With respect to the origin O, the points A, B and C have position vectors given by OA = [1, -4, 2] OB = [-2, 1, 3] OC = [2, 3, 5] (a) Find a vector equation for the line through A and B. (b) Using a scalar product, find the exact value of cos BAС. (c) Hence find the exact area of triangle ABC.

Mark scheme

Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.

1. [M1] Carry out a correct method for finding a vector equation for the line through A and B A complete method. Can be working from any point on AB.
2. [A1] Obtain r = [-3, -4 + λ5, 2 + λ1] OE Must have r = ... not l = .... Accept r_AB =... Accept R = ....
3. [B1] Obtain a direction vector for AC = [1, 7, 3] OE Or CA.
4. [M1] Carry out correct process for evaluating the scalar product of AB and AC or BA and CA -3×1+5×7+1×3=35 Allow for correct answer and no working seen.
5. [M1] Using the correct process for the moduli, divide their scalar product by the product of the moduli Independent M0M1 is possible. ISW finding the angle.
6. [A1] Obtain (cos BAC =) 35/√59 or exact simplified equivalent From correct working. The answer needs to come from using a scalar product. Accept 35/√2065 or 35/(√35√59) or 2065/59. θ = cos^-1 (35/59) without a statement of cos BAC scores A0. ISW finding the angle.
7. [M1] Use area = 1/2 |AB||AC|sin BAC with their |AB| |AC| For "hence", must be using the angle at A. Need not substitute for the trigonometry. Accept any equivalent form for their sin BAC.
8. [M1] Use sin x = √1-cos²x or an equivalent exact method with their cos x (< 1) to obtain an exact value for sin x NB: These two M marks are independent. Might not quote the formula. Could draw a triangle and use Pythagoras, which is equivalent. sin x = 24/59. sin(cos^-1(35/59)) is allowed for the first M1, but not for the second.
9. [A1] Obtain answer √210 from correct working Accept simplified equivalent exact forms, e.g. 1/2 √840. Watch out for fortuitous answers from negative value of the cosine. Do not accept an answer coming from sin(cos^-1(35/59)) with no evidence of the method of evaluation. The answer needs to come from using angle BAC.