Eleven numbers are randomly chosen from the set of integers, `S`, where
`S = {1, 2, 3, 4, ..., 20}`
Prove that the sum of two of the eleven numbers randomly selected must equal 21. (2 marks)
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Combinatorics, EXT1 A1 SM-Bank 11
A multiple choice quiz asks students 4 questions. Each question has three possible answers, a, b or c, and students must attempt each question.
How many students must do the quiz to ensure that at least two sets of answers are identical? (2 marks)
Statistics, SPEC2-NHT 2019 VCAA 6
A paint company claims that the mean time taken for its paint to dry when motor vehicles are repaired is 3.55 hours, with a standard deviation of 0.66 hours.
Assume that the drying time for the paint follows a normal distribution and that the claimed standard deviation value is accurate.
- Let the random variable `barX` represent the mean time taken for the paint to dry for a random sample of 36 motor vehicles.
Write down the mean and standard deviation of `barX`. (2 marks)
At a car crash repair centre, it was found that the mean time taken for the paint company's paint to dry on randomly selected vehicles was 3.85 hours. The management of this crash repair centre was not happy and believed that the claim regarding the mean time taken for the paint to dry was too low. To test the paint company's claim, a statistical test was carried out.
- Write down suitable null and alternative hypotheses `H_0` and `H_1` respectively to test whether the mean time taken for the paint to dry is longer than claimed. (1 mark)
- Write down an expression for the `p` value of the statistical test and evaluate it correct to three decimal places. (2 marks)
- Using a 1% level of significance, state with a reason whether the crash repair centre is justified in believing that the paint company's claim of mean time taken for its paint to dry of 3.55 hours is too low. (1 mark)
- At the 1% level of significance, find the set of sample mean values that would support the conclusion that the mean time taken for the paint to dry exceeded 3.55 hours. Give your answer in hours, correct to three decimal places. (2 marks)
- If the true time taken for the paint to dry is 3.83 hours, find the probability that the paint company's claim is not rejected at the 1% level of significance, assuming the standard deviation for the paint to dry is still 0.66 hours. Give your answer correct to two decimal places. (1 mark)
Mechanics, SPEC2-NHT 2019 VCAA 5
A pallet of bricks weighing 500 kg sits on a rough plane inclined at an angle of `α°` to the horizontal, where `tan(α°) = (7)/(24)`. The pallet is connected by a light inextensible cable that passes over a smooth pulley to a hanging container of mass `m` kilograms in which there is 10 L of water. The pallet of bricks is held in equilibrium by the tension `T` newtons in the cable and a frictional resistance force of 50 `g` newtons acting up and parallel to the plane. Take the weight force exerted by 1 L of water to be `g` newtons.
- Label all forces acting on both the pallet of bricks and the hanging container on the diagram above, when the pallet of bricks is in equilibrium as described. (1 mark)
- Show that the value of `m` is 80. (3 marks)
Suddenly the water is completely emptied from the container and the pallet of bricks begins to slide down the plane. The frictional resistance force of 50 `g` newtons acting up the plane continues to act on the pallet.
- Find the distance, in metres, travelled by the pallet after 10 seconds. (3 marks)
- When the pallet reaches a velocity of `3\ text(ms)^-1`, water is poured back into the container at a constant rate of 2 L per second, which in turn retards the motion of the pallet moving down the plane. Let `t` be the time, in seconds, after the container begins to fill.
- i. Write down, in terms of `t`, an expression for the total mass of the hanging container and the water it contains after `t` seconds. Give your answer in kilograms. (1 mark)
- ii. Show that the acceleration of the pallet down the plane is given by `(text(g)(5 - t))/(t + 290)\ text(ms)^-2` for `t ∈[0, 5)`. (2 marks)
- iii. Find the velocity of the pallet when `t = 4`. Give your answer in metres per second, correct to one decimal place. (2 marks)
Show Answers Only
-
`qquad`
- `text(Proof(Show Worked Solution))`
- `(25 text(g))/(29)`
- i. `80 + 2t`
ii. `text(Proof (Show Worked Solution))`
iii. `3.4\ text(ms)^-1`
Show Worked Solution
a.
b. `text(Resolving vertical forces on container:)`
`T – (m + 10)g = 0 \ …\ (1)`
`text(Resolving forces on plane:)`
`tan α = (7)/(24) \ => \ sin α = (7)/(25)`
`text(Solve for m:)`
| `(m + 10)g` | `= 500 text(g) · (7)/(25) – 50 text(g)` |
| `m + 10` | `= 140 – 50` |
| `:. \ m` | `= 80` |
c. `text(Resolving vertical forces on container:)`
`T – 80 g = 80 a \ …\ (1)`
`text(Resolving forces on plane:)`
`500 g sin α – (T + 50 g) = 500 a`
`90 g – T = 500 a \ …\ (2)`
`text(Add) \ (1) + (2)`
| `10 g` | `= 580 a` |
| `a` | `= (g)/(58)` |
| `s` | `= ut + (1)/(2) at^2` |
| `= 0 + (1)/(2) · (g)/(58) + 10^2` | |
| `= (25g)/(29)` |
d.i. `m = 80 + 2t`
d.ii. `text(Resolving vertical forces on container:)`
`T – (80+2t)g = (80+2t)a \ …\ (1)`
`text(Resolving forces on plane:)`
`90g – T = 500a \ …\ (2)`
`text(Add)\ (1) + (2)`
| `(90 – 80 – 2t)g` | `= (500 + 80 + 2t)a` |
| `(10 – 2t)g` | `=(580 + 2t)a` |
| `a` | `= (g(5 – t))/(t + 290) ms^-2` |
d.iii. `(dv)/(dt) = (g(5 – t))/(t + 290)`
`v = int (dv)/(dt)\ dt = 295 log_e ((t + 290)/(290)) – g t + c`
`text(When)\ t = 0, v = 3\ \ text{(given)} \ => \ c = 3`
`:. \ v = 295 log_e ((t+290)/(290)) – g t + 3`
`:. \ v(4) = 3.4\ text(ms)^-1`
Calculus, SPEC2-NHT 2019 VCAA 3
The vertical cross-section of a barrel is shown above. The radius of the circular base (along the `x`-axis) is 30 cm and the radius of the circular top is 70 cm. The curved sides of the cross-section shown are parts of the parabola with rule `y = (x^2)/(80) - (45)/(4)`. The height of the barrel is 50 cm.
a. i. Show that the volume of the barrel is given by `pi int_0^50 (900 + 80 y)\ dy`. (1 marks)
ii. Find the volume of the barrel in cubic centimetres. (1 marks)
The barrel is initially full of water. Water begins to leak from the bottom of the barrel such that `(dV)/(dt) = (-8000pi sqrth)/(A)` cubic centimetres per second, where after `t` seconds the depth of the water is `h` centimetres, the volume of water remaining in the barrel is `V` cubic centimetres and the uppermost surface area of the water is `A` square centimetres.
b. Show that `(dV)/(dt) = (-400 sqrth)/(4h + 45)`? (2 marks)
c. Find `(dh)/(dt)` in terms of `h`. Express your answer in the form `(-a sqrth)/(pi(b + ch)^2)`, where `a, b` and `c` are positive integers. (3 marks)
d. Using a definite integral in terms of `h`, find the time, in hours, correct to one decimal place, taken for the barrel to empty. (2 marks)
Calculus, SPEC2-NHT 2019 VCAA 2
Consider the function `f` with rule `f(x) = (x^2 + x + 1)/(x^2-1)`.
- State the equations of the asymptotes of the graph of `f`. (2 marks)
- State the coordinates of the stationary points and the point of inflection. Give your answers correct to two decimal places. (2 marks)
- Sketch the graph of `f` from `x = -6` to `x = 6` (endpoint coordinates are not required) on the set of axes below, labeling the turning points and the point of inflection with their coordinates correct to two decimal places. Label the asymptotes with their equations. (3 marks)
Consider the function `f_k` with rule `f_k(x) = (x^2 + x + k)/(x^2-1)` where `k ∈ R`.
- For what values of `k` will `f_k` have no stationary points? (2 marks)
- For what value of `k` will the graph of `f_k` have a point of inflection located on the `y`-axis? (1 marks)
Show Answers Only
| i. `x = 1, x = -1, y = 1` |
|
ii. `(-3.73, 0.87) \ text(and) \ (-0.27, -0.87)` `(-5.52, 0.88)` |
| iii. |
 |
iv. `-2`
v. `-1`
Show Worked Solution
i. `f(x) = 1 + (x + 2)/((x + 1)(x-1))`
`:. \ text(Asymptotes at)\ \ x = 1, x = -1, y = 1`
ii. `f′(x) = (-(x^2 + 4x + 1))/(x^2-1)^2`
`text(Solve:) \ \ f^{′}(x) = 0 \ => \ x = -2 ± sqrt3`
`text(SP’s at) \ (-3.73, 0.87) \ \ text(and)\ \ (-0.27, -0.87)`
`f^{″}(x) = (-2x^3-12x^2-6x-4)/(x^2-1)^3`
`text(Solve:) \ \ f^{″}(x) = 0 \ => \ x = -5.52`
`text(POI at) \ \ (-5.52, 0.88)`
iii.
iv. `f_k^{′}(x) = (-x^2-2(k+1) x-1)/(x^2-1)^2`
`text(If no SP’s,) \ \ Δ < 0`
| `[-2( k + 1)]^2-4(-1)(-1)` | `< 0` |
| `4k^2 + 8k + 4-4` | `< 0` |
| `4k(k + 2)` | `< 0` |
`:. \ -2 < k < 0`
v. `f_k^{″}(x) = (2(x^3 + 3(k + 1) x^2 + 3x + k + 1))/((x^2-1)^3)`
`text(Solve:)\ \ f_k^{″}(x) = 0`
`k = -1`
Statistics, EXT1 S1 EQ-Bank 13
On average, batsmen playing cricket in a T20 competition play a scoring shot two out of every three balls.
In a regular season, a total of 1200 overs that each contain six balls, are bowled.
Estimate how many overs would have at least five scoring shots. (3 marks)
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Statistics, EXT1 S1 EQ-Bank 12
A manufacturer of pool cue tips knows that 4% of pool cue tips produced in its factory need to be scrapped.
A random sample of 10 cue tips produced in the factory is examined.
- What is the probability that exactly 3 cue tips need to be scrapped?
Give your answer correct to 3 decimal places. (1 mark)
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- What is the probability that, at most, 2 cue tips need to be scrapped?
Give your answer correct to 3 decimal places. (2 marks)
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Vectors, EXT1 V1 2019 SPEC2-N 4
A snowboarder at the Winter Olympics leaves a ski jump at an angle of `theta` degrees to the horizontal, rises up in the air, performs various tricks and then lands at a distance down a straight slope that makes an angle of 45° to the horizontal, as shown below.
Let the origin `O` of a cartesian coordinate system be at the point where the snowboarder leaves the jump, with a unit vector in the positive `x` direction being represented by `underset~i` and a unit vector in the positive `y` direction being represented by `underset~j`. Distances are measured in metres and time is measured in seconds.
The position vector of the snowboarder `t` seconds after leaving the jump is given by
`underset~r (t) = (6t - 0.01t^3) underset~i + (6 sqrt3 t - 4.9t^2 + 0.01t^3) underset~j , \ t ≥ 0`
- Find the angle `theta °`. (2 marks)
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- Find the speed, in metres per second, of the snowboarder when she leaves the jump at `O`. (1 mark)
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- Find the maximum height above `O` reached by the snow boarder. Give your answer in metres, correct to one decimal place. (2 marks)
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- Show that the time spent in the air by the snowboarder is `(60(sqrt3 + 1))/(49)` seconds. (3 marks)
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Vectors, SPEC2-NHT 2019 VCAA 4
A snowboarder at the Winter Olympics leaves a ski jump at an angle of `theta` degrees to the horizontal, rises up in the air, performs various tricks and then lands at a distance down a straight slope that makes an angle of 45° to the horizontal, as shown below.
Let the origin `O` of a cartesian coordinate system be at the point where the snowboarder leaves the jump, with a unit vector in the positive `x` direction being represented by `underset~i` and a unit vector in the positive `y` direction being represented by `underset~j`. Distances are measured in metres and time is measured in seconds.
The position vector of the snowboarder `t` seconds after leaving the jump is given by
`underset~r (t) = (6t - 0.01t^3) underset~i + (6 sqrt3 t - 4.9t^2 + 0.01t^3) underset~j , \ t ≥ 0`
- Find the angle `theta °`. (2 marks)
- Find the speed, in metres per second, of the snowboarder when she leaves the jump at `O`. (1 mark)
- Find the maximum height above `O` reached by the snow boarder. Give your answer in metres, correct to one decimal place. (2 marks)
- Show that the time spent in the air by the snowboarder is `(60(sqrt3 + 1))/(49)` seconds. (3 marks)
- Find the total distance the snowboarder travels while airborne. Give your answer in metres, correct to two decimal places. (2 marks)
Complex Numbers, SPEC2-NHT 2019 VCAA 1
In the complex plane, `L` is the with equation `|z + 2| = |z - 1 - sqrt3 i|`.
- Verify that the point (0, 0) lies on `L`. (1 marks)
- Show that the cartesian form of the equation of `L` is `y = - sqrt3 x`. (2 marks)
- The line `L` can also be expressed in the form `|z - 1| = |z - z_1|`, where `z_1 ∈ C`.
Find `z_1` in cartesian form. (2 marks)
- Find, in cartesian form, the points(s) of intersection of `L` and the graph of `|z| = 4`. (2 marks)
- Sketch `L` and the graph of `|z| = 4` on the Argand diagram below. (2 marks)
- Find the area of the sector defined by the part of `L` where `text(Re)(z) ≥ 0`, the graph of `|z| = 4` where `text(Re)(z) ≥ 0`, and imaginary axis where `text(Im)(z) > 0`. (1 marks)
Show Answers Only
- `text(Proof(See Worked Solution))`
- `text(Proof(See Worked Solution))`
- `-(1)/(2) – (sqrt3)/(2) i`
- `(2,-2 sqrt3) text(and) (-2, 2 sqrt3)`
-
- `(20pi)/(3)`
Show Worked Solution
a. `text(Substitute)\ \ z = 0 + 0i\ \ text(into both sides:)`
`text(LHS) = |2| = 2`
`text(RHS) = |-1 – sqrt3i| = sqrt{(-1)^2 + (-sqrt3)^2} = 2`
`:. (0,0)\ \ text(lies on both sides.)`
| b. | `|x + yi + 2|` | `= |x + yi – 1 – sqrt3 i|` |
| `|(x + 2) + yi|` | `= |(x – 1) + (y – sqrt3) i|` | |
| `sqrt(x^2 + 4x + 4 + y^2` | `= sqrt(x^2 – 2x + 1 + y^2 -2 sqrt3 y + 3` | |
| `x^2 + 4x + 4 + y^2` | `= x^2 – 2x + 4 – 2 sqrt3 y + y^2` | |
| `6x` | `= -2 sqrt3 y` | |
| `y` | `= -(3)/(sqrt3) x` | |
| `y` | `= -sqrt3 x` |
c.
`m_text(perp) = (sqrt2)/(3) , text(through)\ (1, 0)`
`y = (sqrt3)/(3) (x – 1)\ …\ L_1`
`text(Intersection) \ L\ text(and) \ L_1,`
`text(Solve:) \ (sqrt3)/3 (x-1) = -sqrt3 x\ \ \ text{(by CAS)}`
`=> x = (1)/(4) , y = -(sqrt3)/(4) \ \ \ text{(point}\ Ptext{)}`
`P(x_1,y_1) \ text(is midpoint of) \ \ z_1 \ text(and) \ \ (1, 0):`
`(x_1 + 1)/(2) = (1)/(4) \ => \ x_1 = -(1)/(2)`
`(y_1 + 0)/(2) = -sqrt3/(4) \ => \ y_1 = -sqrt3/(2)`
`:. \ z_1 = -(1)/(2) – (sqrt3)/(2) i`
d. `|z| = 4 => \ text(circle, centre) \ (0,0), \ text(radius) = 4`
`x^2 + y^2 = 16\ …\ (1)`
`y = – sqrt3 x\ …\ (2)`
`text(Substitute)\ (2) \ text(into) \ (1)`
`x^2 + 3x^2 = 16`
`x = ±2`
`:. \ text(Intersection at)\ (2, – 2 sqrt3) \ text(and) \ (-2, 2 sqrt3)`
e.
f.
| `text(Area)` | `= (5)/(12) xx pi xx 4^2` |
| `= (20pi)/(3)` |
Statistics, SPEC2-NHT 2019 VCAA 18 MC
Consider a random variable `X` with probability density function
`f(x) = {(2x,, 0<= x <= 1),(0,, x < 0\ \ text(or)\ \ x > 1):}`
If a large number of samples, each of size 100, is taken from his distribution, then the distribution of the sample means, `barX`, will be approximately normal with mean `E(barX) = 2/3` and standard deviation `text(sd)(barX)` equal to
- `sqrt2/60`
- `sqrt2/6`
- `1/180`
- `1/18`
- `sqrt2/30`
Vectors, SPEC2-NHT 2019 VCAA 14 MC
The position vector `underset~r(t)` of a mass of 3 kg after `t` seconds, where `t >= 0`, is given by `underset~r(t) = 10tunderset~i + (16t^2 - 4/3t^3)underset~j`.
The force, in newtons, acting on the mass when `t = 2` seconds is
- `16underset~j`
- `32underset~j`
- `48underset~j`
- `30underset~i + 144underset~j`
- `16`
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
- 0
- 1
- 2
- 3
- 4
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)
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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, SPEC2-NHT 2019 VCAA 12 MC
Given that `theta` is the acute angle between the vectors `underset~a = −2underset~i + 2underset~j - underset~k` and `underset~b = −4underset~i + 4underset~j + 7underset~k`, then `sin(2theta)` is equal to
- `2sqrt2`
- `(4sqrt2)/9`
- `(2sqrt2)/9`
- `(2sqrt2)/3`
- `(4sqrt2)/3`
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` |
`=>\ B`
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
- `−2/3(underset~i + underset~j - underset~k)`
- `−2/3(2underset~i - underset~j + 3underset~k)`
- `1/3(8underset~i - underset~j + 7underset~k)`
- `underset~i - 2underset~j + 4underset~k`
- `underset~i + underset~j + 2underset~k`
Calculus, SPEC2-NHT 2019 VCAA 10 MC
Euler's method, with a step size of 0.1, is used to approximate the solution of the differential equation `1/y(dy)/(dx) = cos(x)`, with `y = 2` when `x = 0`.
When `x = 0.2`, the value obtained for `y`, correct to four decimal places, is
- 2.2000
- 2.3089
- 2.3098
- 2.4189
- 2.4199
Calculus, SPEC2-NHT 2019 VCAA 9 MC
With a suitable substitution, `int_1^2sqrt(5x - 1)\ dx` can be expressed as
- `5int_1^2sqrtu\ du`
- `1/5int_1^2sqrtu\ du`
- `5int_4^9sqrtu\ du`
- `1/5int_4^9sqrtu\ du`
- `5int_4^9sqrt(5u - 1)\ du`
Calculus, SPEC2-NHT 2019 VCAA 8 MC
The total area enclosed between the `x`-axis and the graph of `f(x) = |x^3| - x^2 - |x|` is closest to
- −2.015
- −1.008
- 1.008
- 2.015
- 2.824
Calculus, SPEC2-NHT 2019 VCAA 7 MC
The gradient of the line that is perpendicular to the graph of a relation at any point `P(x, y)` is half the gradient of the line joining `P` and the point `Q(−1,1)`.
The relation satisfies the differential equation
- `(dy)/(dx) = (y - 1)/(2(x + 1))`
- `(dy)/(dx) = (2(x + 1))/(y + 1)`
- `(dy)/(dx) = (2(x - 1))/(y + 1)`
- `(dy)/(dx) = (x + 1)/(2(1 - y))`
- `(dy)/(dx) = (2(x + 1))/(1 - y)`
Complex Numbers, SPEC2-NHT 2019 VCAA 4 MC
Which one of the following statements is false for `z_1, z_2 ∈ C`?
- `z^(−1) = (barz)/(|z|^2), z != 0`
- `|z_1 + z_2| > |z_1| + |z_2|`
- `(z_1)/(z_2) = (z_1barz_2)/(|z_2|^2), z_2 != 0`
- `|z_1z_2| = |z_1||z_2|`
- `|(z_1)/(z_2)| = (|z_1|)/(|z_2|), z_2 != 0`
Show Worked Solution
`text(Consider each option:)`
`A.\ \ 1/z xx barz/barz = barz/(|z|^2)`
`B.`
`text(Using vectors:)\ |z_1 + z_2| < |z_1| + |z_2|`
`=>\ text(Incorrect (by triangle inequality))`
`C, D and E\ text(can be shown to be correct.)`
`=>\ B`
Graphs, SPEC2-NHT 2019 VCAA 3 MC
The maximal domain and range of the function `f(x) = acos^(−1)(bx) + c`, where `a`, `b` and `c` are real constants with `a > 0`, `b < 0` and `c > 0`, are respectively
- `[0,pi] and [−a,a]`
- `[0,pi] and [−a + c,a + c]`
- `[−1/b,1/b] and [c, api + c]`
- `[1/b, −1/b] and [c, api + c]`
- `[1/b, −1/b] and [−api + c, api + c]`
Functions, 2ADV F1 SM-Bank 21 MC
A circle with centre `(a,−2)` and radius 5 units has equation
`x^2-6x + y^2 + 4y = b` where `a` and `b` are real constants.
The values of `a` and `b` are respectively
A. −3 and 38
B. 3 and 12
C. −3 and −8
D. 3 and 18
Graphs, SPEC2-NHT 2019 VCAA 2 MC
The curve given by `x = 3sec(t) + 1` and `y = 2tan(t) - 1` can be expressed in cartesian form as
- `((y + 1)^2)/4 - ((x - 1)^2)/9 = 1`
- `((x + 1)^2)/3 - ((y - 1)^2)/2 = 1`
- `((y + 1)^2)/9 - ((x - 1)^2)/4 = 1`
- `((x - 1)^2)/3 + ((y + 1)^2)/2 = 1`
- `((x - 1)^2)/9 - ((y + 1)^2)/4 = 1`
Vectors, EXT1 V1 EQ-Bank 28
A projectile is fired from a canon at ground level with initial velocity `sqrt300` ms−1 at an angle of 30° to the horizontal.
The equations of motion are `(d^2x)/(dt^2) = 0` and `(d^2y)/(dt^2) = −10`.
- Show that `x = 15t`. (1 mark)
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- Show that `y = 5sqrt3t - 5t^2`. (2 marks)
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- Hence find the Cartesian equation for the trajectory of the projectile. (1 mark)
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Show Worked Solution
i.
`dotx = sqrt300\ cos30^@ = 10sqrt3 xx sqrt3/2 = 15\ text(ms)^(−1)`
`x = int 15\ dt = 15t + C_1`
`text(When)\ \ t = 0, \ x = 0 \ => \ C_1 = 0`
`:. x = 15t`
ii. `doty = int −10\ dt = −10t + C_1`
`text(When)\ \ t = 0, \ doty = 10sqrt3 sin30^@ = 5sqrt3 \ => \ C_1 = 5sqrt3`
`:. doty = 5sqrt3 – 10t`
`y = int doty\ dt = 5sqrt3t – 5t^2 + C_2`
`text(When)\ \ t = 0, \ y = 0 \ => \ C_2 = 0`
`:. y = 5sqrt3t – 5t^2`
iii. `x = 15t \ => \ t = x/15`
| `y` | `= 5sqrt3 xx x/15 – 5(x/15)^2` | |
| `:.y` | `=sqrt3/3 x – (x^2)/45` |
Combinatorics, EXT1 A1 EQ-Bank 10
By using the fact that `(1 + x)^11 = (1 + x)^3(1 + x)^8`, show that
`((11),(5)) = ((8),(5)) + ((3),(1))((8),(4)) + ((3),(2))((8),(3)) + ((8),(2))`. (3 marks)
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Combinatorics, EXT1 A1 EQ-Bank 7
Show `\ ^nC_k = \ ^(n-1)C_(k-1) + \ ^(n-1)C_k`. (2 marks)
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Calculus, EXT1 C3 2019 SPEC1-N 9
i. Show that `tan((5pi)/(12)) = sqrt3 + 2`. (2 marks)
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ii.
Hence, find the area bounded by the graph of `f(x) = (2)/(x^2 - 4x + 8)` shown above, the `x`-axis and the lines `x = 0` and `x = 2 sqrt3 +6`. (4 marks)
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Calculus, SPEC1-NHT 2019 VCAA 9
- Show that `tan((5pi)/(12)) = sqrt3 + 2`. (2 marks)
-
Hence, find the area bounded by the graph of `f(x) = (2)/(x^2 - 4x + 8)` shown above, the `x`-axis and the lines `x = 0` and `x = 2 sqrt3 +6`. (4 marks)
Calculus, SPEC1-NHT 2019 VCAA 8
Find the length of the arc of the curve defined by `y = (x^4)/(4) + (1)/(8x^2) + 3` from `x = 1` to `x = 2`. Give your answer in the form `(a)/(b)`, where `a` and `b` are positive integers. (4 marks)
Calculus, SPEC1-NHT 2019 VCAA 7
Given that `3x^2 + 2xy + y^2 = 6`, find `(d^2 y)/(dx^2)` at the point (1, 1). (5 marks)
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Calculus, SPEC1-NHT 2019 VCAA 6
Part of the graph of `y = (2)/(sqrt(x^2-4x+3))`, where `x > 3`, is shown below.
Find the volume of the solid of revolution formed when the graph of `y = (2)/(sqrt(x^2-4x+3))` from `x = 4` to `x = 6` is rotated about the `x`-axis. Give your answer in the form `a log_e(b)` where `a` and `b` are real numbers. (5 marks)
Vectors, EXT2 V1 2019 SPEC2 4
The base of a pyramid is the parallelogram `ABCD` with vertices at points `A(2,−1,3), B(4,−2,1), C(a,b,c)` and `D(4,3,−1)`. The apex (top) of the pyramid is located at `P(4,−4,9)`.
- Find the values of `a, b` and `c`. (2 marks)
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- Find the cosine of the angle between the vectors `overset(->)(AB)` and `overset(->)(AD)`. (2 marks)
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- Find the area of the base of the pyramid. (2 marks)
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Show Worked Solution
| i. | `overset(->)(AB)` | `= (4 – 2)underset~i + (−2 + 1)underset~j + (1 – 3)underset~k` |
| `= 2underset~i – underset~j – 2underset~k` |
`text(S)text(ince)\ ABCD\ text(is a parallelogram)\ => \ overset(->)(AB)= overset(->)(DC)`
`overset(->)(DC) = (a – 4)underset~i + (b – 3)underset~j + (c + 1)underset~k`
`a – 4 = 2 \ => \ a = 6`
`b – 3 = −1 \ => \ b = 2`
`c + 1 = −2 \ => \ c = −3`
ii. `overset(->)(AB) = 2underset~i – underset~j – 2underset~k`
`overset(->)(AD) = 2underset~i + 4underset~j – 4underset~k`
| `cos angleBAD` | `= (overset(->)(AB) · overset(->)(AD))/(|overset(->)(AB)| · |overset(->)(AD)|)` |
| `= (4 – 4 + 8)/(sqrt(4 + 1 + 4) · sqrt(4 + 16 + 16))` | |
| `= 4/9` |
| iii. | `1/2 xx text(Area)_(ABCD)` | `= 1/2 ab sin c` |
| `text(Area)_(ABCD)` | `= |overset(->)(AB)| · |overset(->)(AD)| *sin(cos^(−1)\ 4/9)` |
| `:. text(Area)_(ABCD)` | `= 3 · 6 · sqrt65/9` |
| `= 2sqrt65\ text(u²)` |
Vectors, EXT2 V1 2019 SPEC1-N 5
A triangle has vertices `A(sqrt3 + 1, –2, 4), \ B(1, –2, 3)` and `C(2, –2, sqrt3 + 3)`.
- Find angle `ABC` (3 marks)
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- Find the area of the triangle. (2 marks)
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Vectors, SPEC1-NHT 2019 VCAA 5
A triangle has vertices `A(sqrt3 + 1, –2, 4), \ B(1, –2, 3)` and `C(2, –2, sqrt3 + 3)`.
- Find angle `ABC` (3 marks)
- Find the area of the triangle. (2 marks)
Calculus, SPEC1-NHT 2019 VCAA 4
Evaluate `int_(e^3) ^(e^4) (1)/(x log_e (x))\ dx`. (3 marks)
Statistics, SPEC1-NHT 2019 VCAA 3
The number of cars per day making a U-turn at a particular location is known to be normally distributed with a standard deviation of 17.5. In a sample of 25 randomly selected days, a total of 1450 cars were observed making the U-turn.
- Based on this sample, calculate an approximate 95% confidence interval for the number of cars making the U-turn each day. Use an integer multiple of the standard deviation in your calculations. (3 marks)
- The average number of U-turns made at the location is actually 60 per day.
Find an approximation, correct to two decimal places, for the probability that on 25 randomly selected days the average number of U-turns is less than 53. (1 mark)
Show Worked Solution
a. `barx = (1450)/(25) = 58`
`(σ)/(sqrtn) = (17.5)/(sqrt25) = (35)/(2xx5) = (7)/(2)`
| `text(Limit:)\ ` | `(58 – 2 xx (7)/(2) \ , \ 58 + 2 xx (7)/(2)) ` |
| `= (51, 65)` |
b. `μ = 60 \ , \ (σ)/(sqrtn) = (7)/(2)`
`\ barX ∼ N (60, ((7)/(2))^2)`
| `text(Pr) (barX < 53)` | `= text(Pr) (z <–2)` |
| `= (0.05)/(2)` | |
| `= 0.025` | |
| `≈ 0.02 \ \ text{(round down)}` |
`text{(0.03 was also accepted as a correct answer)}`
Complex Numbers, EXT2 N2 2019 SPEC1-N 1
A cubic polynomial has the form `p(z) = z^3 + bz^2 + cz + d, \ z ∈ C`, where `b, c, d ∈ R`.
Given that a solution of `p(z) = 0` is `z_1 = 3 - 2i` and that `p(–2) = 0`, find the values of `b, c` and `d`. (4 marks)
Complex Numbers, SPEC2-NHT 2019 VCAA 2
A cubic polynomial has the form `p(z) = z^3 + bz^2 + cz + d, \ z ∈ C`, where `b, c, d ∈ R`.
Given that a solution of `p(z) = 0` is `z_1 = 3 - 2i` and that `p(–2) = 0`, find the values of `b, c` and `d`. (4 marks)
Mechanics, SPEC1-NHT 2019 VCAA 1
A 10 kg mass is placed on a rough plane that inclined at 30° to the horizontal, as shown in the diagram below. A force of 40 N is applied to the mass up the slope and parallel to the slope. There is also a frictional resistance force of magnitude `F` that opposes the motion of the mass.
- Find the magnitude of the frictional resistance force, in newtons, acting up the slope if the force is just sufficient to stop the mass from sliding down the slope. (2 marks)
- An additional force of magnitude `P` newtons is applied to the mass up the slope and parallel to the slope. The sum of the additional force and the frictional resistance force of magnitude `F` that now acts down the slope is such that it is just sufficient to stop the mass from sliding up the slope. (2 marks)
Show Worked Solution
a.
| `40 + F` | `= 10g sin30°` |
| `F` | `= 98 xx 0.5 – 40` |
| `= 9\ text(N)` |
b. `text(Frictional force)\ F\ text(acts down slope)`
| `40 + P` | `= 10g sin30° + F` |
| `P` | `= 5g + F – 40` |
| `= 49 + 9 – 40` | |
| `= 18\ text(N)` |
Calculus, 2ADV C4 2006 HSC 2bii
Evaluate `int_0^3 (8x)/(1 + x^2) \ dx`. (3 marks)
Calculus, 2ADV C4 2004 HSC 3bii
Find `int x/(x^2-3)\ dx`. (2 marks)
Calculus, 2ADV C2 SM-Bank 6
Differentiate with respect to `x`:
`log_e x^x`. (2 marks)
Calculus, 2ADV C2 SM-Bank 4
Differentiate `5^(x^2)5x`. (2 marks)
Trigonometry, 2ADV* T1 2011 HSC 24c
A ship sails 6 km from `A` to `B` on a bearing of 121°. It then sails 9 km to `C`. The
size of angle `ABC` is 114°.
Copy the diagram into your writing booklet and show all the information on it.
- What is the bearing of `C` from `B`? (1 mark)
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- Find the distance `AC`. Give your answer correct to the nearest kilometre. (2 marks)
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- What is the bearing of `A` from `C`? Give your answer correct to the nearest degree. (3 marks)
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Show Worked Solution
STRATEGY: This deserves repeating again: Draw North-South parallel lines through major points to make the angle calculations easier!
| i. |  |
`text(Let point)\ D\ text(be due North of point)\ B`
| `/_ABD` | `=180-121\ text{(cointerior with}\ \ /_A text{)}` |
| `=59^@` | |
| `/_DBC` | `=114-59` |
| `=55^@` |
`:. text(Bearing of)\ \ C\ \ text(from)\ \ B\ \ text(is)\ 055^@`
ii. `text(Using cosine rule:)`
| `AC^2` | `=AB^2+BC^2-2xxABxxBCxxcos/_ABC` |
| `=6^2+9^2-2xx6xx9xxcos114^@` | |
| `=160.9275…` | |
| `:.AC` | `=12.685…\ \ \ text{(Noting}\ AC>0 text{)}` |
| `=13\ text(km)\ text{(nearest km)}` |
iii. `text(Need to find)\ /_ACB\ \ \ text{(see diagram)}`
MARKER’S COMMENT: The best responses showed clear working on the diagram.
| `cos/_ACB` | `=(AC^2+BC^2-AB^2)/(2xxACxxBC)` |
| `=((12.685…)^2+9^2-6^2)/(2xx(12.685..)xx9)` | |
| `=0.9018…` | |
| `/_ACB` | `=25.6^@\ text{(to 1 d.p.)}` |
`text(From diagram,)`
`/_BCE=55^@\ text{(alternate angle,}\ DB\ text(||)\ CE text{)}`
`:.\ text(Bearing of)\ A\ text(from)\ C`
| `=180+55+25.6` | |
| `=260.6` | |
| `=261^@\ text{(nearest degree)}` |
Trigonometry, 2ADV* T1 2005 HSC 27c
The bearing of `C` from `A` is 250° and the distance of `C` from `A` is 36 km.
- Explain why `theta` is `110^@`. (1 mark)
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- If `B` is 15 km due north of `A`, calculate the distance of `C` from `B`, correct to the nearest kilometre. (3 marks)
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Show Worked Solution
i. `text(There is 360)^@\ text(about point)\ A`
| `:.theta + 250^@` | `= 360^@` |
| `theta` | `= 110^@` |
| ii. |  |
| `a^2` | `= b^2 + c^2 − 2ab\ cos\ A` |
| `CB^2` | `= 36^2 + 15^2 − 2 xx 36 xx 15 xx cos\ 110^@` |
| `= 1296 + 225 −(text(−369.38…))` | |
| `= 1890.38…` | |
| `:.CB` | `= 43.47…` |
| `= 43\ text{km (nearest km)}` |
Trigonometry, 2ADV* T1 2007 26a
The diagram shows information about the locations of towns `A`, `B` and `Q`.
- It takes Elina 2 hours and 48 minutes to walk directly from Town `A` to Town `Q`.
Calculate her walking speed correct to the nearest km/h. (1 mark)
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- Elina decides, instead, to walk to Town `B` from Town `A` and then to Town `Q`.
Find the distance from Town `A` to Town `B`. Give your answer to the nearest km. (2 marks)
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- Calculate the bearing of Town `Q` from Town `B`. (1 mark)
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Show Worked Solution
i. `text(2 hrs 48 mins) = 168\ text(mins)`
| `text(Speed)\ text{(} A\ text(to)\ Q text{)}` | `= 15/168` |
| `= 0.0892…\ text(km/min)` |
| `text(Speed)\ text{(in km/hr)}` | `= 0.0892… xx 60` |
| `= 5.357…\ text(km/hr)` | |
| `= 5\ text(km/hr)\ text{(nearest km/hr)}` |
| ii. |  |
`text(Using cosine rule)`
| `AB^2` | `= 15^2 + 10^2 – 2 xx 15 xx 10 xx cos 87^@` |
| `= 309.299…` | |
| `AB` | `= 17.586…` |
| `= 18\ text(km)\ text{(nearest km)}` |
`:.\ text(The distance from Town)\ A\ text(to Town)\ B\ text(is 18 km.)`
| iii. |  |
| `/_CAQ` | `= 31^@\ \ \ text{(} text(straight angle at)\ A text{)}` |
| `/_AQD` | `= 31^@\ \ \ text{(} text(alternate angle)\ AC\ text(||)\ DQ text{)}` |
| `/_DQB` | `= 87 – 31 = 56^@` |
| `/_QBE` | `= 56^@\ \ \ text{(} text(alternate angle)\ DQ \ text(||)\ BE text{)}` |
`:.\ text(Bearing of)\ Q\ text(from)\ B`
`= 180 + 56`
`= 236^@`
Vectors, EXT1 V1 EQ-Bank 26
`OABC` is a quadrilateral.
`P`, `Q`, `R` and `S` divide each side of the quadrilateral in half as shown below.
Prove, using vectors, that `PQRS` is a parallelogram. (3 marks)
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Show Worked Solution
`text(Consider diagonal)\ \ overset(->)(OB):`
`overset(->)(OB) = overset(->)(OA) + overset(->)(AB) = overset(->)(OC)+ overset(->)(CB)`
`overset(->)(PQ) = overset(->)(PA) + overset(->)(AQ) = 1/2(overset(->)(OA) + overset(->)(AB)) = 1/2overset(->)(OB)`
`overset(->)(SR) = overset(->)(SC) + overset(->)(CR) = 1/2(overset(->)(OC) + overset(->)(CB)) = 1/2overset(->)(OB)`
`:.overset(->)(PQ) = overset(->)(SR)`
`text(Consider diagonal)\ \ overset(->)(AC):`
`overset(->)(AC) = overset(->)(AB) + overset(->)(BC) = overset(->)(AO) + overset(->)(OC)`
`overset(->)(QR) = overset(->)(QB) + overset(->)(BR) = 1/2(overset(->)(AB) + overset(->)(BC)) = 1/2overset(->)(AC)`
`overset(->)(PS) = overset(->)(PO) + overset(->)(OS) = 1/2(overset(->)(AO) + overset(->)(OC)) = 1/2overset(->)(AC)`
`:. overset(->)(QR) = overset(->)(PS)`
`text(S)text(ince)\ PQRS\ text(has equal opposite sides,)`
`PQRS\ text(is a parallelogram.)`
Vectors, EXT1 V1 EQ-Bank 24
`PQRS` is a parallelogram, where `overset(->)(PQ) = underset~a` and `vec(PS) = underset~b`
Prove, using vectors, that the sum of the squares of the lengths of the diagonals is equal to the sum of the squares of the lengths of the sides. (3 marks)
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Show Worked Solution
`overset(->)(PR) = underset~a + underset~b,\ \ overset(->)(SQ) = underset~a – underset~b`
`overset(->)(PQ) = overset(->)(SR) = underset~a`
`overset(->)(PS) = overset(->)(QR) = underset~b`
`text(Prove)\ \ |underset~a + underset~b|^2 + |underset~a – underset~b|^2 = 2|underset~a|^2 + 2|underset~b|^2`
| `|underset~a + underset~b|^2 + |underset~a – underset~b|^2` | `= (underset~a + underset~b) · (underset~a + underset~b) + (underset~a – underset~b)(underset~a – underset~b)` |
| `= underset~a · underset~a + underset~b · underset~b + 2underset~a · underset~b + underset~a · underset~a + underset~b · underset~b – 2underset~a · underset~b` | |
| `= |underset~a|^2 + |underset~b|^2 + |underset~a|^2 + |underset~b|^2` | |
| `= 2|underset~a|^2 + 2|underset~b|^2\ \ …\ text(as required)` |
Vectors, EXT2, V1 SM-Bank 16
Let `OABCD` be a right square pyramid where `underset ~a = vec(OA),\ underset ~b = vec(OB),\ underset ~c = vec(OC)` and `underset ~d = vec(OD)`.
Show that `underset~a + underset~c = underset~b + underset~d`. (3 marks)
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Show Worked Solution
`text(Let)\ \ A=(p,–p,–k),`
| `underset ~a` | `= overset(->)(OA) = punderset~i – punderset~j – qunderset~k` |
| `underset ~b` | `= overset(->)(OB) = punderset~i + punderset~j – qunderset~k` |
| `underset ~c` | `= overset(->)(OC) = −punderset~i + punderset~j – qunderset~k` |
| `underset ~d` | `= overset(->)(OA) = −punderset~i – punderset~j – qunderset~k` |
| `underset~a + underset~c` | `= −2qunderset~k` |
| `underset ~b + underset ~d` | `= −2qunderset~k` |
`:. underset~a + underset~c = underset~b + underset~d`
Vectors, EXT2 V1 SM-Bank 23
A cube with side length 3 units is pictured below.
- Calculate the magnitude of vector `vec(AG)`. (1 mark)
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- Find the acute angle between the diagonals `vec(AG)` and `vec(BH)`. (3 marks)
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Vectors, EXT2 V1 SM-Bank 22
`ABCDEFGH` are the vertices of a rectangular prism.
- Show that the internal diagonals of the prism, `AG` and `DF`, intersect. (2 marks)
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- Calculate the acute angle, `theta`, between the diagonals, to the nearest minute. (2 marks)
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Vectors, EXT2 V1 EQ-Bank 15
Point `B` sits on the arc of a semi-circle with diameter `AC`.
Using vectors, show `angle ABC` is a right angle. (2 marks)
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Show Worked Solution
`text(Let)\ \ vec (OA) = underset~a, \ text(and)\ \ vec(OC)=underset~c`
`text(Prove) \ overset(->)(AB) ⊥ overset(->)(BC)`
`|underset~a| = |underset~b| = |underset~c|\ \ \ text{(radii)}`
`underset~c = – underset~a `
| `overset(->)(AB) ⋅ overset(->)(BC)` | `= (underset~b-underset~a) (underset~c-underset~b)` |
| `= underset~b · underset~c-|underset~b|^2-underset~a · underset~c + underset~a · underset~b` | |
| `= underset~b · underset~c-|underset~b|^2 + underset~c · underset~c-underset~c · underset~b` | |
| `= |underset~c|^2-|underset~b|^2` | |
| `= 0` |
`:. \ ∠ABC \ text(is a right angle.)`
Vectors, EXT2 V1 SM-Bank 20
`OABD` is a trapezium in which `overset(->)(OA) = underset~a` and `overset(->)(OB) = underset~b`.
`OC` is parallel to `AB` and `DC : CB = 1:2`
Using vectors, express `overset(->)(DA)` in terms of `underset~a` and `underset~b`. (3 marks)
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Show Worked Solution
`overset(->)(DA) = overset(->)(DB) + overset(->)(BA)`
`overset(->)(BA) = underset~a – underset~b`
`text(S) text(ince) \ OC || AB \ \ text(and) \ \ OA || CB`
`=> OABC \ text(is a parallelogram)`
| `overset(->)(OA)` | `= overset(->)(CB) = underset~a` |
| `overset(->)(DC)` | `= (1)/(2) \ underset~a ` |
| `overset(->)(DB)` | `= overset(->)(DC) + overset(->)(CB)` |
| `= (3)/(2) \ underset~a` |
| `:. \ overset(->)(DA)` | `= (3)/(2) \ underset~a + (underset~a – underset~b)` |
| `= (5)/(2) \ underset~a – underset~b` |
Vectors, EXT2 V1 SM-Bank 15
Consider the two vector line equations
`underset~(v_1) = ((1),(4),(−2)) + lambda_1((3),(0),(−1)), qquad underset~(v_2) = ((3),(2),(2)) + lambda_2((4),(2),(−6))`
- Show that `underset~(v_1)` and `underset~(v_2)` intersect and determine the point of intersection . (2 marks)
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- What is the acute angle between the vector lines, to the nearest minute. (2 marks)
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Vectors, EXT2 V1 SM-Bank 2
- Find values of `a`, `b`, `c` and `d` such that `underset~v = ((a),(b)) + 2((c),(d))` is a vector equation of a line that passes through `((3),(1))` and `((−3),(−3))`. (2 marks)
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- Determine whether `underset~u = ((4),(6)) + lambda((−2),(3))` is perpendicular to `underset~v`. (1 mark)
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- Express `underset~u` in Cartessian form. (1 mark)
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Vectors, EXT1 V1 SM-Bank 11 MC
Which pair of line segments intersect at exactly one point
| A. | `{(underset ~u = ((3), (2)) + lambda ((text{−1}),(2)) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((2), (1)) + lambda ((2), (text{−4})) text{,} quad qquad 0 <= lambda <= 1):}` |
| B. | `{(underset ~u = ((4), (1)) + lambda ((3), (text{−1})) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((3), (2)) + lambda ((2), (2)) text{,} quad qquad 0 <= lambda <= 1):}` |
| C. | `{(underset ~u = ((4), (0)) + lambda ((text{−3}),(6)) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((0), (1)) + lambda ((1), (text{−2})) text{,} quad qquad 0 <= lambda <= 1):}` |
| D. | `{(underset ~u = ((0), (2)) + lambda ((3), (text{−2})) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((0), (1)) + lambda ((1), (1)) text{,} quad qquad 0 <= lambda <= 1):}` |
Vectors, EXT2 V1 SM-Bank 10
- Determine the point of intersection of `underset ~a` and `underset ~b` given.
`qquad underset ~a = ((3), (5), (1)) + lambda ((1), (3), (text{−2})),` and
`qquad underset ~b = ((text{−2}), (2), (text{−1})) + mu ((1), (text{−1}), (2))` (2 marks)
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- Determine if the point `(2, text{−2}, 5)` lies on `underset ~b`. (1 mark)
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Vectors, EXT2 V1 SM-Bank 8
Use the vector form of the linear equations
`3x - 2y = 4` and `3y + 2x - 6 = 0`
to show they are perpendicular. (3 marks)
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