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Mathematics Extension 2

Stage 6 · HSC questions by topic

About this subject

Mathematics Extension 2

Atlas organises human-reviewed questions against the NSW Stage 6 Mathematics Extension 2syllabus. Every published question retains its exam source and page.

Syllabus: NESA Stage 6 (2017)
Explore the NESA Stage 6 (2017) syllabus topics

Proof

  • The Nature of Proof
  • Further Proof by Mathematical Induction

Vectors

  • Introduction to three-dimensional vectors
  • Further operations with three-dimensional vectors
  • Vectors and vector equations of lines

Complex Numbers

  • Arithmetic of complex numbers
  • Geometric representation of a complex number
  • Other representations of complex numbers
  • Solving equations with complex numbers
  • Geometrical implications of complex numbers

Calculus

  • Further Integration

Mechanics

  • Simple harmonic motion
  • Modelling motion without resistance
  • Resisted motion
  • Projectiles and resisted motion
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21 questions found

List
2023HSC
Q9

A particle travels along a curve from OOO to EEE in the xyxyxy-plane, as shown in the diagram. The position vector is r\mathbf r, velocity is , and acceleration is . While travelling from to , the particle is always slowing down. Which of the following is consistent with the motion?

r
v\mathbf vv
a\mathbf aa
OOO
EEE

A. r⋅v≤0\mathbf r\cdot\mathbf v\le0r⋅v≤0 and a⋅v≥0\mathbf a\cdot\mathbf v\ge0a⋅v≥0
B. r⋅v≤0\mathbf r\cdot\mathbf v\le0r⋅v≤0 and a⋅v≤0\mathbf a\cdot\mathbf v\le0a⋅v≤0
C. r⋅v≥0\mathbf r\cdot\mathbf v\ge0r⋅v≥0 and a⋅v≥0\mathbf a\cdot\mathbf v\ge0a⋅v≥0
D. r⋅v≥0\mathbf r\cdot\mathbf v\ge0r⋅v≥0 and a⋅v≤0\mathbf a\cdot\mathbf v\le0a⋅v≤0

1 mark
Multiple choice
2023HSC
Q10

Consider three-dimensional vectors a=OA→\mathbf a=\overrightarrow{OA}a=OA, b=OB→\mathbf b=\overrightarrow{OB}b=OB and c=OC→\mathbf c=\overrightarrow{OC}c=OC satisfying a⋅b=1\mathbf a\cdot\mathbf b=1a⋅b=1, b⋅c=2\mathbf b\cdot\mathbf c=2b⋅c=2, and c⋅a=3\mathbf c\cdot\mathbf a=3c⋅a=3. Which statement is true?

A. Two of a,b,c\mathbf a,\mathbf b,\mathbf ca,b,c could be unit vectors.
B. The points A,B,CA,B,CA,B,C could lie on a sphere centred at OOO.
C. For any three-dimensional vector a\mathbf aa, vectors can be found satisfying the conditions. D. For all such vectors, there exist positive reals such that .

1 mark
Multiple choice
2023Trial
Q11 (e) (iii)

Using vector methods, show that the length of the perpendicular from MMM to ON→\overrightarrow{ON}ON is

225 units.\sqrt{\frac{22}{5}}\text{ units}.522​​ units.
3 marks
Short answer
2023HSC
Q11 (b)

Find the angle between the vectors a=i+2j−3k\mathbf a=\mathbf i+2\mathbf j-3\mathbf ka=i+2j−3k and b=−i+4j+2k\mathbf b=-\mathbf i+4\mathbf j+2\mathbf kb=−i+4j+2k, giving your answer to the nearest degree.

3 marks
Short answer
2023Trial
Q11 (d)

Find the equations of a sphere whose centre is at (1,0,1)(1,0,1)(1,0,1) and touches the line

r⃗=(312)+λ(212),λ∈R.\vec r=\begin{pmatrix}3\\1\\2\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\2\end{pmatrix},\qquad \lambda\in\mathbb{R}.r=​3
3 marks
Short answer
2023Trial
Q11 (e) (ii)

Using the diagram and information from part (e), calculate the angle between MD→\overrightarrow{MD}MD and ON→\overrightarrow{ON}ON.

2 marks
Short answer
2023Trial
Q13 (d) (ii)

Another line, ℓ2\ell_2ℓ2​, is defined by the vector equation

r⃗=(012)+λ(a1−1),λ,a∈R.\vec r=\begin{pmatrix}0\\1\\2\end{pmatrix}+\lambda\begin{pmatrix}a\\1\\-1\end{pmatrix},\qquad \lambda,a\in\mathbb{R}.r=​0

Find the possible values of aaa when the angle between ℓ1\ell_1ℓ1​ and ℓ2\ell_2ℓ2​ is π4\frac{\pi}{4}.

2 marks
Short answer
2022Trial
Q1

What is the unit vector in the same direction as a⃗=2i+j−3k\vec a=2\mathbf i+\mathbf j-3\mathbf ka=2i+j−3k?

A. −114[21−3]-\frac1{14}\begin{bmatrix}2\\1\\-3\end{bmatrix}−141​​ B. C. D.

1 mark
Multiple choice
2022HSC
Q9

Let AAA and BBB be two distinct points in three-dimensional space. Let MMM be the midpoint of ABABAB.

Let S1S_1S1​ be the set of all points PPP such that AP→⋅BP→=0\overrightarrow{AP}\cdot\overrightarrow{BP}=0AP⋅BP.

Let S2S_2S2​ be the set of all points NNN such that ∣AN→∣=∣MN→∣|\overrightarrow{AN}|=|\overrightarrow{MN}|∣AN.

The intersection of S1S_1S1​ and S2S_2S2​ is the circle SSS.

What is the radius of the circle SSS?

A. ∣AB→∣2\dfrac{|\overrightarrow{AB}|}{2}2∣AB∣​
B. C. D.

1 mark
Multiple choice
2022Trial
Q11 (d) (ii)

Using the points defined in part (d), find ∣CB→∣|\overrightarrow{CB}|∣CB∣.

2 marks
Short answer
2022HSC
Q11 (d)

A triangle is formed in three-dimensional space with vertices A(1,−1,2)A(1,-1,2)A(1,−1,2), B(0,2,−1)B(0,2,-1)B(0,2,−1) and C(2,1,1)C(2,1,1)C(2,1,1). Find the size of ∠ABC\angle ABC∠ABC, giving your answer to the nearest degree.

3 marks
Short answer
2022Trial
Q11 (d) (iii)

Using the points defined in part (d), what is the size of the acute angle between AB→\overrightarrow{AB}AB and CB→\overrightarrow{CB}CB? Answer correct to the nearest degree.

3 marks
Short answer
2022HSC
Q14 (a) (i)

The two non-parallel vectors u⃗\vec uu and v⃗\vec vv satisfy λu⃗+μv⃗=0⃗\lambda\vec u+\mu\vec v=\vec0λu+μv for some real numbers λ\lambdaλ and μ\muμ. Show that λ=μ=0\lambda=\mu=0λ=μ=0.

2 marks
Short answer
2022Trial
Q15 (c) (ii)

It is given that
c⃗=[140]+t[−115]\vec c=\begin{bmatrix}1\\4\\0\end{bmatrix}+t\begin{bmatrix}-1\\1\\5\end{bmatrix}c=​140​​+t​−115​​
does not intersect r⃗\vec rr. Point TTT is on c⃗\vec cc and SSS is a point on r⃗\vec rr such that TSTSTS is a minimum. By first expressing c⃗\vec cc in parametric form, find an expression for TS→\overrightarrow{TS}TS in terms of ttt.

3 marks
Short answer
2022Trial
Q15 (a) (iii)

Show that the line
(xyz)=(15−16)+δ(4−1−1)\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}15\\-1\\6\end{pmatrix}+\delta\begin{pmatrix}4\\-1\\-1\end{pmatrix}​xyz​​=​15−16​δ​4−1−1​
is also tangent to the sphere
(x−5)2+(y+2)2+(z−3)2=38(x-5)^2+(y+2)^2+(z-3)^2=38(x−5)2+(y+2)2+(
by showing that it is perpendicular to the radius of the sphere at (718)\begin{pmatrix}7\\1\\8\end{pmatrix}​718​.

2 marks
Short answer
2022Trial
Q15 (c) (i)

The line has vector equation
r⃗=λ[131].\vec r=\lambda\begin{bmatrix}1\\3\\1\end{bmatrix}.r=λ​131​​.
For P(x1,y1,z1)P(x_1,y_1,z_1)P(x1​,y1​,z1​), the point QQQ on the line is such that PQPQPQ is a minimum. Using vector projection methods, show that
OQ→=x1+3y1+z111[131].\overrightarrow{OQ}=\frac{x_1+3y_1+z_1}{11}\begin{bmatrix}1\\3\\1\end{bmatrix}.OQ​=

2 marks
Short answer
2021HSC
Q11 (c)

Find the angle between the vectors a⃗=(204)\vec a=\begin{pmatrix}2\\0\\4\end{pmatrix}a=​204​​ and b⃗=(−312)\vec b=\begin{pmatrix}-3\\1\\2\end{pmatrix}b=, giving the angle in degrees correct to 1 decimal place.

3 marks
Short answer
2021HSC
Q16 (a) (ii)

Given vectors a⃗=(a1a2a3)\vec a=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}a=​a1​a2​a3​​​ and b⃗=(b1b2b3)\vec b=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}b=, show that
∣a1b1+a2b2+a3b3∣≤a12+a22+a32b12+b22+b32.|a_1b_1+a_2b_2+a_3b_3|\le\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}.∣a1​b1​+

3 marks
Short answer
2021HSC
Q16 (a) (i)

The point P(x,y,z)P(x,y,z)P(x,y,z) lies on the sphere of radius 111 centred at the origin OOO. Using OP→=xi⃗+yj⃗+zk⃗\overrightarrow{OP}=x\vec i+y\vec j+z\vec kOP=xi+y and the triangle inequality, or otherwise, show that
∣x∣+∣y∣+∣z∣≥1.|x|+|y|+|z|\ge1.∣x∣+∣y∣+∣z∣≥1.

2 marks
Short answer
2021HSC
Q16 (a) (iii)

As in part (i), P(x,y,z)P(x,y,z)P(x,y,z) lies on the unit sphere centred at the origin. Using part (ii), or otherwise, show that
∣x∣+∣y∣+∣z∣≤3.|x|+|y|+|z|\le\sqrt3.∣x∣+∣y∣+∣z∣≤3​.

2 marks
Short answer
2020HSC
Q11 (d)

Consider the two vectors u=−2i−j+3k\mathbf u=-2\mathbf i-\mathbf j+3\mathbf ku=−2i−j+3k and v=pi+j+2k\mathbf v=p\mathbf i+\mathbf j+2\mathbf kv=pi+j+2k. For what values of ppp are u−v\mathbf u-\mathbf vu−v and u+v\mathbf u+\mathbf vu+v perpendicular?

3 marks
Short answer
b,c\mathbf b,\mathbf c
b,c

r,s,tr,s,tr,s,t
ra+sb+tc=0r\mathbf a+s\mathbf b+t\mathbf c=\mathbf0ra+sb+tc=0
1
2
​
​
+
λ​212​​,λ∈
R.
1
2
​
​
+
λ​a1−1​​,λ,a∈
R.
4π​
21−3
​
​

−114[2−13]-\frac1{14}\begin{bmatrix}2\\-1\\3\end{bmatrix}−141​​2−13​​

−114[2−13]-\frac1{\sqrt{14}}\begin{bmatrix}2\\-1\\3\end{bmatrix}−14​1​​2−13​​

114[21−3]\frac1{\sqrt{14}}\begin{bmatrix}2\\1\\-3\end{bmatrix}14​1​​21−3​​
=
0
∣
=
∣MN∣
∣AB→∣4\dfrac{|\overrightarrow{AB}|}{4}
4∣AB∣​

3∣AB→∣2\dfrac{\sqrt3|\overrightarrow{AB}|}{2}23​∣AB∣​

3∣AB→∣4\dfrac{\sqrt3|\overrightarrow{AB}|}{4}43​∣AB∣​
=
0
​
+
​
z
−
3)2=
38
​
11x1​+3y1​+z1​​
​131​​
.
​
−312​
​
​
b1​b2​b3​​
​
a2​
b2​
+
a3​b3​∣≤
a12​+a22​+a32​​b12​+b22​+b32​.
j​
+
zk
​