第12类RD Sharma解决方案–第29章飞机–练习29.3 |套装2

问题11。平面穿过点(1 、、-2、5)，并且垂直于将原点连接到点的线( $3\hat{i}+\hat{j}-\hat{k}$ )。找到平面方程的向量和笛卡尔形式。

解决方案：

As we know that the vector equation of a plane passing through a point $\vec{a}$and normal to $\vec{n}$ is

$(\vec{r}-\vec{a}).\vec{n}=0$          ….(i)

Here, $\vec{a}=\hat{i}-2\hat{j}+5\hat{k}$

$\vec{n}=\overrightarrow{OP}$     

$\overrightarrow{OP}$ = Position vector of P – Position of vector of O

= $(3\hat{i}+\hat{j}-\hat{k})-(0\hat{i}+0\hat{j}+0\hat{k})$

$\vec{n}=(3\hat{i}+\hat{j}-\hat{k})$

Now, put, all these values in equation (i), we get,

$[\vec{r}-(\hat{i}-2\hat{j}+5\hat{k})](3\hat{i}+\hat{j}-\hat{k})=0$

$\vec{r}.(3\hat{i}+\hat{j}-\hat{k})-(\hat{i}-2\hat{j}+5\hat{k})(3\hat{i}+\hat{j}-\hat{k})=0$

$\vec{r}.(3\hat{i}+\hat{j}-\hat{k})-[(1)(3)+(-2)(1)+(5)(-1)]=0$

$\vec{r}.(3\hat{i}+\hat{j}-\hat{k})-[3-2-5]=0$

$\vec{r}.(3\hat{i}+\hat{j}-\hat{k})-[-4]=0$

$\vec{r}.(3\hat{i}+\hat{j}-\hat{k})=-4$ ….(2)

Now put $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ in eq(2), we get

$(x\hat{i}+y\hat{j}+z\hat{k}).(3\hat{i}+\hat{j}-\hat{k})=-4$

(x)(3) + (y)(1) + (z)(−1) = -4

3x + y − z = -4

So, this is the required equation of plane.

问题12。找到将线段连接点(1、2、3)和(3、4、5)一分为二并且与之成直角的平面方程。

解决方案：

As we know that the vector equation of a plane passing through a point $\vec{a}$ and normal to $\vec{n}$ is

$(\vec{r}-\vec{a}).\vec{n}=0$          …..(i)

Here, $\vec{a}$= mid-point of AB

So, $\vec{n}=\frac{\overrightarrow{AB}}{2}$ 

= Position vector of A + Position of vector of B/ 2 

= $\frac{\hat{i}+2\hat{j}+3\hat{k}+3\hat{i}+4\hat{j}+5\hat{k}}{2}$

= $\vec{a}=\frac{4\hat{i}+6\hat{j}+8\hat{k}}{2}$

$\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}$

And, $\vec{n}=\overrightarrow{AB}$

$\overrightarrow{AB}$ = Position vector of B – Position of vector of A

= $(3\hat{i}+4\hat{j}+5\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})$

= $3\hat{i}+4\hat{j}+5\hat{k}-\hat{i}-2\hat{j}-3\hat{k}$

 $\vec{n}=2\hat{i}+2\hat{j}+2\hat{k}$

Now put all these values in eq(1), we get

$\vec{r}-(2\hat{i}+3\hat{j}+4\hat{k})(2\hat{i}+2\hat{j}+2\hat{k})=0$

$\vec{r}(2\hat{i}+2\hat{j}+2\hat{k})-[(2\hat{i}+3\hat{j}+4\hat{k})(2\hat{i}+2\hat{j}+2\hat{k})]=0$

$\vec{r}(2\hat{i}+2\hat{j}+2\hat{k})-[(2)(2)+(3)(2)+(4)(2)]=0$

$\vec{r}(2\hat{i}+2\hat{j}+2\hat{k})-[4+6+8]=0$

$\vec{r}(2\hat{i}+2\hat{j}+2\hat{k})=18$   …..(2)

Now put $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ in eq(2), we get

$(x\hat{i}+y\hat{j}+z\hat{k}).(2\hat{i}+2\hat{j}+2\hat{k})=18$

(x)(2) + (y)(2) + (z)(2) = 18

2x + 2y + 2z = 18

or  we can write as

x + y + z = 9

So, this is the required equation of plane.

问题13：证明以下几对平面的法线彼此垂直：

(i)x – y + z – 2 = 0和3x + 2y – z + 4 = 0

解决方案：

Given equations of planes are

x – y + z – 2 = 0 and 3x + 2y – z + 4 = 0

So first we solve, x – y + z – 2 = 0

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

$\vec{r}.\vec{n_1} =2$            …..(i)

Now we solve, 3x + 2y – z =- 4

$(x\hat{i}+y\hat{j}+z\hat{k})(3\hat{i}+2\hat{j}-\hat{k})=-4$

$\vec{r} (3\hat{i}+2\hat{j}-\hat{k})=-4$

$\vec{r}.\vec{n_2} =-4$          …(ii)

So, from eq(i) and (ii), we conclude that

$\vec{n_1}$is normal to eq(i) and $\vec{n_2}$ is normal to eq(ii)

So, 

$\vec{n_1}\vec{n_2}$=$(\hat{i}-\hat{j}+\hat{k}).(3\hat{i}+2\hat{j}-\hat{k})$

= (1)(3) + (-1)(2) + (1)(-1)

=3 – 2 – 1

= 3 – 3 = 0

Hence,$\vec{n_1}$ is perpendicular to $\vec{n_2}.$

(ii) $\vec{r}.(2\hat{i}-\hat{j}+3\hat{k})=5$和$\vec{r}.(2\hat{i}-2\hat{j}-2\hat{k})=5$

解决方案：

Given equations of planes are

$\vec{r}.(2\hat{i}-\hat{j}+3\hat{k})=5$and $\vec{r}.(2\hat{i}-2\hat{j}-2\hat{k})=5$

So first we solve, $\vec{r}.(2\hat{i}-\hat{j}+3\hat{k})=5$

$\vec{r}.(2\hat{i}-\hat{j}+3\hat{k})=5$

$\vec{r}.\vec{n_1} =5$   …..(1)

Now we solve, $\vec{r}.(2\hat{i}-2\hat{j}-2\hat{k})=5$

$\vec{r}.(2\hat{i}-2\hat{j}-2\hat{k})=5$

$\vec{r}.\vec{n_2} =5$     ……(2)

So, from eq(i) and (ii), we conclude that

$\vec{n_1}$is normal to eq(i) and $\vec{n_2}$ is normal to eq(ii)

So, 

$\vec{n_1}.\vec{n_2}$ = $(2\hat{i}-\hat{j}+3\hat{k})(2\hat{i}-2\hat{j}-2\hat{k})$

= (2)(2) + (-1)(-2) + (3)(-2)

= 4 + 2 – 6

= 6 – 6

= 0

Hence,$\vec{n_1}$ is perpendicular to $\vec{n_2}.$

问题14：证明平面2x + 2y + 2z = 3的法线向量与坐标轴相等地倾斜。

解决方案：

Equation of plane = 2x + 2y + 2z = 3

So,

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

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

$\vec{r}.\vec{n} =d$

So, the normal to the plane $\vec{n}=(2\hat{i}+2\hat{j}+2\hat{k})$

and the direction ratio of $\vec{n} = 2,2,2$

So, the direction cosine of $\vec{n} = \frac{2}{|\vec{n}|} ,\frac{2}{|\vec{n}|},\frac{2}{|\vec{n}|}$ ….(1)

$|\vec{n}|$= √[(2)² + (2)² + (2)²]

= √[4 + 4 + 4]

= √12 = 2√3

Now put the value of |\vec{n}| in eq(1), we get

Direction cosine of |\vec{n}| = $\frac{2}{2√3},\frac{2}{2√3},\frac{2}{2√3}$

= $\frac{1}{√3}, \frac{1}{√3}, \frac{1}{√3}$

So, u = 1/√3, v = 1/√3, W = 1/√3 

Let us assume that the α, β, γ be the angle that normal \vec{n} makes with the coordinate axes.

So, u = cos α = 1/√3 

α = cos^-11/√3                   ….(2)

v = cos β = 1/√3 

β = cos^-11/√3                     ….(3)

w = cos γ = 1/√3 

γ = cos^-11/√3                     ….(4)

So, from equation (2), (3) and (4), we get

α = β = γ

Hence proved that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined with the coordinate axes.

问题15。找到一个垂直于平面12x – 3y + 4y = 1的大小为26个单位的向量。

解决方案：

Given that, 

The equation of plane is = 12x – 3y + 4y = 1

and the magnitude = 26 units

So, 

$(x\hat{i}+y\hat{j}+z\hat{k})(12\hat{i}-3\hat{j}+4\hat{k})=1$

$\vec{r}.\vec{n} =1$

The normal to the plane is

$\vec{n}=(12\hat{i}-3\hat{j}+4\hat{k})$

$|\vec{n}|$ = √[(12)² + (-3)² + (4)²]

= √[144 + 9 + 16]

= √169 = 13

Hence, the unit vector $\hat{n}$ = $(12\hat{i}-3\hat{j}+4\hat{k})/13$

$=\frac{12}{13}\hat{i}-\frac{3}{13}\hat{j}+\frac{4}{13}\hat{k}$

Now we find a vector normal to the plane with magnitude

So, 

26 = 26 $\vec{n}$

= 26 $(\frac{12}{13}\hat{i}-\frac{3}{13}\hat{j}+\frac{4}{13}\hat{k})$

= $24\hat{i}-6\hat{j}+8\hat{k}$

So, this is the required vector

问题16.如果从(4，-1，2)绘制的线在点(-10，5，4)处以直角与平面相交，请找到该平面的方程。

解决方案：

As we know that the vector equation of a plane passing through a point $\vec{a}$ and normal to $\vec{n}$ is

$(\vec{r}-\vec{a}).\vec{n}$       …..(i)

Here, $\vec{a}$ = position vector of B

So, $\vec{a}=(-10\hat{i}+5\hat{j}+4\hat{k})$

and $\vec{n}=\overrightarrow{AB}$

$\overrightarrow{AB}$ = Position vector of B – Position of vector of A

= $(-10\hat{i}+5\hat{j}+4\hat{k})-(4\hat{i}-\hat{j}+2\hat{k})$

= $-10\hat{i}+5\hat{j}+4\hat{k}-4\hat{i}+\hat{j}-2\hat{k}$

$\vec{n} = -14\hat{i}+6\hat{j}+2\hat{k}$

Now put all these values in eq(1), we get

$[\vec{r}-(-10\hat{i}+5\hat{j}+4\hat{k})].(-14\hat{i}+6\hat{j}+2\hat{k})=0$

$\vec{r}(-14\hat{i}+6\hat{j}+2\hat{k})-(-10\hat{i}+5\hat{j}+4\hat{k})(-14\hat{i}+6\hat{j}+2\hat{k})=0$

$\vec{r}(-14\hat{i}+6\hat{j}+2\hat{k})-[(-10)(-14)+(5)(6)+(4)(2)]=0$  

$\vec{r}(-14\hat{i}+6\hat{j}+2\hat{k})-[140+30+8]=0$

$\vec{r}(-14\hat{i}+6\hat{j}+2\hat{k})=178$  …..(2)

Now put $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ in eq(2), we get

$(x\hat{i}+y\hat{j}+z\hat{k}).(-14\hat{i}+6\hat{j}+2\hat{k})=178$

(x)(-14) + (y)(6) + (z)(2) = 178

-14x + 6y + 3z = 178

Or we can write as 

7x – 2y – z = -89  

So, this is the required equation of plane.

问题17。找到平面方程，该平面将直角连接点(-1、2、3)和(3，-5、6)的线段一分为二。

解决方案：

Let assume that point (-1, 2, 3) is A point and point (3, -5, 6) is B point and C be the line mid-point of line segment AB

As we know that the vector equation of a plane passing through a point $\vec{a}$ and normal to $\vec{n}$ is

$(\vec{r}-\vec{a}).\vec{n}=0$        …(i)

Here, $\vec{a}$= Position vector of C

So, $\vec{a}=\frac{\overrightarrow{AB}}{2}$     [Because c is the mid point of line AB]             

 $\vec{a}=\frac{-\hat{i}+2\hat{j}+3\hat{k}+3\hat{i}-5\hat{k}+6\hat{k}}{2}$

$=\frac{2\hat{i}}{2}-\frac{3\hat{j}}{2}+\frac{9\hat{k}}{2}$

$\vec{a}=\hat{i}-\frac{3}{2}\hat{j}+\frac{9}{2}\hat{k}$

Now, $\vec{n}=\overrightarrow{AB}$

$\overrightarrow{AB}$ = Position vector of B- Position vector of A

=$(3\hat{i}-5\hat{j}+6\hat{k})-(-\hat{i}+2\hat{j}+3\hat{k})$

= $3\hat{i}-5\hat{j}+6\hat{k}+\hat{i}-2\hat{j}-3\hat{k}$

= $4\hat{i}-7\hat{j}+3\hat{k}$

$\vec{n}=4\hat{i}-7\hat{j}+3\hat{k}$

Now put all these values in eq(1), we get

$[\vec{r}-(\hat{i}-\frac{3}{2}\hat{j}+\frac{9}{2}\hat{k})](4\hat{i}-7\hat{j}+3\hat{k})=0$

$\vec{r}(4\hat{i}-7\hat{j}+3\hat{k})-(\hat{i}-\frac{3}{2}\hat{j}+\frac{9}{2}\hat{k})(4\hat{i}-7\hat{j}+3\hat{k})=0$

$\vec{r}(4\hat{i}-7\hat{j}+3\hat{k})$ = 28 ….(2)

Now put $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ in eq(2), we get

$(x\hat{i}+y\hat{j}+z\hat{k}).(4\hat{i}-7\hat{j}+3\hat{k})= 28$

(x)(4) + (y)(-7) + (z)(3) = 28

4x – 7y + 3z = 28

So, this is the required equation of plane.

问题18。找到通过点(5，2，-4)并垂直于线径比为2、3，-1的平面的矢量和笛卡尔方程。

解决方案：

According to the given question

$\vec{a}=5\hat{i}+2\hat{j}-4\hat{k}$

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

As we know that the vector equation of a plane passing through a point $\vec{a}$ and normal to $\vec{n}$ is

$(\vec{r}-\vec{a}).\vec{n}=0$

So, 

$[\vec{r}-(5\hat{i}+2\hat{j}-4\hat{k})].(2\hat{i}+3\hat{j}-\hat{k})=0$

$\vec{r}(2\hat{i}+3\hat{j}-\hat{k})-(5\hat{i}+2\hat{j}-4\hat{k}).(2\hat{i}+3\hat{j}-\hat{k})=0$

$\vec{r}(2\hat{i}+3\hat{j}-\hat{k})-10-6-4=0$

$\vec{r}(2\hat{i}+3\hat{j}-\hat{k})=20$  ….(1)

For cartesian equation:

Put $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ in eq(1), we get

$(x\hat{i}+y\hat{j}+z\hat{k}).(2\hat{i}+3\hat{j}-\hat{k})=20$

(x)(2) + (y)(3) + (z)(-1) = 20

2x + 3y -z = 20

So, this is the required equation of plane.

问题19.如果O是原点，而P的坐标是(1、2，-3)，则找到通过P并垂直于OP的平面的方程。

解决方案：

According to the question, a normal pass through point O(0, 0, 0) and P (1, 2, -3)

So, $\vec{n}=\hat{i}+2\hat{j}-3\hat{k}$

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

As we know that the vector equation of a plane passing through a point $\vec{a}$ and normal to $\vec{n}$ is

$(\vec{r}-\vec{a}).\vec{n}=0$

So,

$[\vec{r}-(\hat{i}+2\hat{j}-3\hat{k})].(\hat{i}+2\hat{j}-3\hat{k})=0$

$\vec{r}(\hat{i}+2\hat{j}-3\hat{k})-(\hat{i}+2\hat{j}-3\hat{k})(\hat{i}+2\hat{j}-3\hat{k})=0$

$\vec{r}(\hat{i}+2\hat{j}-3\hat{k})-1-4-9=0$

$\vec{r}(\hat{i}+2\hat{j}-3\hat{k})=14$

For cartesian equation:

Put $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ in eq(1), we get

$(x\hat{i}+y\hat{j}+z\hat{k}).(\hat{i}+2\hat{j}-3\hat{k})=14$

(x)(1) + (y)(2) + (z)(-3) = 14

x + 2y – 3z = 14

So, this is the required equation of plane.

问题20.如果O是原点，而A的坐标是(a，b，c)。求出OA的方向余弦，以及与OA成直角的A方向的平面方程。

解决方案：

According to the question it is given that, O is the origin and the coordinates of A are (a, b, c) 

So, $\overrightarrow{OA}=a\hat{i}+b\hat{j}+c\hat{k}$

Since, the direction ratios of OA are proportional to a, b, c

So, the direction cosines are:

$\frac{a}{\sqrt{a^2+b^2+c^2}},\frac{b}{\sqrt{a^2+b^2+c^2}},\frac{c}{\sqrt{a^2+b^2+c^2}}$

So the equation of the line is,

$[(x\hat{i}+y\hat{j}+z\hat{k})-(a\hat{i}+b\hat{j}+c\hat{k})].(a\hat{i}+b\hat{j}+c\hat{k})=0$

ax + by + cz = a² + b² + c²