Exercise practice: Three Dimensional Geometry

5.1 Exercise practice: the chapter's exercise sections

The chapter has two numbered exercise sections and one miscellaneous exercise. The worked set below covers every question type the board repeats. The first move in each is to pull out a point and a direction.

5.1.1 Exercise 11.1

E1.1 Find the direction cosines of the line joining the points (1, 2, 3) and (4, 5, 6). The direction ratios are (3, 3, 3), whose length is 3 root 3, so the direction cosines are each 1 divided by root 3.

E1.2 Show that the line through (1, -1, 2) and (3, 4, -2) is perpendicular to the line through (0, 3, 2) and (3, 5, 6). The two direction vectors are (2, 5, -4) and (3, 2, 4), and their dot product is zero, so the lines are perpendicular.

E1.3 Show that the line through (4, 7, 8) and (2, 3, 4) is parallel to the line through (-1, -2, 1) and (1, 2, 5). The directions are (-2, -4, -4) and (2, 4, 4), which are proportional with ratio minus one. Parallel directions mean parallel lines.

E1.4 Find the angle between the lines with direction ratios (1, 2, 3) and (2, -2, 1). The dot product is 2 - 4 + 3, which is 1, and the lengths are root 14 and 3, so the cosine of the acute angle is 1 divided by 3 root 14.

E1.5 Find the direction cosines of the line 6x - 12 = 3y + 9 = 2z - 2. Write the line in ratio form to get direction ratios proportional to (1, 2, 3); the length is root 14, so the cosines are 1, 2 and 3 each divided by root 14.

5.1.2 Exercise 11.2

E2.1 Find the vector equation of the line through (1, 2, 3) parallel to 3 i + 2 j - 2 k. The answer is r = i + 2 j + 3 k plus lambda times (3 i + 2 j - 2 k).

E2.2 Find the cartesian equation of the line through (-2, 4, -5) parallel to the line (x + 3)/3 = (y - 4)/5 = (z + 8)/6. The direction ratios are (3, 5, 6), so the answer is (x + 2)/3 = (y - 4)/5 = (z + 5)/6.

E2.3 Write the vector form of the line (x - 5)/3 = (y + 4)/7 = (z - 6)/2. The point is (5, -4, 6) and the direction is 3 i + 7 j + 2 k, so r = 5 i - 4 j + 6 k plus lambda times (3 i + 7 j + 2 k).

E2.4 The directions (2, -2, 1) and (1, 2, 2) belong to two lines. Find the angle between them. The dot product is 2 - 4 + 2, which is zero, so the lines are perpendicular and the angle is ninety degrees.

E2.5 Find the shortest distance between the lines r = (i + 2 j + k) + lambda(i - j + k) and r = (2 i - j - k) + mu(2 i + j + 2 k). This is the pair used in the PYQ section; the direction vectors are (1, -1, 1) and (2, 1, 2), their cross product is (-3, 0, 3), and the triple product with the joining vector gives a distance that reduces to a multiple of 1 divided by root 2 after solving the parameter relation.

E2.6 Find the distance between the parallel lines r = (i - j) + lambda(2 i + j - k) and r = (2 i + j) + mu(2 i + j - k). The joining vector is (1, 2) as a 3D vector (1, 2, 0) and the cross product with (2, 1, -1) is (-2, 1, -3), whose magnitude is root 14. The direction length is root 6, so the distance is root 14 divided by root 6, which simplifies to root 21 divided by 3.

E2.7 Find the value of lambda so that the lines (1 - x)/3 = (7y - 14)/(2 lambda) = (z - 3)/2 and (7 - 7x)/(3 lambda) = (y - 5)/1 = (6 - z)/5 are perpendicular. Convert both to direction ratio form, then set the dot product of the two direction triples equal to zero; the resulting linear equation gives lambda.

E2.8 Two lines with directions (7, -5, 1) and (1, 2, 3) pass through the points (5, -2, 0) and the origin respectively. Prove that they intersect, and find the point. Equate the general points and solve the three coordinate equations; a consistent solution gives the intersection point, and a contradiction would prove the lines are skew.

E2.9 Find the vector equation of the line through the point (1, 2, 3) and perpendicular to both i + j + k and i - j + k. The direction is the cross product (2, 0, -2), which simplifies to (1, 0, -1), so the answer is r = i + 2 j + 3 k plus lambda(1, 0, -1) written with unit vectors.

E2.10 Find the distance of the point (1, 2, 3) from the line through (0, 2, 0) with direction 3 i + 2 j + 6 k. The vector from the line point to the given point is (1, 0, 3), and its cross product with the direction is (-6, 3, 2) with magnitude 7. The direction length is 7, so the distance is 1 unit, and the foot of the perpendicular can be found from the same computation.

5.1.3 Miscellaneous exercise

M1 Find the angle between the line r = (2 i + j - k) + lambda(3 i - j) and the line joining (1, 1, 1) and (3, 2, 4). The directions are (3, -1, 0) and (2, 1, 3), whose dot product is 5, so the cosine is 5 divided by the product of root 10 and root 14.

M2 Find the shortest distance between the lines r = (i + 2 j + 3 k) + lambda(2 i + 3 j + 4 k) and r = (2 i + 4 j + 5 k) + mu(3 i + 4 j + 5 k), and state whether they are skew. Compute the cross product of the directions, then the triple product with the joining vector. A non-zero value proves the lines are skew and gives the distance; if the cross product were zero they would be parallel.

M3 Three lines carry directions proportional to (1, 2, 3), (2, 3, 1) and (3, 1, 2). Examine whether any pair is perpendicular. Check the three pairwise dot products: 2 + 6 + 3, 3 + 2 + 6 and 6 + 3 + 2 are all non-zero, so no pair is perpendicular; the exercise shows how a single dot product settles such a claim.