LogoLogo
    Logo
    • TutoringSchools
    1. Home
    2. IB
    3. Mathematics Analysis and Approaches (AA)
    4. Questions

    Question
    HLPaper 1

    Consider the three planes

    π1:2x−y+z=4{\pi}_1: 2x - y + z = 4π1​:2x−y+z=4

    π2:x−2y+3z=5{\pi}_2: x - 2y + 3z = 5π2​:x−2y+3z=5

    π3:−9x+3y−2z=32{\pi}_3: -9x + 3y - 2z = 32π3​:−9x+3y−2z=32

    1.

    Show that the three planes do not intersect.

    [4]
    Verified
    Solution

    METHOD 1 attempt to eliminate a variable M1 obtain a pair of equations in two variables

    EITHER −3x+z=−3-3x+z=-3−3x+z=−3 A1 −3x+z=44-3x+z=44−3x+z=44A1

    OR −5x+y=−7-5x+y=-7−5x+y=−7A1 −5x+y=40-5x+y=40−5x+y=40A1

    OR 3x−z=33x-z=33x−z=3 A1 3x−z=−7953x-z=-\frac{79}{5}3x−z=−579​A1

    THEN the two lines are parallel (−3≠44-3 \neq 44−3=44 or −7≠40-7 \neq 40−7=40 or 3≠−7953 \neq -\frac{79}{5}3=−579​)R1

    Note: There are other possible pairs of equations in two variables. To obtain the final R1, at least the initial *M1 *must have been awarded.

    hence the three planes do not intersectAG

    METHOD 2 vector product of the two normals =(−1−5−3)=\begin{pmatrix}-1 \\ -5 \\ -3\end{pmatrix}=​−1−5−3​​ (or equivalent)A1 r=(1−20)+λ(153)\mathbf{r}=\begin{pmatrix}1 \\ -2 \\ 0\end{pmatrix}+\lambda\begin{pmatrix}1 \\ 5 \\ 3\end{pmatrix}r=​1−20​​+λ​153​​ (or equivalent)A1

    Note: Award A0 if "r=\mathbf{r}=r=" is missing. Subsequent marks may still be awarded.

    Attempt to substitute 1+λ,−2+5λ,3λ1+\lambda,-2+5\lambda,3\lambda1+λ,−2+5λ,3λ in ∏3\prod_3∏3​M1 −9(1+λ)+3(−2+5λ)−2(3λ)=32-9(1+\lambda)+3(-2+5\lambda)-2(3\lambda)=32−9(1+λ)+3(−2+5λ)−2(3λ)=32 −15=32-15=32−15=32,a contradictionR1 hence the three planes do not intersectAG

    METHOD 3 attempt to eliminate a variable M1 −3y+5z=6-3y+5z=6−3y+5z=6A1 −3y+5z=100-3y+5z=100−3y+5z=100A1 0=940=940=94,a contradictionR1

    Note: Accept other equivalent alternatives. Accept other valid methods. To obtain the final R1, at least the initial *M1 *must have been awarded.

    hence the three planes do not intersectAG

    [4 marks]

    2.

    Verify that the point P(1,−2,0)P(1,-2,0)P(1,−2,0) lies on both Π1\Pi_1Π1​ and Π2\Pi_2Π2​.

    [1]
    Verified
    Solution

    ∏1:2+2+0=4{\prod_{1}:2+2+0=4}∏1​:2+2+0=4 and ∏2:1+4+0=5{\prod_{2}:1+4+0=5}∏2​:1+4+0=5A1

    [1 mark]

    3.

    Find a vector equation of LLL, the line of intersection of Π1{\Pi}_1Π1​ and Π2{\Pi}_2Π2​.

    [4]
    Verified
    Solution

    METHOD 1 attempt to find the vector product of the two normals M1 2−1+1×(−2−3)2 - 1 + 1 \times ( - 2 - 3 )2−1+1×(−2−3) =−1−5−3= - 1 - 5 - 3=−1−5−3 A1 r=(1−20)+λ(153)\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix}r=​1−20​​+λ​153​​ A1A1

    Note: Award A1A0 if "r=\mathbf{r} =r=" is missing. Accept any multiple of the direction vector. Working for (b)(ii) may be seen in part (a) Method 2. In this case penalize lack of "r=\mathbf{r} =r=" only once.

    METHOD 2 attempt to eliminate a variable fromΠ1\Pi_1Π1​ andΠ2\Pi_2Π2​ M1 3x−z=33x - z = 33x−z=3 OR3y−5z=−63y - 5z = -63y−5z=−6 OR5x−y=75x - y = 75x−y=7 Letx=tx = tx=t substitutingx=tx = tx=tin3x−z=33x - z = 33x−z=3to obtain z=−3+3tz = -3 + 3tz=−3+3t andy=5t−7y = 5t - 7y=5t−7(for all three variables in parametric form) A1 r=(0−7−3)+λ(153)\mathbf{r} = \begin{pmatrix} 0 \\ -7 \\ -3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 5 \\ 3 \end{pmatrix}r=​0−7−3​​+λ​153​​ A1A1

    **Note:**Award A1A0 if "r=\mathbf{r} =r=" is missing. Accept any multiple of the direction vector.Accept other position vectors which satisfy both the planesΠ1\Pi_1Π1​ andΠ2\Pi_2Π2​.

    [4 marks]

    4.

    Find the distance between LLL and Π3\Pi_3Π3​.

    [6]
    Verified
    Solution

    METHOD 1 the line connecting LLL andΠ3\Pi_3Π3​ is given by L1L_1L1​ attempt to substitute position and direction vector to formL1L_1L1​ (M1) s=(1−20)+t(−93−2)\mathbf{s}=\begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}+t\begin{pmatrix} -9 \\ 3 \\ -2 \end{pmatrix}s=​1−20​​+t​−93−2​​A1 substitute(1−9t−2+3t−2t)\begin{pmatrix} 1-9t \\ -2+3t \\ -2t \end{pmatrix}​1−9t−2+3t−2t​​ inΠ3\Pi_3Π3​ M1 −9(1−9t)+3(−2+3t)−2(−2t)=32-9(1-9t)+3(-2+3t)-2(-2t)=32−9(1−9t)+3(−2+3t)−2(−2t)=32 94t=47⇒t=1294t=47 \Rightarrow t=\frac{1}{2}94t=47⇒t=21​A1 attempt to find distance between(1−20)\begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}​1−20​​ and their point(−72−12−1)\begin{pmatrix} -\frac{7}{2} \\ -\frac{1}{2} \\ -1 \end{pmatrix}​−27​−21​−1​​ (M1) =∣(1−20)+12(−93−2)−(1−20)∣=12(−9)2+32+(−2)2=\left| \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}+\frac{1}{2}\begin{pmatrix} -9 \\ 3 \\ -2 \end{pmatrix}-\begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} \right|=\frac{1}{2}\sqrt{(-9)^2+3^2+(-2)^2}=​​1−20​​+21​​−93−2​​−​1−20​​​=21​(−9)2+32+(−2)2​ =942=\frac{\sqrt{94}}{2}=294​​A1

    METHOD 2 unit normal vector equation ofΠ3\Pi_3Π3​ is given by−9x+3y+2z81+9+4\frac{-9x+3y+2z}{\sqrt{81+9+4}}81+9+4​−9x+3y+2z​ (M1) =3294=\frac{32}{\sqrt{94}}=94​32​A1 letΠ4\Pi_4Π4​be the plane parallel toΠ3\Pi_3Π3​and passing through PPP, then the normal vector equation ofΠ4\Pi_4Π4​is given by (−932)⋅(xyz)=−15\begin{pmatrix} -9 \\ 3 \\ 2 \end{pmatrix}\cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix}=-15​−932​​⋅​xyz​​=−15 M1

    unit normal vector equation ofΠ4\Pi_4Π4​is given by −9x+3y+2z81+9+4=−1594\frac{-9x+3y+2z}{\sqrt{81+9+4}}=\frac{-15}{\sqrt{94}}81+9+4​−9x+3y+2z​=94​−15​A1 distance between the planes is3294−−1594\frac{32}{\sqrt{94}}-\frac{-15}{\sqrt{94}}94​32​−94​−15​ (M1) =4794=\frac{47}{\sqrt{94}}=94​47​A1

    [6 marks]

    Sign up for free to view this answer

    Still stuck?

    Get step-by-step solutions with Jojo AI

    FreeJojo AI

    Want more practice questions for Mathematics Analysis and Approaches (AA)?

    Related topics


    Footer

    General

    • About us
    • Mission
    • Tutoring
    • Blog
    • State of learning surveyNew

    • Trustpilot
    • Contact us
    • Join us We're hiring!

    Features

    • Jojo AI
    • Questionbank
    • Study notes
    • Flashcards
    • Test builder
    • Exam mode
    • Coursework
    • IB grade calculator

    Legal

    • Terms and conditions
    • Privacy policy
    • Cookie policy
    • Trust Center

    IB

    • Biology (New syllabus)
    • Business Management
    • Chemistry (New syllabus)
    • Chinese A Lang & Lit
    • Chinese B
    • Computer Science (CS)
    • Design Technology (DT)
    • Digital Society (DS)
    • Economics
    • English B
    • View more...
    Logo

    © 2022 - 2025 RevisionDojo (MyDojo Inc)

    RevisionDojo was developed independently of the IBO and as such is not endorsed by it in any way.

    RedditInstagramTikTokDiscord
    GDPR compliant