第 11 类 NCERT 解决方案-第 12 章三维几何介绍 - 练习 12.3

问题1：求出分割线段的点的坐标，该线段连接点（–2,3,5）和（1,–4,6）的比率。

(i) 2:3 内部

解决方案：

By using Section formula,

$(x, y, z) = \mathbf{(\frac{(mx_2 + nx_1)}{(m + n)}, \frac{(my_2 + ny_1)}{(m + n)}, \frac{(mz_2 + nz_1)}{(m + n)})}$

Upon comparing we get,

x1 = -2, y1 = 3, z1 = 5

x2 = 1, y2 = -4, z2 = 6 and

m = 2, n = 3

So, the coordinates of the point which divides the line segment joining the points P (– 2, 3, 5) and Q (1, – 4, 6) in the ratio 2 : 3 internally is given by:

((2 * 1 + 3 * -2)/(2 + 3), (2 * -4 + 3 * 3)/(2 + 3), (2 * 6 + 3 * 5)/(2 + 3))

((-4/5), (1/5), (27/5))

Hence, the coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) is (-4/5, 1/5, 27/5)

(ii) 2:3 外部

解决方案：

By using Section formula,

$(x, y, z) = \mathbf{(\frac{(mx_2 - nx_1)}{(m - n)}, \frac{(my_2 - ny_1)}{(m - n)}, \frac{(mz_2 - nz_1)}{(m - n)})}$

Upon comparing we get,

x1 = -2, y1 = 3, z1 = 5;

x2 = 1, y2 = -4, z2 = 6 and

m = 2, n = 3

So, the coordinates of the point which divides the line segment joining the points P (– 2, 3, 5) and Q (1, – 4, 6) in the ratio 2: 3 externally is given by:

((2 * 1 – 3 * -2)/(2 + 3), (2 * -4 – 3 * 3)/(2 + 3), (2 * 6 – 3 * 5)/(2 + 3))

((2 + 6)/(-1), (-8 – 9)/(-1), (12 – 15)/(-1)

(-8), (17), (3)

∴ The co-ordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) is (-8, 17, 3).

问题 2：假设 P (3, 2, – 4), Q (5, 4, – 6) 和 R (9, 8, –10) 共线。求 Q 除以 PR 的比率。

解决方案：

Let us consider Q divides PR in the ratio k: 1.

By using Section formula,

$(x, y, z) = \mathbf{(\frac{(mx_2 + nx_1)}{(m + n)}, \frac{(my_2 + ny_1)}{(m + n)}, \frac{(mz_2 + nz_1)}{(m + n)})}$

Upon comparing we get,

x1 = 3, y1 = 2, z1 = -4;

x2 = 9, y2 = 8, z2 = -10 and

m = k, n = 1

So, we have

((k * 9 + 1 * 3)/(k + 1), (k * 8 + 1 * 2)/(k + 1), (-10 * k + 1 * -4)/(k + 1) = (5, 4, -6)

(9k + 3)/(k + 1) = 5, (8k + 2)/(k + 1) = 4, (-10k + -4)/(k + 1) = -6

 (8k + 2)/(k + 1) = 4

8k + 2 = 4k + 4

4k = 2

k = 1/2

Hence, the ratio in which Q divides PR is 1: 2

问题 3：求 YZ 平面分割由点 (–2, 4, 7) 和 (3, –5, 8) 连接形成的线段的比率。

解决方案：

Let the line segment formed by joining the points P (-2, 4, 7) and Q (3, -5, 8) be PQ.

We know that any point on the YZ-plane is of the form (0, y, z).

So now, let R (0, y, z) divides the line segment PQ in the ratio k: 1.

Then,

Upon comparing we get,

x1 = -2, y1 = 4, z1 = 7;

x2 = 3, y2 = -5, z2 = 8 and

m = k, n = 1

By using the Section formula,

$(x, y, z) = \mathbf{(\frac{(mx_2 + nx_1)}{(m + n)}, \frac{(my_2 + ny_1)}{(m + n)}, \frac{(mz_2 + nz_1)}{(m + n)})}$

So we have,

((3k – 2)/(k + 1), (-5k + 4)/(k + 1), (8k + 7)/(k + 1)) = (0, y, z)

3k – 2 = 0

3k = 2

k = 2/3

Hence, the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8) is 2:3.

问题 4：使用截面公式，证明点 A (2, –3, 4)、B (–1, 2, 1) 和 C (0, 1/3, 2) 共线。

解决方案：

Let the point P divides AB in the ratio k: 1.

Upon comparing we get,

x1 = 2, y1 = -3, z1 = 4;

x2 = -1, y2 = 2, z2 = 1 and

m = k, n = 1

By using Section formula,

$(x, y, z) = \mathbf{(\frac{(mx_2 + nx_1)}{(m + n)}, \frac{(my_2 + ny_1)}{(m + n)}, \frac{(mz_2 + nz_1)}{(m + n)})}$

So we have,

The coordinates of P are:

((-k + 2)/(k + 1), (2k – 3)/(k + 1), (k + 4)/(k + 1))

Now, we check if for some value of k, the point coincides with the point C.

Put (-k + 2)/(k + 1) = 0

-k + 2 = 0

k = 2

When k = 2, then (2k – 3)/(k + 1) = (2(2) – 3)/(2 + 1)

= (4 – 3)/3

= 1/3

And, (k + 4)/(k +1) = (2 + 4)/(2 +1)

= 6/3

= 2

∴ C (0, 1/3, 2) is a point which divides AB in the ratio 2: 1 and is same as P.

Hence, A, B, C are collinear.

问题 5：求三等分连接点 P (4, 2, – 6) 和 Q (10, –16, 6) 的线段的点的坐标。

解决方案：

Let A (x1, y1, z1) and B (x2, y2, z2) trisect the line segment joining the points P (4, 2, -6) and Q (10, -16, 6).

A divides the line segment PQ in the ratio 1: 2.

Upon comparing we get,

x1 = 4, y1 = 2, z1 = -6;

x2 = 10, y2 = -16, z2 = 6 and

m = 1, n = 2

By using the Section formula,

$(x, y, z) = \mathbf{(\frac{(mx_2 + nx_1)}{(m + n)}, \frac{(my_2 + ny_1)}{(m + n)}, \frac{(mz_2 + nz_1)}{(m + n)})}$

So we have,

The coordinates of A are:

((1*10 + 2*4)/(1 + 2), (1*(-16) + 2*2)/(1 + 2), (1*6 + 2*(-6))/(1 + 2)

((18/3), (-12/3), (-6/3))

The coordinates of A are (6, -4, -2)

Similarly, we know that B divides the line segment PQ in the ratio 2: 1.

Upon comparing we get,

x1 = 4, y1 = 2, z1 = -6;

x2 = 10, y2 = -16, z2 = 6 and

m = 2, n = 1

By using the Section formula,

$(x, y, z) = \mathbf{(\frac{(mx_2 + nx_1)}{(m + n)}, \frac{(my_2 + ny_1)}{(m + n)}, \frac{(mz_2 + nz_1)}{(m + n)})}$

So we have,

The coordinates of B are:

((2 * 10 + 1 * 4)/(2 + 1), (2 * -16 + 1 * 2)/(2 + 1), (2 * 6 + 1 * -6)/(2 + 1)

The coordinates of B are (8, -10, 2)

∴ The coordinates of the points which trisect the line segment joining the points P (4, 2, – 6) and Q (10, –16, 6) are (6, -4, -2) and (8, -10, 2).