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Vectors, SPEC2 2024 VCAA 14 MC

Consider the vectors \(\underset{\sim}{r}\) and \(\underset{\sim}{s}\)  where  \(\abs{\underset{\sim}{ r }}=9\)  and  \(\underset{\sim}{ s }=2 \underset{\sim}{ i }-2 \underset{\sim}{ j }+\underset{\sim}{k}\).

If the vector resolute of \(\underset{\sim}{r}\) in the direction of \(\underset{\sim}{s}\) is equal to  \(-4 \underset{\sim}{i}+4 \underset{\sim}{j}-2 \underset{\sim}{k}\),  then the scalar resolute of \(\underset{\sim}{s}\) in the direction of \(\underset{\sim}{r}\) is equal to

  1. \(-18\)
  2. \(-2\)
  3. \(2\)
  4. \(3\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Vector resolute of \(\underset{\sim}{r}\) in the direction of \(\underset{\sim}{s}\):}\)

\(\underset{\sim}{s}=\dfrac{\underset{\sim}{r} \cdot \underset{\sim}{s}}{\abs{\underset{\sim}{s}}^2} \underset{\sim}{s}=-4 \underset{\sim}{i}+4 \underset{\sim}{j}-2 \underset{\sim}{k}\)

\(\abs{\underset{\sim}{s}}=\sqrt{4+4+1}=3\)

\(\dfrac{\underset{\sim}{r} \cdot \underset{\sim}{s}}{3^2}(2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k})=-4 \underset{\sim}{i}+4 \underset{\sim}{j}-2 \underset{\sim}{k}\)

\(\dfrac{\underset{\sim}{r} \cdot \underset{\sim}{s}}{9}=-2 \ \ \Rightarrow \ \ \underset{\sim}{r} \cdot \underset{\sim}{s}=-18\)

\(\text{Scalar resolute of \(\underset{\sim}{s}\) in the direction of \(\underset{\sim}{r}\):}\)

\(\dfrac{\underset{\sim}{r} \cdot \underset{\sim}{s}}{\abs{\underset{\sim}{r}}}=\dfrac{-18}{9}=-2\)
 

\(\Rightarrow B\)

♦♦ Mean mark 36%.

Filed Under: Basic Concepts and Calculations Tagged With: Band 5, smc-1176-40-Vector resolute

Vectors, SPEC2 2021 VCAA 13 MC

The scalar resolute of vector  `underset~a`  in the direction of vector  `underset~b`  is  `-4`.

If  `underset~b = – sqrt3 underset~i`,  the vector resolute of  `underset~a`  in the direction of  `underset~b`  is

  1. `-4underset~i`
  2. `-3underset~i`
  3. `1/sqrt3 underset~i`
  4. `3underset~i`
  5. `4underset~i`
Show Answers Only

`E`

Show Worked Solution

`text(Scalar resolute of)\ underset~a\ text(is direction of)\ underset~b = -4`

♦ Mean mark 48%.

`underset~a · overset^underset~b = -4`

`underset~b = -sqrt3 underset~i \ => \ overset^underset~b = -underset~i`

`text(Vector resolute of)\ underset~a\ text(in direction of)\ underset~b`

`(underset~a · overset^underset~b)overset^underset~b = 4underset~i`

`=>\ E`

Filed Under: Basic Concepts and Calculations Tagged With: Band 5, smc-1176-40-Vector resolute

Vectors, SPEC2 2020 VCAA 14 MC

The magnitude of the component of the force  `underset~F = underset~i + 6underset~j - 18underset~k`  that acts in the direction  `underset~d = 2underset~i - 3underset~j - 6underset~k`  is

  1. `92/19`
  2. `92/7`
  3. `124/7`
  4. `92/11`
  5. `18/7`
Show Answers Only

`B`

Show Worked Solution
`|underset~d|` `= sqrt(2^2 + (−3)^2 + (−6)^2) = 7`
`underset~overset^d` `= 1/7(2underset~i – 3underset~j – 6underset~k)`
`underset~F · underset~overset^d` `= 1/7((1),(6),(−18))((2),(−3),(−6))`
  `= 92/7`

 
`=>B`

Filed Under: Basic Concepts and Calculations Tagged With: Band 4, smc-1176-40-Vector resolute

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)

    --- 6 WORK AREA LINES (style=lined) ---

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

    --- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

  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`

Filed Under: Basic Concepts and Calculations Tagged With: Band 4, Band 5, smc-1176-30-Perpendicular vectors, smc-1176-40-Vector resolute

Vectors, SPEC2-NHT 2019 VCAA 11 MC

The vector resolute of  `underset~a = 2underset~i - underset~j + 3underset~k`  that is perpendicular to  `underset~b = underset~i + underset~j - underset~k`  is

  1. `−2/3(underset~i + underset~j - underset~k)`
  2. `−2/3(2underset~i - underset~j + 3underset~k)`
  3. `1/3(8underset~i - underset~j + 7underset~k)`
  4. `underset~i - 2underset~j + 4underset~k`
  5. `underset~i + underset~j + 2underset~k`
Show Answers Only

`C`

Show Worked Solution

`underset~a = ((2),(−1),(3)),\ \ underset~b = ((1),(1),(−1))`

`|underset~b| = sqrt3`

`underset~a · underset~b = 2 – 1 – 3 = −2`
 

`text(Vector resolute)\ underset~a\ text(onto)\ underset~b`

`= (underset~a · underset~overset^b)underset~overset^b`

`= −2/sqrt3 xx 1/sqrt3((1),(1),(−1))`

`= −2/3((1),(1),(−1))`
 

`text(Vector resolute)\ underset~a ⊥ underset~b`

`= underset~a – (underset~a · underset~overset^b)underset~overset^b`

`= ((2),(−1),(3)) + 2/3((1),(1),(−1))`

`= 1/3((8),(−1),(7))`

`=>\ C`

Filed Under: Basic Concepts and Calculations Tagged With: Band 4, smc-1176-40-Vector resolute

Vectors, SPEC2 2019 VCAA 12 MC

The vector resolute of  `underset~i + underset~j - underset~k`  in the direction of  `m underset~i + n underset~j + p underset~k`  is  `2underset~i - 3underset~j + underset~k`,  where  `m, n`  and  `p`  are real constants.

The values of  `m, n`  and  `p`  can be found by solving the equations.

  1. `(m(m + n - p))/(m^2 + n^2 + p^2) = 2, \ (n(m + n - p))/(m^2 + n^2 + p^2) = −3 and (p(m + n - p))/(m^2 + n^2 + p^2) = 1`
  2. `(m(m + n - p))/(m^2 + n^2 + p^2) = 1, \ (n(m + n - p))/(m^2 + n^2 + p^2) = 1 and (p(m + n - p))/(m^2 + n^2 + p^2) = −1`
  3. `m + n - p = 6, \ m + n - p = −9 and m + n - p = −3`
  4. `m + n - p = 3m, \ m + n - p = 3n and m + n - p = −3p`
  5. `m + n - p = 2sqrt3, \ m + n - p = −3sqrt3 and m + n - p = sqrt3`
Show Answers Only

`A`

Show Worked Solution

`underset~a = underset~i + underset~j – underset~k`

`underset~b = m underset~i + n underset~j + p underset~k`

`overset^b = (m underset~i + n underset~j + p underset~k)/sqrt(m^2 + n^2 + p^2)`

`text(Vector resolute)`

`= (underset~a · overset^b)overset^b`

`= (m + n – p)/(m^2 + n^2 + p^2) (m underset~i + n underset~j + p underset~k)`

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

`=>A`

Filed Under: Basic Concepts and Calculations Tagged With: Band 5, smc-1176-40-Vector resolute

Vectors, SPEC2 2012 VCAA 17 MC

If  `underset ~u = 2 underset ~i - 2 underset ~j + underset ~k`  and  `underset ~v = 3 underset ~i - 6 underset ~j + 2 underset ~k`, the vector resolute of  `underset ~v`  in the direction of  `underset ~u`  is

A.   `20/49(3 underset ~i - 6 underset ~j + 2 underset ~k)`

B.   `20/3(2 underset ~i - 2underset ~j + underset ~k)`

C.   `20/7(3 underset ~i - 6 underset ~j + 2 underset ~k)`

D.   `20/9(2 underset ~i - 2 underset ~j + underset ~k)`

E.   `1/9(−2 underset ~i + 2 underset ~j - underset ~k)`

Show Answers Only

`D`

Show Worked Solution

`tildeu = 2tildei – 2tildej + tildek,qquadtildev = 3tildei – 6tilde j + 2tildek`

`|\ tildeu\ |` `= sqrt(2^2 + (−2)^2 + 1^2)=3`
`hatu` `= = 1/3(2tildei – 2tildej + tildek)`

 

`tildev*hatu\ \ (text(Scalar resolute of)\ tildev\ text(in the direction of)\ tildeu)`

`= (3tildei – 6tildej + 2tildek)* 1/3(2tildei – 2tildej + tildek)`

`= 20/3`
 

`(tildev*hatu)hatu\ \ (text(Vector resolute of)\ tildev\ text(in the direction of)\ tildeu)`

`= 20/3 xx 1/3(2tildei – 2tildej + tildek)`

`= 20/9(2tildei – 2tildej + tildek)`
 

`=> D`

Filed Under: Basic Concepts and Calculations Tagged With: Band 4, smc-1176-40-Vector resolute

Vectors, SPEC2 2015 VCAA 15 MC

The component of the force  `underset~F = aunderset~i + bunderset~j`, where `a` and `b` are non-zero real constants, in the direction of the vector  `underset~w = underset~i + underset~j`, is

A.   `((a + b)/2)underset~w`

B.   `underset~F/(a + b)`

C.   `((a + b)/(a^2 + b^2))underset~F`

D.   `(a + b)underset~w`

E.   `((a + b)/sqrt2)underset~w` 

Show Answers Only

`A`

Show Worked Solution
`hatw` `= tildew/sqrt(1+1)`
  `= (tildei + tildej)/sqrt2`

`tildeF*hatw = (a + b)/sqrt2`

♦ Mean mark 49%.

`(tildeF*hatw)hatw` `= ((a + b)/sqrt2) tildew/sqrt2`
  `= ((a + b)/2) tildew`

 
`=> A`

Filed Under: Basic Concepts and Calculations Tagged With: Band 5, smc-1176-40-Vector resolute

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)

    --- 5 WORK AREA LINES (style=lined) ---

  2. Find the value of  `d`  if the vectors are linearly dependent.  (2 marks)

    --- 6 WORK AREA LINES (style=lined) ---

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`  

Filed Under: Basic Concepts and Calculations Tagged With: Band 4, smc-1176-40-Vector resolute, smc-1176-50-Linear dependence

Vectors, SPEC2-NHT 2017 VCAA 12 MC

If  `underset ~u = 3 underset ~i + 6 underset ~j - 2 underset ~k`  and  `underset ~v = 2 underset ~i + 2 underset ~j - underset ~k`, then the vector resolute of  `underset ~u`  in the direction of  `underset ~v`  is

A.   `20/7 (3 underset ~i + 6 underset ~j - 2 underset ~k)`

B.   `20/9 (2 underset ~i + 2 underset ~j - underset ~k)`

C.   `20/49 (3 underset ~i + 6 underset ~j - 2 underset ~k)`

D.   `20/3 (2 underset ~i + 2 underset ~j - underset ~k)`

E.   `3/7 (3 underset ~i + 6 underset ~j - 2 underset ~k)`

Show Answers Only

`B`

Show Worked Solution

`text(Using CAS:)`

`(underset ~u ⋅ underset ~hat v) underset ~ hat v`

`= 20/9 (2 underset ~i + 2 underset ~j – underset ~k)`

 
`=>   B`

Filed Under: Basic Concepts and Calculations Tagged With: Band 5, smc-1176-40-Vector resolute

Vectors, SPEC1-NHT 2018 VCAA 2

Let  `underset ~a = 3 underset ~i - 2 underset ~j + m underset ~k`  and  `underset ~b = 2 underset ~i - underset ~j + 3 underset ~k`, where  `m in R`.

Find the value(s) of `m` such that the magnitude of the vector resolute of  `underset ~a` parallel to `underset ~b`  is equal to  `sqrt 14`.  (3 marks)

Show Answers Only

`m = (-22)/3, 2`

Show Worked Solution

`underset ~a ⋅ underset ~ hat b = (3 xx 2 +(-2 xx – 1) + m xx 3)/sqrt(2^2 + (-1)^2 + 3^2) = +- sqrt 14`

`(8 + 3m)/sqrt 14` `= +- sqrt 14`
`8 + 3m` `= +- 14`
`3m` `= -8 +- 14`
`m` `= (-8 +- 14)/3`
`:. m` `= – 22/3, 2`

Filed Under: Basic Concepts and Calculations Tagged With: Band 4, smc-1176-40-Vector resolute

Vectors, SPEC2 2018 VCAA 14 MC

The scalar resolute of  `underset ~a = 3 underset ~i - 2 underset ~k`  in the direction of  `underset ~b = - underset ~i + 2 underset ~j + 3 underset ~k`  is

A.   `-(9 sqrt 13)/13`

B.   `-9/14(-underset ~i + 2 underset ~j + 3 underset ~k)`

C.   `-(9 sqrt 14)/14`

D.   `-9/13 (3 underset ~i - 2 underset ~k)`

E.   `- sqrt 14/2` 

Show Answers Only

`C`

Show Worked Solution
`underset ~a ⋅ underset ~hat b`

`= (3 xx (-1) + 0 xx 2 + (-2) xx 3)/sqrt((-1)^2 + 2^2 + 3^2)`
  `= (-3-6)/sqrt 14`
  `= -9/sqrt 14 xx sqrt 14/sqrt 14`
  `= -(9 sqrt 14)/14`

 
`=>  C`

Filed Under: Basic Concepts and Calculations Tagged With: Band 4, smc-1176-40-Vector resolute

Vectors, SPEC2 2017 VCAA 13 MC

Given the vectors  `underset~a = 3underset~i - 4underset~j + 12underset~k`  and  `underset~b = 2underset~i + 2underset~j - underset~k`, the vector resolute of  `underset~a` in the direction of  `underset~b`  is

  1.  `−14/3`
  2.  `−14/3(2underset~i - 2underset~j - underset~k)`
  3.  `−14/9(2underset~i + 2underset~j - underset~k)`
  4.  `−14/13`
  5.  `−14/169(3underset~i - 4underset~j + 12underset~k)`
Show Answers Only

`C`

Show Worked Solution
`(underset~a · overset^(underset~b))overset^(underset~b)` `= (3 xx 2 + −4 xx 2 + 12 xx −1)/sqrt(2^2 + 2^2 +(−1)^2) xx (2underset~i + 2underset~j – underset~k)/sqrt(2^2 + 2^2 + (−1)^2)`
  `= – 14/9 (2underset~i + 2underset~j – underset~k)`

`=> C`

Filed Under: Basic Concepts and Calculations Tagged With: Band 3, smc-1176-40-Vector resolute

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