A-LevelMathematicsVectors9709/33 · May/June 2018 · Q1011 Marks

The points A and B have position vectors 2i + j + 3k and 4i + j + k respectively. The line l has equation r = 4i + 6j + μ(i + 2j – 2k).

(i)Show that l does not intersect the line passing through A and B.[5]
(ii)The point P, with parameter t, lies on l and is such that angle PAB is equal to 120°. Show that 3t² + 8t + 4 = 0. Hence find the position vector of P.[6]

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

M1 Carry out a correct method for finding a vector equation for AB
A1 Obtain r = 2i + j +3k+λ(2i – 2k), or equivalent
M1(dep*) Equate pair(s) of components AB and l and solve for λ or μ
A1 Obtain correct answer for λ or μ
A1 Verify that all three component equations are not satisfied
B1 State or imply a direction vector for AP has components (2+t, 5 + 2t, – 3 – 2t)
M1 State or imply that cos 120° equals the scalar product of AP and AB divided by the product of their moduli
M1 Carry out the correct processes for finding the scalar product and the product of the moduli in terms of t, and obtain an equation in terms of t
A1 Obtain the given equation correctly
M1 Solve the quadratic and use a root to find a position vector for P
A1 Obtain position vector 2i + 2j + 4k from t = – 2, having rejected the root t = 2/3

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.