The points A, B and C have position vectors OA =−2i+j+4k, OB = 5i+2j and OC = 8i+5j-3k, where O is the origin. The line l₁ passes through B and C.
- (a)Find a vector equation for l₁.[3]
- (b)The line l₂ has equation r =-2i+j+4k+µ(3i+j – 2k). Find the coordinates of the point of intersection of l₁ and l₂.[4]
- (c)The point D on l₂ is such that AB = BD. Find the position vector of D.[5]
Mark scheme
Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.
- [B1] Correct direction vector seen or implied (BC = 3i + 3j - 3k). Condone BC = -3i-3j+3k.
- [M1] Use a correct method to form a vector equation. Allow for the RHS with no LHS.
- [A1] Obtain r = 5i + 2j + λ(i + j - k). ISW. Must have r = ... or y = ..., not l₁ = ... Or, equivalent vector form, e.g. r = 8i + 5j-3k + α(i+j-k) or r = 5i + 2j + λ(3i + 3j-3k). Condone a column vector with i, j, k.
- [B1FT] Use components to form two relevant equations in 2 unknowns. For their l₁ B0 if they use the same unknown for both lines. Two components of (5+λ, 2+λ, -λ) = (-2+3μ, 1+μ, 4-2μ) seen or implied.
- [M1] Solve 2 relevant equations in 2 unknowns for λ or μ. For their l₁.
- [A1] Obtain λ = 2 or μ=3. Or equivalent e.g. using BC as direction vector gives λ = 2/3.
- [A1] Obtain (7, 4, -2). Accept position vector. Condone a column vector with i, j, k. SC: B1 M1 A1 A1 if one component of their line is incorrect but they do not use that component.
- [B1] State AB = √7²+1²+4² (= √66). Or (AB)² = 66 Condone a sign error in AB.
- [B1] State BD in component form. (-7+3r, -1+r, 4-2r) or equivalent.
- [M1] AB = BD ⇒ (3r−7)² + (r-1)² + (-2r+4)² = 66. Or equivalent equation in one unknown for their AB and their BD ≠ OD. If you never see a correct form and they go direct to 9r² + 49 + r² + 1..... then M0.
- [A1] Obtain r = 60/14 = 30/7. Correct only. Ignore r = 0 if seen.
- [A1] Obtain (50/7)i + (37/7)j - (2/7)k. Must be a vector. Condone if also have OD = OA.
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