第 11 课 NCERT 解决方案-第 12 章三维几何介绍 - 练习 12.2

问题 1：求以下两对点之间的距离：

(i) (2, 3, 5) 和 (4, 3, 1)

解决方案：

Let P be (2, 3, 5) and Q be (4, 3, 1)

Now, by using the distance formula,

Length of distance PQ = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = 2, y1 = 3, z1 = 5

x2 = 4, y2 = 3, z2 = 1

Length of distance PQ = √[(4 – 2)² + (3 – 3)² + (1 – 5)²]

= √[(2)² + (0)² + (-4)²]

= √[4 + 0 + 16]

= √20

= 2√5

∴ The length of distance PQ is 2√5 units.

(ii) (–3, 7, 2) 和 (2, 4, –1)

解决方案：

Let P be (– 3, 7, 2) and Q be (2, 4, – 1)

Now, by using the distance formula,

Length of distance PQ = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = – 3, y1 = 7, z1 = 2

x2 = 2, y2 = 4, z2 = – 1

Length of distance PQ = √[(2 – (-3))² + (4 – 7)² + (-1 – 2)²]

= √[(5)² + (-3)² + (-3)²]

= √[25 + 9 + 9]

= √43

∴ The length of distance PQ is √43 units.

(iii) (–1, 3, – 4) 和 (1, –3, 4)

解决方案：

Let P be (– 1, 3, – 4) and Q be (1, – 3, 4)

Now, by using the distance formula,

Length of distance PQ = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = – 1, y1 = 3, z1 = – 4

x2 = 1, y2 = – 3, z2 = 4

Length of distance PQ = √[(1 – (-1))² + (-3 – 3)² + (4 – (-4))²]

= √[(2)² + (-6)² + (8)²]

= √[4 + 36 + 64]

= √104

= 2√26

∴ The length of distance PQ is 2√26 units.

(iv) (2, –1, 3) 和 (–2, 1, 3)

解决方案：

Let P be (2, – 1, 3) and Q be (– 2, 1, 3)

Now, by using the distance formula,

Length of distance PQ = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = 2, y1 = – 1, z1 = 3

x2 = – 2, y2 = 1, z2 = 3

Length of distance PQ = √[(-2 – 2)² + (1 – (-1))² + (3 – 3)²]

= √[(-4)² + (2)² + (0)²]

= √[16 + 4 + 0]

= √20

= 2√5

∴ The required distance is 2√5 units.

问题 2：证明点 (–2, 3, 5), (1, 2, 3) 和 (7, 0, –1) 共线。

解决方案：

If three points are collinear, then they lie on a line.

Firstly let us calculate distance between the 3 points

i.e. PQ, QR and PR

P ≡ (– 2, 3, 5) and Q ≡ (1, 2, 3)

Now, by using the distance formula,

Length of distance PQ = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = – 2, y1 = 3, z1 = 5

x2 = 1, y2 = 2, z2 = 3

Length of distance PQ = √[(1 – (-2))² + (2 – 3)² + (3 – 5)²]

= √[(3)2 + (-1)2 + (-2)2]

= √[9 + 1 + 4]

= √14

Length of distance PQ is √14

Q ≡ (1, 2, 3) and R ≡ (7, 0, – 1)

Now, by using the distance formula,

Length of distance QR = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = 1, y1 = 2, z1 = 3

x2 = 7, y2 = 0, z2 = – 1

Length of distance QR = √[(7 – 1)² + (0 – 2)² + (-1 – 3)²]

= √[(6)² + (-2)² + (-4)²]

= √[36 + 4 + 16]

= √56

= 2√14

Length of distance QR is 2√14

P ≡ (– 2, 3, 5) and R ≡ (7, 0, – 1)

Now, by using the distance formula,

Length of distance PR= √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = – 2, y1 = 3, z1 = 5

x2 = 7, y2 = 0, z2 = – 1

Length of distance PR = √[(7 – (-2))² + (0 – 3)² + (-1 – 5)²]

= √[(9)² + (-3)² + (-6)²]

= √[81 + 9 + 36]

= √126

= 3√14

Length of distance PR is 3√14

Thus, PQ = √14, QR = 2√14 and PR = 3√14

So, PQ + QR = √14 + 2√14

= 3√14

= PR

∴ The points P, Q and R are collinear.

问题 3：验证以下内容：

(i) (0, 7, –10), (1, 6, – 6) 和 (4, 9, – 6) 是等腰三角形的顶点。

解决方案：

(0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.

Let us consider the points be

P(0, 7, –10), Q(1, 6, – 6) and R(4, 9, – 6)

If any 2 sides are equal, hence it will be an isosceles triangle

So firstly let us calculate the distance of PQ, QR

P ≡ (0, 7, – 10) and Q ≡ (1, 6, – 6)

Now, by using the distance formula,

Length of distance PQ = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = 0, y1 = 7, z1 = – 10

x2 = 1, y2 = 6, z2 = – 6

Length of distance PQ = √[(1 – 0)² + (6 – 7)² + (-6 – (-10))²]

= √[(1)² + (-1)² + (4)²]

= √[1 + 1 + 16]

= √18

Calculating QR

Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)

Now, by using the distance formula,

Length of distance QR = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = 1, y1 = 6, z1 = – 6

x2 = 4, y2 = 9, z2 = – 6

Length of distance QR = √[(4 – 1)² + (9 – 6)² + (-6 – (-6))²]

= √[(3)² + (3)² + (-6+6)²]

= √[9 + 9 + 0]

= √18

Hence, 

Length of distance PQ = Length of distance QR i.e 

√18 = √18

∴ Length of 2 sides are equal

∴ PQR is an isosceles triangle.

(ii) (0, 7, 10), (–1, 6, 6) 和 (– 4, 9, 6) 是直角三角形的顶点。

解决方案：

 (0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right-angled triangle.

Let the points be

P(0, 7, 10), Q(– 1, 6, 6) & R(– 4, 9, 6)

Firstly let us calculate the distance of PQ, OR and PR

Calculating PQ

P ≡ (0, 7, 10) and Q ≡ (– 1, 6, 6)

Now, by using the distance formula,

Length of distance PQ = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = 0, y1 = 7, z1 = 10

x2 = – 1, y2 = 6, z2 = 6

Length of distance PQ = √[(-1 – 0)² + (6 – 7)² + (6 – 10)²]

= √[(-1)² + (-1)² + (-4)²]

= √[1 + 1 + 16]

= √18

Length of distance PQ is √18cm

Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)

Now, by using the distance formula,

Length of distance QR = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = 1, y1 = 6, z1 = – 6

x2 = 4, y2 = 9, z2 = – 6

Length of distance QR = √[(4 – 1)² + (9 – 6)² + (-6 – (-6))²]

= √[(3)² + (3)² + (-6+6)²]

= √[9 + 9 + 0]

= √18

Length of distance QR is √18cm

P ≡ (0, 7, 10) and R ≡ (– 4, 9, 6)

Now, by using the distance formula,

Length of distance PR = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = 0, y1 = 7, z1 = 10

x2 = – 4, y2 = 9, z2 = 6

Length of distance PR = √[(-4 – 0)² + (9 – 7)² + (6 – 10)²]

= √[(-4)² + (2)² + (-4)²]

= √[16 + 4 + 16]

= √36

Length of distance PR is √36cm

Now,

PQ² + QR² = 18 + 18

= 36

= PR²

By using converse of Pythagoras theorem,

∴ The given vertices P, Q & R are the vertices of a right-angled triangle at Q

(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) 和 (2, –3, 4) 是平行四边形的顶点。

解决方案 ：

(–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

Let the points be: A(–1, 2, 1), B(1, –2, 5), C(4, –7, 8) & D(2, –3, 4)

ABCD can be vertices of parallelogram only if opposite sides are equal.

i.e. AB = CD and BC = AD

Firstly let us calculate the distance

A ≡ (– 1, 2, 1) and B ≡ (1, – 2, 5)

Now, by using the distance formula,

Length of distance AB = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = – 1, y1 = 2, z1 = 1

x2 = 1, y2 = – 2, z2 = 5

Length of distance AB = √[(1 – (-1))2 + (-2 – 2)² + (5 – 1)²]

= √[(2)² + (-4)² + (4)²]

= √[4 + 16 + 16]

= √36

= 6

Length of distance AB is 6cm

B ≡ (1, – 2, 5) and C ≡ (4, – 7, 8)

Now, by using the distance formula,

Length of distance BC = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = 1, y1 = – 2, z1 = 5

x2 = 4, y2 = – 7, z2 = 8

Length of distance BC = √[(4 – 1)² + (-7 – (-2))² + (8 – 5)²]

= √[(3)²

 + (-5)² + (3)²]

= √[9 + 25 + 9]

= √43

Length of distance BC is √43cm

C ≡ (4, – 7, 8) and D ≡ (2, – 3, 4)

Now, by using the distance formula,

Length of distance CD = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = 4, y1 = – 7, z1 = 8

x2 = 2, y2 = – 3, z2 = 4

Length of distance CD = √[(2 – 4)² + (-3 – (-7))² + (4 – 8)²]

= √[(-2)² + (4)² + (-4)²]

= √[4 + 16 + 16]

= √36

= 6

Length of distance CD is 6cm

D ≡ (2, – 3, 4) and A ≡ (– 1, 2, 1)

By using the formula,

Length of distance DA = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = 2, y1 = – 3, z1 = 4

x2 = – 1, y2 = 2, z2 = 1

Length of distance DA = √[(-1 – 2)² + (2 – (-3))² + (1 – 4)²]

= √[(-3)² + (5)² + (-3)²]

= √[9 + 25 + 9]

= √43

Length of distance DA is √43cm

Since AB = CD and BC = DA (given)

So, In ABCD both pairs of opposite sides are equal

∴ ABCD is a parallelogram

问题 4：求与点 (1, 2, 3) 和 (3, 2, –1) 等距的点集的方程。

解决方案：

Let A (1, 2, 3) & B (3, 2, – 1)

Let point P be (x, y, z)

Since it is given that point P(x, y, z) is equal distance from point A(1, 2, 3) & B(3, 2, – 1) i.e. PA = PB

P ≡ (x, y, z) and A ≡ (1, 2, 3)

Now, by using the distance formula,

Now, by using the distance formula, PA = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = x, y1 = y, z1 = z

x2 = 1, y2 = 2, z2 = 3

Length of distance PA = √[(1 – x)² + (2 – y)² + (3 – z)²]

P ≡ (x, y, z) and B ≡ (3, 2, – 1)

Now, by using the distance formula,

Length of distance PB = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = x, y1 = y, z1 = z

x2 = 3, y2 = 2, z2 = – 1

Length of distance PB = √[(3 – x)² + (2 – y)² + (-1 – z)²]

Since PA = PB

Square on both the sides, we get

PA² = PB²

(1 – x)² + (2 – y)² + (3 – z)² = (3 – x)² + (2 – y)² + (– 1 – z)²

(1 + x² – 2x) + (4 + y² – 4y) + (9 + z² – 6z)

(9 + x² – 6x) + (4 + y² – 4y) + (1 + z² + 2z)

– 2x – 4y – 6z + 14 = – 6x – 4y + 2z + 14

4x – 8z = 0

x – 2z = 0

∴ The required equation is x – 2z = 0

问题 5：求点集 P 的方程，其与 A (4, 0, 0) 和 B (– 4, 0, 0) 的距离之和等于 10。

解决方案：

Let A (4, 0, 0) & B (– 4, 0, 0)

Let the coordinates of point P be (x, y, z)

Calculating PA

P ≡ (x, y, z) and A ≡ (4, 0, 0)

Now, by using the distance formula,

Length of distance PA = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = x, y1 = y, z1 = z

x2 = 4, y2 = 0, z2 = 0

Length of distance PA = √[(4– x)² + (0 – y)² + (0 – z)²]

Calculating PB

P ≡ (x, y, z) and B ≡ (– 4, 0, 0)

Now, by using the distance formula,

Length of distance PB = √[(x2 – x1)² + (y2 – y1)² + (z2 – z1)²]

So here,

x1 = x, y1 = y, z1 = z

x2 = – 4, y2 = 0, z2 = 0

Length of distance PB = √[(-4– x)² + (0 – y)² + (0 – z)²]

Now it is given that:

PA + PB = 10

PA = 10 – PB

Square on both the sides, we get

PA² = (10 – PB)²

PA² = 100 + PB² – 20 PB

(4 – x)² + (0 – y)² + (0 – z)²

100 + (– 4 – x)² + (0 – y)² + (0 – z)² – 20 PB

(16 + x² – 8x) + (y²) + (z²)

100 + (16 + x² + 8x) + (y²) + (z²) – 20 PB

20 PB = 16x + 100

5 PB = (4x + 25)

Square on both the sides again, we get

25 PB² = 16x² + 200x + 625

25 [(– 4 – x)² + (0 – y)² + (0 – z)²] = 16x² + 200x + 625

25 [x² + y² + z² + 8x + 16] = 16x² + 200x + 625

25x² + 25y² + 25z² + 200x + 400 = 16x² + 200x + 625

9x² + 25y² + 25z² – 225 = 0

∴ The required equation is 9x² + 25y² + 25z² – 225 = 0