The lines l and m have vector equations l: r = 2i+j-3k+λ(-i+2k) and m: r = 2i+j-3k+µ(2i−j+5k). Lines l and m intersect at the point P.
- (a)State the coordinates of P.[1]
- (b)Find the exact value of the cosine of the acute angle between l and m.[3]
- (c)The point A on line l has coordinates (0,1,1). The point B on line m has coordinates (0,2, -8). Find the exact area of triangle APB.[3]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.
- [B1] (a) P(2, 1, -3). Accept x = 2, y = 1, z = -3.
- [M1] (b) Use the correct method to find the scalar product of the direction vectors. (-1)(2) + (0)(-1) + (2)(5) = 8.
- [M1] (b) Divide the scalar product by the product of the moduli to obtain ±cosθ using consistent vectors throughout. (their 8) / (√((-1)²+0²+2²) * √(2²+(-1)²+5²)).
- [A1] (b) Obtain cos θ = 8/(5√6) OE.
- [B1] (c) For Method 2: PA = <-2, 0, 4> (A is (0,1,1) and P is (2,1,-3)), PB = <-2, 1, -5> (B is (0,2,-8) and P is (2,1,-3)). Obtain PA × PB = -4i-18j-2k.
- [M1] (c) Area = 1/2 |PA × PB| = 1/2 √((-4)²+(-18)²+(-2)²). Correct method for the exact area of the triangle.
- [A1] (c) Obtain Area = √86. Or simplified exact equivalent. ISW
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