第 12 课 RD Sharma 解决方案 - 第 29 章飞机 - 练习 29.7

问题 1. 求下列平面的标量积形式的向量方程$(\vec{r}.\vec{n}=d):$

解决方案：

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

Here, $\vec{r}=\{2\hat{i}-\hat{k}\}+λ \hat{i}+µ\{\hat{i}-2\hat{j}-\hat{k}\}$

We know that, $\vec{r}=\vec{a}+λ\vec{b}+μ\vec{c}$ represent a plane passing through a point having position vector $\vec{a}$ and parallel to vectors  $\vec{b}$and .$\vec{c}$

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

The given plane is perpendicular to a vector

$\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&0&0\\1&-2&-1\end{vmatrix}\\ =\hat{i}(0-0)-\hat{j}(-1-0)+\hat{k}(-2-0)\\ =0\hat{i}+\hat{j}+2\hat{k}\\ \vec{n}=\hat{j}-2\hat{k}$

We know that vector equation of plane in scalar product form is,

$\vec{r}.\vec{n}=\vec{a}.\vec{n}$—(Equation-1)

Put $\vec{n}$ and $\vec{a}$ in (Equation-1),

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

The equation is required form is,

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

(ii) $\vec{r}=(1+s-t)\hat{i}+(2-s)\hat{j}+(3-2s+2t)\hat{k}$

Here,$\vec{r}=(1+s-t)\hat{i}+(2-s)\hat{j}+(3-2s+2t)\hat{k}\\ \vec{r}=(1+2\hat{j}+3\hat{k})+s(\hat{i}-\hat{j}-2\hat{k})+t(-\hat{i}+2\hat{k})\\$

We know that, $\vec{r}=\vec{a}+λ\vec{b}+μ\vec{c}$ represent a plane passing through a point having position vector $\vec{a}$and parallel to vectors $\vec{b}$ and $\vec{c}$

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

The given plane is perpendicular to a vector

$\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&-2\\-1&0&2\end{vmatrix}\\ =\hat{i}(-2-0)-\hat{j}(2-2)+\hat{k}(0-1)\\ \vec{n}=-2\hat{i}-\hat{k}$

We know that, vector equation of a plane is scalar product is,

$\vec{r}.\vec{n}=\vec{a}.\vec{n}$ —(Equation-1)

Put value of $\vec{a}$ and $\vec{n}$ in (Equation-1)

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

Multiplying both the sides by (-1),

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

The equation in the required form,

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

(iii) $\vec{r}=(\hat{i}+\hat{j})+λ(\hat{i}+2\hat{j}-\hat{k})+μ(-\hat{i}+\hat{j}-2\hat{k})$

Given, equation of plane,

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

We know that, $\vec{r}=\vec{a}+λ\vec{b}+μ\vec{c}$ is the equation of a plane passing through point $\vec{a}$ and parallel to $\vec{b}$ and $\vec{c}$.

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

The given plane is perpendicular to a vector

$\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&-1\\-1&1&-2\end{vmatrix}\\ =\hat{i}(-4+1)-\hat{j}(-2-1)+\hat{k}(1+2)\\ -3\hat{i}+3\hat{j}+3\hat{k}$

We know that, equation of plane in scalar product form is given by,  

$\vec{r}.\vec{n}=\vec{a}.\vec{n}\\ \vec{r}.(-3\vec{i}+3\vec{j}+3\hat{k})=(\hat{i}+\hat{j})(-3\vec{i}+3\vec{j}+3\hat{k})\\ =(1)(-3)+(1)(3)+(0)(3)\\ =-3+3\\ \vec{r}.(-3\vec{i}+3\vec{j}+3\hat{k})=0$

Dividing by 3, we get

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

Equation in required form is,

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

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

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

Plane is passing through $(\hat{i}-\hat{j})$ and parallel to b $b(\hat{i}+\hat{j}+\hat{k})$ and $c(4\hat{i}-2\hat{j}+3\hat{k})$

$\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&1\\4&-2&3\end{vmatrix}\\ n=5i+j-6k\\ \vec{r}.n=(i-j)(5i+j-6k)=5-1=4\\ \vec{r}.(5i+j-6k)=4$

问题 2. 求下列平面方程的笛卡尔形式：

解决方案：

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

Here, given equation of plane is,

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

We know that, $\vec{r}=\vec{a}+λ\vec{b}+μ\vec{c}$ represents the equation of a plane passing through a vector $\vec{a}$ and parallel to vector $\vec{b}$ and $\vec{c}$ . 

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

Given plane is perpendicular to vector

$\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&1&2\\1&2&1\end{vmatrix}\\ =\hat{i}(1-4)-\hat{j}(-1-2)+\hat{k}(-2-1)\\ \vec{n}=-3\hat{i}+3\hat{j}-3\hat{k}$

We know that, equation of plane in the scalar product form,

$\vec{r}.\vec{n}=\vec{a}.\vec{n}$ —Equation-1

Put the value of $\vec{a}$ and $\vec{n}$ in Equation-1,

$\vec{r}.(\hat{i}-\hat{j})=(\hat{i}-\hat{j})(-3\hat{i}+3\hat{j}-3\hat{k})\\ \vec{r}.(-3\hat{i}+3\hat{j}-3\hat{k})=(1)(-3)+(-1)(3)+(0)(-3)\\ =-3-3+0\\ \vec{r}.(-3\hat{i}+3\hat{j}-3\hat{k})=-6$

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

(x)(-3) + (y)(3) + (z)(-3) = -6

-3x + 3y – 3z = -6

Dividing by (-3), we get

x – y + z = 2

Equation in required form is,

x – y + z = 2

(ii) $\vec{r}=(1+s+t)\hat{i}+(2-s+t)\hat{j}+(3-2s+2t)\hat{k}$

Given, equation of plane,

$\vec{r}=(1+s+t)\hat{i}+(2-s+t)\hat{j}+(3-2s+2t)\hat{k}\\ =(\hat{i}+2\hat{j}+3\hat{k})+s(\hat{i}-\hat{j}-2\hat{k})+t(\hat{i}+\hat{j}+2\hat{k})$

We know that, $\vec{r}=\vec{a}+λ\vec{b}+μ\vec{c}$  represents the equation of a plane passing through the vector $\vec{a}$ and parallel to vector $\vec{b}$ and $\vec{c}$

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

The given plane is perpendicular to vector

$\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&-2\\1&1&2\end{vmatrix}\\ =\hat{i}(-2+2)-\hat{j}(2+2)+\hat{k}(1+1)\\ =0(\hat{i})-4\hat{j}+2\hat{k}\\ \vec{n}=-4\hat{j}+2\hat{k}$

We know that, equation of plane in scalar product form is given by,

$\vec{r}.\vec{n}=\vec{a}.\vec{n}$ —Equation-1

Put, the value of $\vec{a}$ and $\vec{n}$ in equation-1

$\vec{r}.(-4\hat{j}+2\hat{k})=(\hat{i}+2\hat{j}+3\hat{k})(-4\hat{j}+2\hat{k})\\ \vec{r}.(-4\hat{j}+2\hat{k})=(1)(0)+(2)(-4)+(3)(2)\\ =0-8+6\\ \vec{r}.(-4\hat{j}+2\hat{k})=-2$

Put $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

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

(x)(0) + (y)(-4) + (z)(2) = -2

-4y + 2z = -2

The equation in required form is,

2y – z = 1

问题 3. 求下列平面的非参数形式的向量方程：

解决方案：

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

Given, equation of plane is,

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

We know that, $\vec{r}=\vec{a}+λ\vec{b}+μ\vec{c}$ represents the equation of a plane passing through a point $\vec{a}$ and parallel to vector $\vec{b}$ and $\vec{c}$ .

Given,

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

The given plane is perpendicular to

$\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&0&2\\-2&-1&1\end{vmatrix}\\ =\hat{i}(0+2)-\hat{j}(1+4)+\hat{k}(1-0)\\ \vec{n}=2\hat{i}-5\hat{j}-\hat{k}$

Vector equation of plane in non-parametric form is.

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

= (0)(2) + (3)(-5) + (0)(-1)

= 0 – 15 + 0

$\vec{r}.(2\hat{i}-5\hat{j}-\hat{k})=-15\\ \vec{r}.(2\hat{i}-5\hat{j}-\hat{k})+15=0$

The required form of equation is,

$\vec{r}.(2\hat{i}-5\hat{j}-\hat{k})+15=0$

(ii) $\vec{r}=(2\hat{i}+2\hat{j}-\hat{k})+λ(\hat{i}+2\hat{j}+3\hat{k})+μ(5\hat{i}-2\hat{j}+7\hat{k})\\$

Given, equation of plane is,

$\vec{r}=(2\hat{i}+2\hat{j}-\hat{k})+λ(\hat{i}+2\hat{j}+3\hat{k})+μ(5\hat{i}-2\hat{j}+7\hat{k})\\$

We know that, $\vec{r}=\vec{a}+λ\vec{b}+μ\vec{c}$ represents the equation of a plane passing through a vector $\vec{a}$ and parallel to vector $\vec{b}$ and $\vec{c}$ .

Here,

$\vec{a}=2\hat{i}+2\hat{j}-\hat{k}\\ \vec{b}=\hat{i}+2\hat{j}+3\hat{k}\\ \vec{c}=5\hat{i}-2\hat{j}+7\hat{k}$

The given plane is perpendicular to vector

$\vec{n}=\vec{b}\times\vec{c}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&3\\5&-2&7\end{vmatrix}\\ =\hat{i}(14+6)-\hat{j}(7-15)+\hat{k}(-2-10)\\ \vec{n}=20\hat{i}+8\hat{j}-12\hat{k}$

We know that, equation of a plane in non-parametric form is given by,

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

$\vec{r}.(20\hat{i}+8\hat{j}-12\hat{k})=(2\hat{i}+2\hat{j}-\hat{k})(20\hat{i}+8\hat{j}-12\hat{k})\\$

= (2)(20) + (2)(8) – (-1)(-12)

=40 + 16 + 12

$\vec{r}.(20\hat{i}+8\hat{j}-12\hat{k})=68$

Dividing by 4,

$\vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=17$

Equation of plane in required form is,

$\vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=17$