Page 3 - LESSON NOTE

The x-coordinate, y-coordinate, and z-coordinate of point (4, 2, –5) are positive, positive, and

negative respectively. Therefore, this point lies in octant V

The x-coordinate, y-coordinate, and z-coordinate of point (–4, 2, –5) are negative, positive, and

negative respectively. Therefore, this point lies in octant VI.

The x-coordinate, y-coordinate, and z-coordinate of point (–4, 2, 5) are negative, positive, and

positive respectively. Therefore, this point lies in octant II.

The x-coordinate, y-coordinate, and z-coordinate of point (–3, –1, 6) are negative, negative,

and positive respectively. Therefore, this point lies in octant III.

The x-coordinate, y-coordinate, and z-coordinate of point (2, –4, –7) are positive, negative, and

negative respectively. Therefore, this point lies in octant VIII.

3. Find the distance between (2, 3, 5) and (4, 3, 1)

Ans:- Distance between points (2, 3, 5) and (4, 3, 1)

2
2
2
= √{(4 – 2) + (3 – 3) + (1 – 5) }
2
2
= √{2 + 0 + (-4) }
= √(4 + 16) = √20 = 2√5
4. Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

Ans:- Let points (–2, 3, 5), (1, 2, 3), and (7, 0, –1) be denoted by P, Q, and R respectively.

Points P, Q, and R are collinear if they lie on a line.Now,

2
2
2
PQ = √{(1 + 2) + (2 - 3) + (3 – 5) }
2
2
2
= √{3 + (-1) +(-2) }
= √(9 + 1 + 4) = √14
2
2
2
QR = √{(7 - 1) + (0 - 2) + (-1 – 3) }
2
2
2
= √{6 + (-2) +(-4) }
= √(36 + 4 + 16) = √56 = 2√14
2
2
2
PR = √{(7 + 2) + (0 - 3) + (-1 – 5) }
2
2
2
= √{9 + (-3) +(-6) }
= √(81 + 9 + 36)= √126 = 3√14
Here, PQ + QR = √14 + 2√14 = 3√14 = PR
Hence, points P(–2, 3, 5), Q(1, 2, 3), and R(7, 0, –1) are collinear.