第12类RD Sharma解决方案–第28章空间中的直线–练习28.1 |套装2

问题11.找到线的方向余弦$\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$ 。另外，将其简化为矢量形式

解决方案：

Given:

$\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$

$\frac{x-4}{-2}=\frac{y}{6}=\frac{1-z}{-3}=\lambda$

x = -2λ + 4, y = 6λ, z = -3λ + 1

So,

$x\hat{i}+y\hat{j}+z\hat{k}=(-2λ+4)\hat{i}+(6λ)\hat{j}+(-3λ+1)\hat{k}\\ \vec{r}=(4\hat{i}+\hat{k})+λ(-2\hat{i}+6\hat{j}-3\hat{k})$

Direction ratios of the line are = -2, 6, -3

Direction cosines of the lines are,

$\frac{a}{\sqrt{a^2+b^2+c^2}},\ \frac{b}{\sqrt{a^2+b^2+c^2}},\ \frac{c}{\sqrt{a^2+b^2+c^2}}\\ \frac{-2}{\sqrt{(-2)^2+(6)^2+(-3)^2}},\ \frac{6}{\sqrt{(-2)^2+6^2+(-3)^2}},\ \frac{-3}{\sqrt{(-2)^2+6^2+(-3)^2}}\\ \frac{-2}{7},\ \frac{6}{7},\ \frac{-3}{7}$

问题12.线的笛卡尔方程为x = ay + b，z = cy + d。找到其方向比率，并将其简化为矢量形式。

解决方案：

x = ay + b

z = cy + d

$\frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}=\lambda$

So, DR’s of line are (a, 1, c)

From above equation, we can write

x = aλ + b

y = λ

z = cλ + d

So vector equation of line is

$x\hat{i}+y\hat{j}+z\hat{k}=(b\hat{i}+d\hat{k})+λ(a\hat{i}+\hat{j}+c\hat{k})$

问题13.找到通过具有位置矢量的点的直线的矢量方程$\hat{i}-2\hat{j}-3\hat{k}$并平行于将点与位置向量连接在一起的线$\hat{i}-\hat{j}+4\hat{k}$和$2\hat{i}+\hat{j}+2\hat{k}$ 。同样，找到该方程式的笛卡尔等效项。

解决方案：

We know that, equation of a line passing through$\vec{a}$and parallel to vector$\vec{b}$is

$\vec{r}=\vec{a}+λ\vec{b}$……. (i)

Here,

$\vec{a}=\hat{i}-2\hat{j}-3\hat{k}$

and, \vec{b} = line joining$(\hat{i}-\hat{j}+4\hat{k})$and$(2\hat{i}+\hat{j}+2\hat{k})$

$=(2\hat{i}+\hat{j}+2\hat{k})-(\hat{i}-\hat{j}+4\hat{k})\\ =2\hat{i}-\hat{i}+\hat{j}+\hat{j}+2\hat{k}-4\hat{k}\\ =\hat{i}+2\hat{j}-2\hat{k}$

Equation of the line is

$\vec{r}=(\hat{i}-2\hat{j}-3\hat{k})+\lambda(\hat{i}+2\hat{j}-2\hat{k})$

For Cartesian form of equation put$x\hat{i}+y\hat{j}+z\hat{k}$

$x\hat{i}+y\hat{j}+z\hat{k}=(1+\lambda)+\hat{i}(-2+2\lambda)\hat{j}+(-3-2\lambda)\hat{k}$

Equating coefficients of$\hat{i},\ \hat{j},\ \hat{k}$

x = 1 + λ, y = -2 + 2λ, z = -3 – 2λ

$\frac{x-1}{1}=λ,\ \frac{y+2}{2}=λ,\ \frac{z+3}{-2}=λ\\ \frac{x-1}{1}=\frac{y+2}{2}=\frac{z+3}{-2}$

问题14.找到线上的点$\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$与点P(1、3、3)的距离为5个单位。

解决方案：

Given, line is$\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=\lambda$

General points Q on line is (3λ – 2, 2λ -1), 2λ + 3)

Distance of points P from Q =$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

PQ =$\sqrt{(3λ-2-1)^2+(2λ-1-3)^2+(2λ+3-3)^2}$

(5)² = (3λ -3)² + (2λ – 4) + (2λ)²

25 = 9λ² + 9 – 18λ + 4λ² + 16 – 16λ + 4λ²

17λ² – 34λ = 0

17λ (λ – 2) = 0

λ = 0 or 2

So, points on the line are (3(0) – 2, 2(0) – 1, 2(0) + 3)

(3(2) – 2, 2(2) – 1, 2(2) + 3)

= (-2, -1, 3), (4, 3, 7)

问题15：证明点的位置向量是$-2\hat{i}+3\hat{j},\ \hat{i} + 2\hat{j}+3\hat{k}$和$7\hat{i}-\hat{k}$是共线的。

解决方案：

Let the given points are A,B.C with position vectors$\vec{a},\ \vec{b},\ \vec{c}$respectively.

$\vec{a}=2\hat{i}+3\hat{j},\ \vec{b}=\hat{i}+2\hat{j}+3\hat{k},\ \vec{c}=7\hat{i}-\hat{k}$

We know that, equation of a line passing through$\vec{a}$and$\vec{b}$are

$\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})\\ =(-2\hat{i}+3\hat{j})+λ((\hat{i}+2\hat{j}+3\hat{k})-(-2\hat{i}+3\hat{j}))\\ =(-2\hat{i}+3\hat{j})+\lambda(\hat{i}+2\hat{j}+3\hat{k}+2\hat{i}-3\hat{j})\\ \vec{r}=(-2\hat{i}+3\hat{j})+\lambda(3\hat{i}-\hat{j}+3\hat{k})\ \ \ \ .....(i)$

If A, B, C are collinear then$\vec{c}$must satisfy equation (i)

$7\hat{i}-\hat{k}=(-2+3\lambda)\hat{i}+(3-\lambda)\hat{j}+(3\lambda)\hat{k}$

Equation the coefficients of$\vec{i},\ \vec{j},\ \vec{k}$

-2 + 3 = 7 , λ = 3

3 – λ = 0 , λ = 3

3λ = -1 , λ =$-\frac{1}{3}$

Since, value of λ are not equal, so,

Given points are collinear.

问题16.找到通过点(1、2、3)并与该线平行的线的笛卡尔方程和矢量方程$\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}$

解决方案：

We know that, equation of a line passing through a point (x₁, y₁, z₁) and having direction ratios proportional to a, b, c is

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\ \ \ \ .....(i)$

Here,

(x₁, y₁, z₁) = (1, 2, 3) and

Given line$\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}$

$\frac{x+2}{-1}=\frac{y+3}{7}=\frac{z-3}{\frac{3}{2}}$

Its parallel to the required line, so

a = μ , b = 7μ, c =$\frac{3}{2}$μ

So, equation of required line using equation (i) is,

$\frac{x-1}{-μ }=\frac{y-2}{7μ }=\frac{z-3}{\frac{3}{2}μ }\\ \frac{x-1}{-1}=\frac{y-2}{7}=\frac{z-3}{\frac{3}{2}}$

Multiplying the denominators by 2

$\frac{x-1}{-2}=\frac{y-2}{14}=\frac{z-3}{3}=λ$

x = -2λ + 1, y = 14λ + 2, z = 3λ + 3

So, vector form of the equation of required line,

$x\hat{i}+y\hat{j}+z\hat{k}=(-2λ+1)\hat{i}+(14λ+2)\hat{j}+(3λ+3)\hat{k}\\ \vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+λ(-2\hat{i}+14\hat{j}+3\hat{k})$

问题17.直线的笛卡尔方程为3x + 1 = 6y – 2 = 1 – z。找到它穿过的固定点，其方向比率以及其矢量方程。

解决方案：

Given equation of line is,

3x + 1 = 6y -2 = 1 – z

Dividing all by 6

$=\frac{3x+1}{6}=\frac{6y-2}{6}=\frac{1-z}{6}\\ =\frac{3x}{6}+\frac{1}{6}=\frac{6y}{6}-\frac{2}{6}=\frac{1}{6}-\frac{z}{6}\\ =\frac{1}{2}x+\frac{1}{6}=y-\frac{1}{3}=-\frac{z}{6}+\frac{1}{6}\\ =\frac{1}{2}\left(x+\frac{1}{3}\right)=1\left(y-\frac{1}{3}\right)=\frac{1}{6}(z-1)\\ =\frac{x+\frac{1}{3}}{2}=\frac{y-\frac{1}{3}}{1}=\frac{z-1}{-6}=λ$

Comparing it with equation of line equation of line passing through (x,₁ y₁, z₁) and the direction ratios a, b, c,

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ (x_1,\ y_1,\ z_1)=\left(-\frac{1}{3},\ \frac{1}{3},\ 1\right)$

a = 2, b = 1, -6

So, direction ratios of the line are -2, 1, -6

From equation (i)

$x = \left(2λ-\frac{1}{3}\right),\ y=\left(λ+\frac{1}{3}\right),\ z=(-6λ+1)$

So, vector equation of the given line is,

$x\hat{i}+y\hat{j}+z\hat{k}=\left(2λ-\frac{1}{3}\right)\hat{i}+\left(λ+\frac{1}{3}\right)\hat{j}+(-6λ+1)\hat{k}\\ \vec{r}=\left(-\frac{1}{3}\hat{i}+\frac{1}{3}\hat{j}+\hat{k}\right)+λ(2\hat{i}+\hat{j}-6\hat{k})$