第 12 课 RD Sharma 解决方案 - 第 29 章飞机 - 练习 29.9

问题1.求点的距离$2\hat{i}-\hat{j}-4\hat{k}$从飞机上$\vec{r}.(3\hat{i}-4\hat{j}+12\hat{k})-9=0.$

解决方案：

As we know the distance of a point $\vec{a}$ from a plane $\vec{r}.\vec{n}=d$ is given by:

$D=|\frac{\vec{a}\vec{n}-d}{|\vec{n}|}|units$

Here, $a = 2\hat{i}-\hat{j}-4\hat{k}$ and $\vec{r}.(3\hat{i}-4\hat{j}+12\hat{k})-9=0$ is the plane.

Hence, $D=|\frac{(2\hat{i}-\hat{j}-4\hat{k})(3\hat{i}-4\hat{j}+12\hat{k})-9}{\sqrt{(3)^2+(-4)^2+(12)^2}}|$

⇒ $D=|\frac{(2)(3)+(-1)(-4)+(-4)(12)-9}{\sqrt{9+16+144}}|$

= |-47/13| units 

= 47/13 units 

Hence, the distance of the point from the plane is 47/13 units.

问题 2. 证明点$\hat{i}-\hat{j}+3\hat{k}$和$3\hat{i}+3\hat{j}+3\hat{k}$与飞机等距$\vec{r}.(5\hat{i}+2\hat{j}-7\hat{k})+9=0.$

解决方案：

As we know the distance of a point $\vec{a}$ from a plane $\vec{r}.\vec{n}=d$ is given by:

$D=|\frac{\vec{a}\vec{n}-d}{|\vec{n}|}|units$

Let D₁ be the distance of $\hat{i}-\hat{j}+3\hat{k}$ from the plane $\vec{r}.(5\hat{i}+2\hat{j}-7\hat{k})+9=0.$

⇒ $D_1=|\frac{(\hat{i}-\hat{j}+3\hat{k})(5\hat{i}+2\hat{j}-7\hat{k})+9}{\sqrt{5)^2+(2)^2+(-7)^2}}|$

= $|\frac{5-2-21+9}{\sqrt{78}}|$

= 9/√78 units                  …….(1)

Now, let D₂ be the distance between point $3\hat{i}+3\hat{j}+3\hat{k}$ and the plane $\vec{r}.(5\hat{i}+2\hat{j}-7\hat{k})+9=0$.

⇒ $D_2=|\frac{(3\hat{i}+3\hat{j}+3\hat{k})(5\hat{i}+2\hat{j}-7\hat{k})+9)}{\sqrt{5)^2+(2)^2+(-7)^2}}|$

= $|\frac{15+6-21+9}{\sqrt{78}}|$

= 9/√78 units                    ……(2)

From eq(1) and (2), we have

The given points are equidistant from the given plane.

问题 3. 求点 (2, 3, -5) 到平面 x + 2y - 2z - 9 = 0 的距离。

解决方案：

As we know the distance is given by:

$|\frac{ax_1+by_1+cz_1+d}{\sqrt{(a)^2+(b)^2+(c)^2}}|$

⇒ $D=|\frac{2+(2)(3)-2(-5)-9}{\sqrt{(1)^2+(2)^2+(-2)^2}}|$

= $|\frac{2+6+10-9}{\sqrt{1+4+4}}|$

= 9/√9 

D = 3 units.

问题 4. 找出平行于平面 x + 2y - 2z + 8 = 0 的平面方程，这些平面与点 (2, 1, 1) 的距离为 2 个单位。

解决方案：

The equation of a plane parallel to the given plane is x + 2y − 2z + p = 0.

As we know that the distance between a point and plane is given by:

$D=|\frac{ax_1+by_1+cz_1+d}{\sqrt{(a)^2+(b)^2+(c)^2}}|$

Given, D = 2 units. Hence,

$2=|\frac{2+2-2+p}{\sqrt{1+4+4}}|$

⇒ $2=|\frac{2+p}{\sqrt9}|$

On squaring both sides, we have

$4=\frac{(2+p)^2}{9}$

⇒ 36 = (2 + p)²

⇒ 2 + p = 6           or         2 + p = −6

⇒ p = 4                 or           p = −8

Hence, the equations of the required planes are:

x + 2y − 2z + 4 = 0 and x + 2y − 2z − 8 = 0.

问题 5. 证明点 (1, 1, 1) 和 (-3, 0, 1) 与平面 3x + 4y − 12z +13 = 0 等距。

解决方案：

We know that the distance between a point and plane is given by:

$D=|\frac{ax_1+by_1+cz_1+d}{\sqrt{(a)^2+(b)^2+(c)^2}}|$

Let D₁ be the distance of the point (1,1,1) from the plane.

⇒ $D_1=|\frac{(3)(1)+(4)(1)-(12)(1)+13}{\sqrt{(3)^2+(4)^2+(-12)^2}}|$

= 8/13 units 

Let D₂ be the distance of the point.

⇒ $D_2=|\frac{(3)(-3)+(4)(0)-(12)(1)+13}{\sqrt{(3)^2+(4)^2+(-12)^2}}|$

= 8/13 units 

Hence, the points are equidistant from the plane.

问题 6. 找出平行于平面 x - 2y + 2z - 3 = 0 的平面方程，这些平面距离点 (2, 1, 1) 有一个单位距离。

解决方案：

Equation of a plane parallel to the given plane is x − 2y + 2z + p = 0.

As we know that the distance between a point and plane is given by:

$D=|\frac{ax_1+by_1+cz_1+d}{\sqrt{(a)^2+(b)^2+(c)^2}}|$

Given, D = 1 units. Hence,

$1=|\frac{1+2-2+p}{\sqrt{1+4+4}}|$

⇒ $1=|\frac{1+p}{\sqrt9}|$

On squaring both sides, we have

$1=\frac{(1+p)^2}{3}$

⇒ 9 = (1 + p)²

⇒ 1 + p = 3           or         1 + p = −3

⇒ p = 2                 or           p = −4

Hence, the equations of the required planes are:

x − 2y + 2z + 2 = 0 and x − 2y + 2z − 4 = 0.

问题 7. 求点 (2, 3, 5) 到 xy 平面的距离。

解决方案：

As we know the distance of the point from the plane is given by:

D = $|\frac{ax_1+by_1+cz_1+d}{\sqrt{(a)^2+(b)^2+(c)^2}}|$

= $|\frac{(2)(0)+(3)(0)+(5)(1)+0}{\sqrt{(0)^2+(0)^2+(1)^2}}|$

= $|\frac{0+0+5}{1}|$

D = 5 units

问题 8. 求点 (3, 3, 3) 到平面的距离$\vec{r}.(5\hat{i}+2\hat{j}+3\hat{k})+9=0.$

解决方案：

As we know the distance of a point $\vec{a}$ from a plane $\vec{r}.\vec{n}=d$ is given by:

D = $|\frac{\vec{a}\vec{n}-d}{|\vec{n}|}|units$

= $|\frac{(3\hat{i}+3\hat{j}+3\hat{k})(5\hat{i}+2\hat{j}+3\hat{k})+9}{\sqrt{25+4+49}}|$

= $|\frac{15+6-21+9}{\sqrt{78}}|$

D = 9/√78 units.

问题 9. 如果点 (1, 1, 1) 到原点和平面 x - y + z + p = 0 的距离的乘积为 5，求 p。

解决方案：

The distance of the point (1, 1, 1) from the origin is $\sqrt{3}.$

Distance of (1, 1, 1) from the plane is $|\frac{1+p}{\sqrt{3}}|$

Given: $\sqrt{3}.|\frac{1+p}{\sqrt{3}}|=5$

⇒ |1 + p| = 5

⇒ p = 4 or −6.

问题 10. 求与平面 3x - 4y + 12 = 6 和 4x + 3z = 7 等距的所有点的集合的方程。

解决方案：

$D_1=|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}|$

  $=|\frac{3x_1-4y_1+12z_1-6}{\sqrt{3^2+)(-4)^2+12^2}}|$

= $|\frac{3x_1-4y_1+12z_1-6}{13}|$

Now, $D_2=|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}|$

= $|\frac{4x_1+3z_1-7}{5}|$

Given, D₁ = D₂

⇒ $|\frac{3x_1-4y_1+12z_1-6}{13}|=|\frac{4x_1+3z_1-7}{5}|$

Hence, the equations become

37x₁ + 20y₁ − 21z₁ − 61 = 0 and  67x₁ + 20y₁ + 99z₁ − 121 = 0.

问题 11. 求点 (7, 2, 4) 与由点 A(-2, -3, 5) 和 C(5, 3, -3) 确定的平面之间的距离。

解决方案：

The equations of the plane are given as:

−4a − 8b + 8c = 0 and 3a − 2b + 0c = 0

Solving the above set using cross multiplication method, we get

$\frac{a}{(-8)(0)-(-2)(8)}=\frac{b}{3(-8)-(-4)(0)}=\frac{c}{(-4)(-2)-(-3)(-8)}=p$

⇒ $\frac{a}{0+16}=\frac{b}{24+0}=\frac{c}{8+24}=p$

⇒ $\frac{a}{16}=\frac{b}{24}=\frac{c}{32}=p$

⇒ $\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=p$

⇒ a = 2p, b = 3p, c = 4p

Thus, the equation of the plane becomes 2x + 3y + 4z − 7 = 0.

and, distance = √29 units.

问题 12. 平面在坐标轴上分别截距 -6、3、4。从它的原点找到垂线的长度。

解决方案：

Since the plane makes intercepts −6, 3, 4, the equation becomes:

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

Let p be the distance of perpendicular from the origin to the plane.

⇒ $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$

⇒ $\frac{1}{p^2}=\frac{1}{(-6)^2}+\frac{1}{3^2}+\frac{1}{4^2}$

⇒ $\frac{1}{p^2}=\frac{4+16+9}{144}$

⇒ 1/p² = 29/144 

⇒ p² = 144/29

⇒ p = 12/√29 units.