IBDP Maths AA: Topic : SL 3.1: The distance between two points : IB style Questions HL Paper 1

Ans:

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

EITHER

-3x + z = -3 and
-3x + z = 44

OR

-5x + y = -7 and
-5x + y = 40

OR

3x – z = 3 and
3x – z = \(\frac{79}{5}\)

THEN

the two lines are parallel ( – 3 ≠ 44 or – 7 ≠ 40 or 3 ≠ -\(\frac{79}{5}\)

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 intersect

METHOD 2

vector product of the two normals = \(\begin{pmatrix}-1\\-5\\-3\end{pmatrix}\) (or equivalent)

\(r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}\)      (or equivalent)

Note: Award A0 if “r =” is missing. Subsequent marks may still be awarded.

Attempt to substitute (1 + λ, -2 + 5λ, 3λ) in ∏₃

-9(1 + λ) + 3(-2+5λ) – 2(3λ) = 32

− 15 = 32, a contradiction
hence the three planes do not intersect

(b) (i)     ∏₁ : 2 + 2 + 0 = 4 and ∏₂ : 1 + 4 + 0 = 5

      (ii) METHOD 1
attempt to find the vector product of the two normals

\(\begin{pmatrix}2\\-1 \\1 \end{pmatrix} \times \begin{pmatrix}1\\-2 \\3 \end{pmatrix}\)

\(=\begin{pmatrix}-1\\-5 \\-3 \end{pmatrix}\)

\(r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}\) 

Note: Award A1A0 if “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 =” only once.

METHOD 2
attempt to eliminate a variable from ∏₁ and ∏₂

3x – z = 3 OR 3y – 5z = -6  OR 5x – y = 7
Let x = t
substituting x = t in 3x – z = 3 to obtain
z = -3 + 3t and y = 5t – 7 (for all three variables in parametric form)

\(r = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + \lambda \begin{pmatrix}1\\5 \\3 \end{pmatrix}\) 

Note: Award A1A0 if “r =” is missing.
Accept any multiple of the direction vector. Accept other position vectors
which satisfy both the planes ∏₁ and ∏₂ .

(c) METHOD 1
the line connecting L and ∏₃ is given by L1
attempt to substitute position and direction vector to form L1

\(s = \begin{pmatrix}1\\-2 \\0 \end{pmatrix} + t \begin{pmatrix}-9\\3 \\-2 \end{pmatrix}\)
substitute (1 – 9t, – 2 + 3t, – 2t) in ∏₃
-9(1-9t) + 3(-2+3t) – 2(-2t) = 32
\(94t = 47\Rightarrow t=\frac{1}{2}\)

attempt to find distance between (1, -2,0) and their point \(\begin{pmatrix}-\frac{7}{2}, & -\frac{1}{2}, & -1 \end{pmatrix}\)

\(=\left | \begin{pmatrix}1\\-2 \\0 \end{pmatrix} +\frac{1}{2}\begin{pmatrix}-9\\3 \\-2 \end{pmatrix}-\begin{pmatrix}1\\-2 \\0 \end{pmatrix}\right | = \frac{1}{2}\sqrt{(-9)^{2}+3^{2}+(-2)^{2}}\)

\(=\frac{\sqrt{94}}{2}\)

METHOD 2

unit normal vector equation of ∏₃ is given by

\(=\frac{32}{\sqrt{94}}\)