第12类RD Sharma解决方案–第29章飞机–练习29.5

问题1.找到通过点(1、1、1)，(1，-1、1)和(-7，-3，-5)的平面的矢量方程

解决方案：

Given that, plane is passing through

(1, 1, 1), (1, -1, 1) and (-7, -3, -5)

We know that, equation of plane passing through 3 points,

$\begin{vmatrix}x-x_1 & y-y_1 & z-z_1\\ x_2-x_1 & y_2-y_1 & z_2-z_1\\ x_3-x_1 & y_3-y_1 & z_3-z_1\end{vmatrix}=0$

$\begin{vmatrix}x-1 & y-1 & z-1\\ 1-1 & -1-1 & 1-1\\ -7-1 & -3-1 & -5-1\end{vmatrix}=0$

$\begin{vmatrix}x-1 & y-1 & z-1\\ 0 & -2 & 0\\ -8 & -4 & -6\end{vmatrix}=0$

(x – 1)(12 – 0) – (y – 1)(0 – 0) + (z – 1)(0 – 16) = 0

(x – 1)(12) – (y – 1)(0) + (z – 1)(-16) = 0

12x – 12 – 0 – 16z + 16 = 0

12x – 16z + 4 = 0

Dividing by 4,

3x – 4z + 1 = 0

$(x\hat{i}+y\hat{j}+z\hat{k})(3\hat{i}+0\hat{j}-4\hat{k})+1=0\\ \vec{r}.(3\hat{i}-4\hat{k})+1=0$

Equation of the required plane,

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

问题2。找到通过点P(2，5，-3)，Q(-2，-3，5)和R(5，3，-3)的平面的矢量方程。

解决方案：

Let  P(2, 5, -3), Q(-2, -3, 5) and R(5, 3, -3) be the three points on a plane having position vectors $\vec{p},\vec{q}\ and\ \vec{s}$ respectively. Then the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ are in the same plane. Therefore, $\overrightarrow{PQ}\times\overrightarrow{PR}$ is a vector perpendicular to the plane.

Let  = $\vec{n} = \overrightarrow{PQ}\times\overrightarrow{PR}$

$\overrightarrow{PQ}=(-2-2)\hat{i}+(-3-5)\hat{j}+(5-(-3))\hat{k}\\ \overrightarrow{PQ}=-4\hat{i}-8\hat{j}+8\hat{k}$ 

Similarly,

$\overrightarrow{PR}=(5-2)\hat{i}+(3-5)\hat{j}+(-3-(-3))\hat{k}\\ \overrightarrow{PQ}=3\hat{i}-2\hat{j}+0\hat{k}$

Thus 

$\vec{n}=\overrightarrow{PQ}\times\overrightarrow{PR}\\ =\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ -4 & -8 & 8\\ 3 & -2 & 0\end{vmatrix}\\ =16\hat{i}+24\hat{j}+32\hat{k}$

The plane passes through the point P with position vector $\vec{p}=2\hat{i}+5\hat{j}-3\hat{k}$

Thus, its vector equation is 

$(\vec{r}-(2\hat{i}+5\hat{j}-3\hat{k})).(16\hat{i}+24\hat{j}+32\hat{k})=0\\$
$(\vec{r}-(2\hat{i}+5\hat{j}-3\hat{k})).(16\hat{i}+24\hat{j}+32\hat{k})=0\\ ⇒\vec{r}.(16\hat{i}+24\hat{j}+32\hat{k})-(32+120-96)=0\\ ⇒\vec{r}.(16\hat{i}+24\hat{j}+32\hat{k})-56=0\\ ⇒\vec{r}.(16\hat{i}+24\hat{j}+32\hat{k})=56\\ ⇒\vec{r}.(2\hat{i}+3\hat{j}+4\hat{k})=7$

问题3.找到通过点A(a，0，0)，B(0，b，0)和C(0，0，c)的平面的矢量方程。将其缩小为正常形式。如果飞机ABC距原点的距离为p，则证明$\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$

解决方案：

Let A(a, 0, 0), B(0, b, 0) and C(0, 0, c) be three points on a plane having their position vectors $\vec{a},\vec{b}\ and\ \vec{c}$ respectively. Then vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are in the same plane. Therefore, $\overrightarrow{AB}\times\overrightarrow{AC}$ is a vector perpendicular to the plane.

Let $\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC}$

$\overrightarrow{AB}=(0-a)\hat{i}+(b-0)\hat{j}+(0-0)\hat{k}\\ \overrightarrow{AB}=-a\hat{i}+b\hat{j}+0\hat{k}$

Similarly,

$\overrightarrow{AC}=(0-a)\hat{i}+(0-0)\hat{j}+(c-0)\hat{k}\\ \overrightarrow{AC}=-a\hat{i}+0\hat{j}+c\hat{k}$

Thus

$\vec{n} = \overrightarrow{AB}\times\overrightarrow{AC}$

      $\hat{i}\ \ \ \hat{j}\ \ \ \hat{k}$

= |  -a  b  0 |

       -a  0  c

$\vec{n}=bc\hat{i}+ac\hat{j}+ab\hat{k}$

$⇒\hat{n}=\frac{bc\hat{i}+ac\hat{j}+ab\hat{k}}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}$

The plane passes through the point P with position vector $\vec{a}=a\hat{i}+0\hat{j}+0\hat{k}$

Thus, the vector equation in the normal form is

$\{\vec{r}-\left(a\hat{i}+0\hat{j}+0\hat{k}\right)\}.\left(\frac{bc\hat{i}+ac\hat{j}+ab\hat{k}}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}\right)=0\\ ⇒\vec{r}.\frac{(bc\hat{i}+ac\hat{j}+ab\hat{k})}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}=\frac{abc}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}\\ ⇒\vec{r}.\frac{(bc\hat{i}+ac\hat{j}+ab\hat{k})}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}=\frac{1}{\sqrt{\frac{b^2c^2+a^2c^2+a^2b^2}{a^2b^2c^2}}}\\ ⇒\vec{r}.\frac{(bc\hat{i}+ac\hat{j}+ab\hat{k})}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}...(1)$

The vector equation of a plane normal to the unit vector $\hat{n}$ and at a distance ‘d’ from the origin is $\vec{r}.\hat{n}=d$ ….(2).

Given that the plane is at a distance ‘p’ from the origin.

Comparing equations (1) and (2), we have,

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

问题4.找到通过点(1、1，-1)，(6、4，-5)和(-4，-2、3)的平面的矢量方程。

解决方案：

Let P(1, 1, -1), Q(6, 4, -5) and R(-4, -2, 3) be three points on a plane having position vectors $\vec{p},\ \vec{q}\ and\ \vec{s}$ respectively. Then the vectors $\overrightarrow{PQ}\ and\ \overrightarrow{PR}$ are in the same plane. Therefore, $\overrightarrow{PQ}\times\overrightarrow{PR}$ is a vector perpendicular to the plane.

Let $\vec{n}=\overrightarrow{PQ}\times\overrightarrow{PR}$

$\overrightarrow{PQ}=(6-1)\hat{i}+(4-1)\hat{j}+(-5-(-1))\hat{k}\\ \overrightarrow{PQ}=5\hat{i}+3\hat{j}-4\hat{k}$

Similarly,

$\overrightarrow{PR}=(-4-1)\hat{i}+(-2-1)\hat{j}+(3-(-1))\hat{k}\\ \overrightarrow{PR}=-5\hat{i}-3\hat{j}+4\hat{k}$

Thus

Here, $\overrightarrow{PQ}=-\overrightarrow{PR}$

Therefore, the given points are collinear.

Thus, $\vec{n}=a\hat{i}+b\hat{j}+c\hat{k}$ where, 5a + 3b – 4c = 0

The plane passes through the point P with position vector $\vec{p}=\hat{i}+\hat{j}-\hat{k}$

Thus, its vector equation is

$\{\vec{r}-(\hat{i}+\hat{j}-\hat{k})\}.(a\hat{i}+b\hat{j}+c\hat{k})=0$ , where, 5a + 3b – 4c = 0

问题5.找到通过点的平面的矢量方程

$3\hat{i}+4\hat{j}+2\hat{k},\ 2\hat{i}-2\hat{j}-\hat{k}\ \ and\ \ 7\hat{i}+6\hat{k}$

解决方案：

Let A, B, C be the points with position vector $(3\hat{i}+4\hat{j}+2\hat{k}),(2\hat{i}-2\hat{j}-\hat{k})\ and\ (7\hat{i}+6\hat{k})$

respectively. Then

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

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

$\overrightarrow{BC}$  = Position vector of C – Position vector of B

$=(7\hat{i}+6\hat{k})-(2\hat{i}-2\hat{j}-\hat{k})\\ =7\hat{i}+6\hat{k}-2\hat{i}+2\hat{j}+\hat{k}\\ =5\hat{i}+2\hat{j}+7\hat{k}$

A vector normal to A, B, C is a vector perpendicular to $\overrightarrow{AB}\times\overrightarrow{BC}$

$\vec{n}=\overrightarrow{AB}\times\overrightarrow{BC}\\ =\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ -1&-6&-3\\ 5&2&7\end{vmatrix}\\ \vec{n}=\hat{i}(-42+6)-\hat{j}(-7+15)+\hat{k}(-2+30)\\ =-36\hat{i}-8\hat{j}+28\hat{k}$

As we know that, equation of a plane passing through vector $\vec{a}$ and perpendicular to vector $\vec{n}$ is given by,

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

Put $\vec{a}$ and $\vec{n}$ in equation (1)

$\vec{r}.(-36\hat{i}-8\hat{j}+28\hat{k})=(3\hat{i}+4\hat{j}+2\hat{k})(-36\hat{i}-8\hat{j}+28\hat{k})$

= (3)(-36) + (4)(-8) + (2)(28)

= -108 – 32 + 56

= -140 + 56

$\vec{r}.(-36\hat{i}-8\hat{j}+28\hat{k})$ = -84

Dividing by (-4), we will get

$\vec{r}.(9\hat{i}+2\hat{j}-7\hat{k})=21$

Equation of required plane is,

$\vec{r}.(9\hat{i}+2\hat{j}-7\hat{k})=21$