Exercise 11.1
A point is on the x-axis. What are its y-coordinate and z-coordinates?
Answer
A point on the x-axis has y = 0 and z = 0. The point is of the form (x, 0, 0).
A point is in the XZ-plane. What can you say about its y-coordinate?
Answer
A point in the XZ-plane has y-coordinate = 0.
Name the octants in which the following points lie
Answer
(1, 2, 3): all positive → Octant I
(4, –2, 3): x+, y–, z+ → Octant IV
(4, –2, –5): x+, y–, z– → Octant VIII
(4, 2, –5): x+, y+, z– → Octant V
(–4, 2, –5): x–, y+, z– → Octant VI
(–4, 2, 5): x–, y+, z+ → Octant II
(–3, –1, 6): x–, y–, z+ → Octant III
(–2, –4, –7): all negative → Octant VII
Fill in the blanks
Answer
(i) The x-axis and y-axis taken together determine a plane known as XY-plane.
(ii) The coordinates of points in the XY-plane are of the form (x, y, 0).
(iii) Coordinate planes divide the space into eight octants.
Exercise 11.2
Find the distance between the following pairs of points
Answer
Distance = √[(x₂–x₁)² + (y₂–y₁)² + (z₂–z₁)²]
(i) (2,3,5) and (4,3,1): √[(4–2)²+(3–3)²+(1–5)²] = √[4+0+16] = √20 = 2√5
(ii) (–3,7,2) and (2,4,–1): √[(2+3)²+(4–7)²+(–1–2)²] = √[25+9+9] = √43
(iii) (–1,3,–4) and (1,–3,4): √[(1+1)²+(–3–3)²+(4+4)²] = √[4+36+64] = √104 = 2√26
(iv) (2,–1,3) and (–2,1,3): √[(–2–2)²+(1+1)²+(3–3)²] = √[16+4+0] = √20 = 2√5
Show that (–2,3,5), (1,2,3), (7,0,–1) are collinear.
Answer
Let A=(–2,3,5), B=(1,2,3), C=(7,0,–1).
AB = √[(1+2)²+(2–3)²+(3–5)²] = √[9+1+4] = √14
BC = √[(7–1)²+(0–2)²+(–1–3)²] = √[36+4+16] = √56 = 2√14
AC = √[(7+2)²+(0–3)²+(–1–5)²] = √[81+9+36] = √126 = 3√14
AB + BC = √14 + 2√14 = 3√14 = AC. Therefore A, B, C are collinear. ∎
Q3. Verify
(i) (0,7,–10), (1,6,–6), (4,9,–6) are vertices of an isosceles triangle.
Ans. Let A=(0,7,–10), B=(1,6,–6), C=(4,9,–6).
AB = √[1+1+16] = √18 = 3√2
BC = √[9+9+0] = √18 = 3√2
AC = √[16+4+16] = √36 = 6
AB = BC ≠ AC. Therefore isosceles triangle. ∎
(ii) (0,7,10), (–1,6,6), (–4,9,6) are vertices of a right angled triangle.
Ans. Let A=(0,7,10), B=(–1,6,6), C=(–4,9,6).
AB² = 1+1+16 = 18; BC² = 9+9+0 = 18; AC² = 16+4+16 = 36
AB² + BC² = 18+18 = 36 = AC². Therefore right angle at B. ∎
(iii) (–1,2,1), (1,–2,5), (4,–7,8), (2,–3,4) are vertices of a parallelogram.
Ans. Let A=(–1,2,1), B=(1,–2,5), C=(4,–7,8), D=(2,–3,4).
AB = √[4+16+16] = 6; CD = √[4+16+16] = 6. AB=CD.
BC = √[9+25+9] = √43; AD = √[9+25+9] = √43. BC=AD.
Since opposite sides are equal, ABCD is a parallelogram. ∎
Find equation of set of points equidistant from (1,2,3) and (3,2,–1).
Answer
Let P=(x,y,z). PA=PB:
(x–1)²+(y–2)²+(z–3)² = (x–3)²+(y–2)²+(z+1)²
x²–2x+1+z²–6z+9 = x²–6x+9+z²+2z+1
4x–8z = 0 → x–2z = 0 i.e., x = 2z
Find equation of set of points P such that PA + PB = 10, where A=(4,0,0) and B=(–4,0,0).
Answer
This is an ellipse. 2a=10
⇒ a=5; 2c=8
⇒ c=4; b²=25–16=9.
Equation: x²/25 + y²/9 + z²/9 = 1
Miscellaneous Exercise
Three vertices of parallelogram ABCD: A(3,–1,2), B(1,2,–4), C(–1,1,2). Find D.
Answer
In parallelogram, diagonals bisect each other. Midpoint of AC = midpoint of BD.
Midpoint of AC = ((3–1)/2, (–1+1)/2, (2+2)/2) = (1, 0, 2)
Midpoint of BD = ((1+x)/2, (2+y)/2, (–4+z)/2) = (1,0,2)
1+x=2 → x=1; 2+y=0 → y=–2; –4+z=4 → z=8
D = (1, –2, 8)
Find lengths of medians of triangle A(0,0,6), B(0,4,0), C(6,0,0).
Answer
Midpoint of BC = D = (3,2,0). Median AD = √[9+4+36] = 7
Midpoint of AC = E = (3,0,3). Median BE = √[9+16+9] = √34
Midpoint of AB = F = (0,2,3). Median CF = √[36+4+9] = 7
Origin is centroid of PQR with P(2a,2,6), Q(–4,3b,–10), R(8,14,2c). Find a, b, c.
Answer
Centroid = ((2a–4+8)/3, (2+3b+14)/3, (6–10+2c)/3) = (0,0,0)
2a+4=0
⇒ a=–2; 3b+16=0
⇒ b=–16/3; 2c–4=0
⇒ c=2
A(3,4,5), B(–1,3,–7). Find equation of set of points P such that PA²+PB²=k².
Answer
PA² = (x–3)²+(y–4)²+(z–5)²
PB² = (x+1)²+(y–3)²+(z+7)²
PA² + PB² = 2x²+2y²+2z²–4x–14y+4z + (9+16+25+1+9+49)
= 2x²+2y²+2z²–4x–14y+4z+109 = k²
2(x²+y²+z²) – 4x – 14y + 4z = k²–109
Important Extra Questions
Find the coordinates of the point which divides the line segment joining A(2,–1,3) and B(–4,3,1) in ratio 2:1 internally.
Answer
Section formula: P = ((m·x₂+n·x₁)/(m+n), (m·y₂+n·y₁)/(m+n), (m·z₂+n·z₁)/(m+n))
= ((2×(–4)+1×2)/3, (2×3+1×(–1))/3, (2×1+1×3)/3)
= (–8+2)/3, (6–1)/3, (2+3)/3) = (–6/3, 5/3, 5/3) = (–2, 5/3, 5/3)
Find the centroid of triangle with vertices (3,–5,7), (–1,7,–6), (1,1,2).
Answer
Centroid = ((3–1+1)/3, (–5+7+1)/3, (7–6+2)/3) = (3/3, 3/3, 3/3) = (1,1,1)
Show that points (0,–1,–1), (4,5,1), (3,9,4), (–4,4,4) are coplanar.
Answer
Find vectors from first point: AB=(4,6,2), AC=(3,10,5), AD=(–4,5,5).
Scalar triple product = |4 6 2; 3 10 5; –4 5 5|
= 4(50–25) – 6(15+20) + 2(15+40) = 4(25)–6(35)+2(55) = 100–210+110 = 0
Since scalar triple product = 0, points are coplanar. ∎
Find the ratio in which the YZ-plane divides the line joining A(2,4,5) and B(3,5,–4).
Answer
Let YZ-plane divide in ratio k:1. x-coordinate of division point = (3k+2)/(k+1) = 0 (on YZ-plane means x=0).
3k+2=0
⇒ k=–2/3. So ratio is 2:3 externally.
If distance between (a,2,1) and (–1,1,1) is 5, find a.
Answer
√[(a+1)²+1+0] = 5
⇒ (a+1)²+1 = 25
⇒ (a+1)²=24
⇒ a+1=±2√6
⇒ a=–1±2√6