A-Level Vector geometry in three dimensions: worked solution

Question

A drone flies straight to a rooftop along l: r = (2, 1, 0) + lambda(3, -2, 1), the rooftop being p: 2x + y - z = 7. On reaching it, the drone takes a new path m, the acute angle between m and the rooftop's normal equalling that between l and the same normal, with l, m and the normal coplanar. Find, in exact non-trigonometric form, the cosine of the angle between l and m.

Step 1. Read off the two vectors that matter. The direction of l is the vector multiplied by lambda: d = (3, −2, 1). The normal to the rooftop plane 2x + y − z = 7 is read off the coefficients of x, y and z: n = (2, 1, −1). Step 2. Let theta be the acute angle between l and the normal, and use the scalar product formula cos theta = (d . n)/(|d| |n|). d . n = 3(2) + (−2)(1) + 1(−1) = 6 − 2 − 1 = 3 |d| = sqrt(3² + (−2)² + 1²) = sqrt(9 + 4 + 1) = sqrt14 |n| = sqrt(2² + 1² + (−1)²) = sqrt(4 + 1 + 1) = sqrt6 cos theta = 3/(sqrt14 sqrt6) This is positive, so theta is already the acute angle and no modulus adjustment is needed. Step 3. Locate m relative to l. The new path m makes the same acute angle theta with the normal, and l, m and the normal all lie in one plane. So the normal sits between l and m and bisects the angle between them. Angle between l and m = theta + theta = 2theta. Step 4. Get cos 2theta without ever finding theta, using the double angle formula cos 2theta = 2cos²theta − 1. This is what makes the answer exact and non-trigonometric. cos²theta = (3/(sqrt14 sqrt6))² = 9/(14 × 6) = 9/84 = 3/28 cos 2theta = 2(3/28) − 1 = 6/28 − 1 = 3/14 − 1 = 3/14 − 14/14 = −11/14 So the cosine of the angle between l and m is −11/14. (The negative sign simply says that, measured between the two direction vectors as written, the angle is obtuse.)