)`  

 
`text(Mult)\ \ (1) xx 4`

`-4 = 8m + 12n\ \ …\ \ (1)′`

`text(Add)\ \  (2) + (1)′`

`2` `= 14n`
`n` `= 1/7`

 
`text(Substitute)\ \ n = 1/7\ \ text{into (1):}`

`-1` `= 2m + 3/7`
`2m` `= -10/7`
`m` `= -5/7`

 
`text(Substitute)\ \ m, n\ \ text{into (3):}`

`-3` `= – 25/7 + |1 – p^2| · 1/7`
`-21` `= -25 + |1 – p^2|`
`|1-p^2|` `= 4`
`1-p^2` `= ±4`
`p^2` `= 5\ \ \ (p^2 != -3)`
`p` `= ±sqrt5`

 
`:.\ text(Vectors are linearly independent for)\ \ p ∈ R\ text(\) {±sqrt5}`

Vectors, SPEC2 2020 VCAA 13 MC

The vectors  `underset~a = underset~i + 2underset~j - underset~k, \ underset~b = lambdaunderset~i + 3underset~j + 2underset~k`  and  `underset~c = underset~i + underset~k`  will be linearly dependent when the value of `lambda` is

  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
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`E`

Show Worked Solution

`text(Linearly dependent:)\ \ underset~b = m underset~a + n underset~c`

`m + n` `= lambda\ …\ (1)`
`2m` `= 3\ …\ (2)`
`−m + n` `= 2\ …\ (3)`

 
`m = 3/2\ \ text{(using (2))}`

`n = 2 + 3/2 = 7/2\ \ text{(using (3))}`

`:. lambda = 3/2 + 7/2 = 5`

`=>E`

Vectors, SPEC2-NHT 2019 NHT 13 MC

For the vectors  `underset~a = underset~i + 3underset~j - underset~k`,  `underset~b = −underset~i - 4underset~j + 2underset~k`  and  `underset~c = −underset~i - 6underset~j + lambdaunderset~k`  to be linearly dependent, the value of  `lambda`  must be

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4
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`E`

Show Worked Solution

`text(Linearly dependent if)\ ∃ lambda, m, n,\ text(such that)`

`((−1),(−6),(lambda)) = m((1),(3),(−1)) + n((−1),(−4),(2))`

`text{Solve (by CAS):}`

`lambda = 4, m = 2, n = 3`

`=>\ E`

Vectors, SPEC1 2019 VCAA 6

Find value of  `d`  for which the vectors  `underset~a = 2underset~i - 3underset~j + 4underset~k`,  `underset~b = −2underset~i + 4underset~j - 8underset~k`  and  `underset~c = −6underset~i + 2underset~j + dunderset~k`  are linearly dependent.  (3 marks)

Show Answers Only

`16`

Show Worked Solution

`text(Linear dependent:)\ \ m underset~a + n underset~b = underset~c`

`2m – 2n` `= −6\ \ …\ (1)`
`−3m + 4n` `= −2\ \ …\ (2)`
`4m – 8n` `= d\ \ …\ (3)`

 
`(1) xx 2`

`4m – 4n = −12\ \ …\ (1′)`

`(2) + (1′)`

`m = −10`

`text(Substitute)\ \ m = −10\ \ text(into (1))`

`−20 – 2n` `= −6`
`n` `= −7`

 

`:.d` `= 4 xx −10 – 8 xx −7`
  `= 16`

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)
Show Answers Only
  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 2013 VCAA 17 MC

Consider the four vectors  `underset~a = underset~j + 3underset~k`,  `underset~b = underset~i - 4underset~k`,  `underset~c = 3underset~j - k`  and  `underset~d = 2underset~j + underset~k`.

Which one of the following is a linearly dependent set of vectors?

  1. `{underset~a, underset~b, underset~c}`
  2. `{underset~a, underset~c, underset~d}`
  3. `{underset~a, underset~b, underset~d}`
  4. `{underset~b, underset~c, underset~d}`
  5. `{underset~a, underset~b}`
Show Answers Only

`B`

Show Worked Solution

`underset ~b\ \ text(is the only vector which has an)\ \ i\ \ text(component.)`
 

`text(The other three vectors are on the same plane are therefore`

`text(linearly dependent.)`

`=> B`

Vectors, SPEC2 2016 VCAA 11 MC

Let  `underset ~a = 3 underset ~i + 2 underset ~j + alpha underset ~k`  and  `underset ~b = 4 underset ~i - underset ~j + alpha^2 underset ~k`, where  `alpha`  is a real constant.

If the scalar resolute of  `underset ~a`  in the direction of  `underset ~b`  is  `74/sqrt 273`,  then  `alpha`  equals

A.   1

B.   2

C.   3

D.   4

E.   5 

Show Answers Only

`D`

Show Worked Solution
`underset ~a ⋅ underset ~ hat b` `= (3 xx 4 + 2 xx -1 + alpha xx alpha^2)/sqrt(4^2 + 1^2 + alpha^4)`
`74/sqrt 273` `= (12 – 2 + alpha^3)/ sqrt(17 + alpha^4)`
`74 sqrt(17 + alpha^4)` `=sqrt 273 (10 + alpha^3)`

 

`alpha = 1:\ \ text(LHS:)\ sqrt 273 xx 11 = 74 sqrt 18\ \ text(RHS)`  ✖

`alpha = 2:\ \ sqrt 273(18) = 74 sqrt 33`  ✖

`alpha = 3:\ \ sqrt 273 xx 37 = 74 sqrt 98`  ✖

`alpha = 4:\ \ sqrt 273 xx 74 = 74 sqrt 273`  ✔

`=>  D`

Vectors SPEC1 2016 VCAA 5

Consider the vectors  `underset ~a = 3 underset ~i + 5 underset ~j - 2 underset ~k,\ underset ~b = underset ~i - 2 underset ~j + 3 underset ~k`  and  `underset ~c = underset ~i + d\  underset ~k`, where  `d`  is a real constant.

  1. Find the vector resolute of  `underset ~a`  in the direction of  `underset ~b`.  (2 marks)
  2. Find the value of  `d`  if the vectors are linearly dependent.  (2 marks)
Show Answers Only
  1.  `-13/14 (underset ~i – 2 underset ~j + 3 underset ~k)`
  2.  `d = 1`
Show Worked Solution

a.   `underset ~hat b = 1/sqrt14(underset ~i – 2 underset ~j + 3 underset ~k)`

`text(Scalar resolute)\ \ (underset ~a ⋅ underset ~hat b)`

`=1/sqrt14 (3xx1 – 5xx2 -2xx3)`

`=-13/sqrt14`
 

`text(Vector resolute)`

`(underset ~a ⋅ underset ~hat b) underset ~hat b` `= -13/sqrt14 xx 1/sqrt14(underset ~i – 2 underset ~j + 3 underset ~k)`  
   `=-13/14 (underset ~i – 2 underset ~j + 3 underset ~k)`  

 

b.    `text(If linearly dependent)\ \ =>\ \ alpha underset ~a + beta underset ~b = underset ~c`

`(3 alpha + beta) underset ~i + (5 alpha – 2 beta) underset ~j + (-2 alpha + 3 beta) underset ~k = underset ~i + d\ underset ~k`

`3 alpha + beta` `= 1\ \ …\ (1)`
`5 alpha – 2 beta` `=0\ \ …\ (2)`
`3 beta – 2 alpha` `= d\ \ …\ (3)`

 
`2 xx (1) + (2):`

`11 alpha = 2 => alpha = 2/11`

`text(Substitute)\ \ alpha = 2/11\ \ text{into (1):}`

`6/11 + beta = 1 \ => \ beta = 5/11`

`:.d` `=3(5/11) – 2(2/11)`  
  `=15/11 – 4/11`  
  `=1`  

Vectors, SPEC2-NHT 2017 VCAA 14 MC

Given that the vectors  `underset ~a = underset ~i + underset ~j - underset ~k,\ underset ~b = 2 underset ~i - underset ~j + 2 underset ~k`  and  `underset ~c = lambda underset ~i - underset ~j + 4 underset ~k`  are linearly dependent, the value of  `lambda`  is

A.  –10

B.  –8

C.    2

D.    4

E.    8

Show Answers Only

`E`

Show Worked Solution

`alpha underset ~a + beta underset ~j = underset ~c`

`(1):\ \ (alpha + 2 beta) = lambda,`

`(2):\ \ alpha – beta = -1,`

`(3):\ \ -alpha + 2 beta = 4`
 

`(2) + (3):`

`beta = 3`
 

`text(Substitute)\ \ beta = 3\ \ text(into)\ \ (2):`

`alpha – 3` `= -1`
`alpha` `= 2`

 
`text(Substitute)\ \ alpha = 2,\ beta = 3\ \ text(into)\ \ (1):`

`lambda` `= 2 + 6`
  `= 8`

 
`=>   E`

Vectors, SPEC2 2017 VCAA 11 MC

The vectors  `underset~a = 2underset~i + 3underset~j + d underset~k, \ underset~b = underset~i + underset~j - 4underset~k`  and  `underset~c = 2underset~i + underset~j - 2underset~k`  where `d` is a real constant, are linearly dependent if

  1. `d = −10`
  2. `d ∈ R\ text(\)\ {−14}`
  3. `d = −14`
  4. `d = R\ text(\)\ {−10}`
  5. `d ∈ R`
Show Answers Only

`C`

Show Worked Solution

`alphaunderset~a + betaunderset~b = underset~c`

`2alpha + beta = 2\ \ …\ (1)`

`3alpha + beta = 1\ \ …\ (2)`

`dalpha + 4beta = −2\ \ …\ (3)`

 
`(2) – (1):`

`alpha = −1`

`=> beta = 2+2=4`
 

`text(Substitute)\ \ alpha = −1, beta = 4\ \ text{into (3):}`

`-d – 16` `= −2`
`d` `= −14`

 
`=> C`