Q5. Verify that (−1, 2, 1), (1, −2, 5), (4, −7, 8) and (2, −3, 4) are the vertices of a parallelogram (NCERT Exercise 11.2, Q3 (iii)).
Call them A, B, C and D. The mid-point of AC is (3/2, −5/2, 9/2) and the mid-point of BD is (3/2, −5/2, 9/2). The diagonals bisect each other, so ABCD is a parallelogram. Its sides are AB = CD = 6 and BC = DA = √43; the equal sides alone would not prove it.
Q6. Find the equation of the set of points P the sum of whose distances from A(4, 0, 0) and B(−4, 0, 0) is 10 (NCERT Exercise 11.2, Q5).
Write r for √((x + 4)2 + y2 + z2), the distance PB. PA + PB = 10 gives
√((x − 4)2 + y2 + z2) = 10 − r.
Squaring, with r2 = (x + 4)2 + y2 + z2, and cancelling y2 + z2:
(x − 4)2 = 100 − 20r + (x + 4)2, so
20r = 100 + 16x, that is, 5r = 4x + 25.
Squaring again:
9x2 + 25y2 + 25z2 = 225.
Every plane through the x-axis cuts this surface in an ellipse with foci A and B and 2a = 10. So the surface is the ellipse x2/25 + y2/9 = 1 turned about the x-axis (an ellipsoid).
Q7. A and B are (3, 4, 5) and (−1, 3, −7). Find the set of points P with PA2 + PB2 = k2, where k is a constant (NCERT Miscellaneous Exercise on Chapter 11, Q4).
As in Example 4, with k2 in place of 2k2, then divided by 2:
x2 + y2 + z2 − 2x − 7y + 2z = (k2 − 109)/2.
Q8. Show that A(−1, 2, 0), B(1, 5, 6) and C(5, 11, 18) are collinear. (NCERT Exercise 11.2, Q2 uses the points of Example 2.)
AB = √(4 + 9 + 36) = √49 = 7, BC = √(16 + 36 + 144) = √196 = 14 and AC = √(36 + 81 + 324) = √441 = 21. Since AB + BC = 7 + 14 = 21 = AC, the three points are collinear, with B between A and C.
Q9. How far is (3, −4, 12) from the origin, from the x-axis and from the z-axis?
From the origin √(9 + 16 + 144) = 13; from the x-axis √(16 + 144) = 4√10; from the z-axis √(9 + 16) = 5.
Q10. The distance between (5, −1, 7) and (x, 5, 1) is 9. Find x.
(x − 5)2 + 36 + 36 = 81, so (x − 5)2 = 9 and x = 2 or 8.
Q11. Find the point on the y-axis that is equidistant from (3, 1, 2) and (5, 5, 2).
Take (0, y, 0): 9 + (y − 1)2 + 4 = 25 + (y − 5)2 + 4, so 8y − 24 = 16 and y = 5. The point is (0, 5, 0), √29 from each.
Q12. Are (1, 2, 3), (2, 3, 4) and (4, 5, 6) collinear?
The distances are √3, √12 = 2√3 and √27 = 3√3, and √3 + 2√3 = 3√3, so yes: the middle point (2, 3, 4) lies between the other two.