A-Level Vector geometry in three dimensions

From the earlier part, l: r = (1, 3, -2) + α(1, -1, 4), i.e. l = i + 3j - 2k + α(i - j + 4k). Step 1: write a GENERAL point P on l in terms of the parameter α. Every point on l has this form, so finding the required points means finding the right values of α. OP = (1 + α, 3 - α, -2 + 4α) Step 2: form the displacement vector from B(2, -3, 7) to P by subtracting position vectors, BP = OP - OB. BP = (1 + α - 2, 3 - α - (-3), -2 + 4α - 7) = (α - 1, 6 - α, 4α - 9) Step 3: use the distance condition. |BP| = 3√2, and it is easier to square both sides than to work with a square root: |BP|² = (3√2)² = 9 × 2 = 18 (α - 1)² + (6 - α)² + (4α - 9)² = 18 Step 4: expand each bracket separately. (α - 1)² = α² - 2α + 1 (6 - α)² = 36 - 12α + α² (4α - 9)² = 16α² - 72α + 81 Add them: α² terms: 1 + 1 + 16 = 18α² α terms: -2α - 12α - 72α = -86α constants: 1 + 36 + 81 = 118 so 18α² - 86α + 118 = 18 18α² - 86α + 100 = 0 Divide through by 2: 9α² - 43α + 50 = 0 Step 5: solve the quadratic. The discriminant is (-43)² - 4(9)(50) = 1849 - 1800 = 49 and √49 = 7, so α = (43 ± 7)/(2 × 9) = (43 ± 7)/18 α = 50/18 = 25/9 or α = 36/18 = 2 Step 6: substitute each value of α back into OP. α = 2: OP = (1 + 2, 3 - 2, -2 + 4(2)) = (3, 1, 6) α = 25/9: 1 + 25/9 = 9/9 + 25/9 = 34/9 3 - 25/9 = 27/9 - 25/9 = 2/9 -2 + 4(25/9) = -18/9 + 100/9 = 82/9 OP = (34/9, 2/9, 82/9) So the two points on l at a distance 3√2 from B have position vectors (3, 1, 6) and (34/9, 2/9, 82/9).

Step 1 - set up the notation and see what the given relation says. Take O as the origin, so OA = a, OB = b and OC = c. CB = OB - OC = b - c, and the question gives a = (3/5)(b - c) = (3/5)CB. So OA is a scalar multiple of CB, meaning OA is PARALLEL to CB - that is what makes OABC a trapezium, with OA : CB = 3 : 5. Step 2 - rearrange the given relation so that b is expressed in terms of a and c. This reduces the whole problem to just the two independent vectors a and c. a = (3/5)(b - c) (5/3)a = b - c b = c + (5/3)a Step 3 - write the vector equation of line OC. It passes through the origin with direction c: r = lambda c, where lambda is a real parameter. Step 4 - write the vector equation of line AB. It passes through A with direction AB = b - a: r = a + mu(b - a), where mu is a real parameter. Substitute b = c + (5/3)a to remove b: b - a = c + (5/3)a - a = c + (2/3)a r = a + mu[c + (2/3)a] Collect the a terms: r = [1 + (2/3)mu] a + mu c Step 5 - D lies on BOTH lines, so equate the two expressions for r. lambda c = [1 + (2/3)mu] a + mu c a and c are two non-parallel sides meeting at O, so they are independent vectors: the only way this can hold is if the coefficient of a matches on both sides and the coefficient of c matches on both sides. Coefficient of a: 0 = 1 + (2/3)mu (2/3)mu = -1 mu = -3/2 Coefficient of c: lambda = mu = -3/2 Step 6 - substitute back into the equation of OC. OD = lambda c = -(3/2)c The negative sign is not a mistake: it says D lies on CO produced beyond O, which is exactly where the two non-parallel sides of a trapezium meet when extended.

The plane p has equation r·n = 0, so n is a normal vector to p and the plane passes through the origin (r = 0 satisfies it). Step 1: use the fact that CF is perpendicular to the plane. A vector perpendicular to the plane is parallel to the normal n, so CF = λn for some scalar λ. Travelling from O to C and then from C to F: OF = OC + CF = c + λn Step 2: use the fact that F lies in the plane. Every point of the plane satisfies r·n = 0, so putting r = OF: (c + λn)·n = 0 Expand using the distributive rule for the scalar product: c·n + λ(n·n) = 0 and n·n = |n|², so c·n + λ|n|² = 0 Step 3: solve for λ and substitute back. λ = −(c·n)/|n|² OF = c + λn = c − ((c·n)/|n|²) n