第 12 类 RD Sharma 解决方案 - 第 29 章飞机 - 练习 29.15 |设置 2

问题 8. 求平面 2x – y + z + 3 = 0 中点 (1, 3, 4) 的图像。

解决方案：

According to the question we have to find the image of point P(1, 3, 4)

in the plane 2x – y + z +3 = 0

Now let us assume that Q be the image of the point.

Here, the direction ratios of normal to plane = 2, -1, 1

The direction ratios of PQ which is parallel to the normal to the plane 

is proportional to 2, -1, 1 and the line PQ is passing through point P(1, 3, 4).

Thus, equation of the line PQ is: 

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{-1}=λ$

Now, the general point on the line PQ = (2λ + 1, -λ + 3, λ + 4)

Let Q = (2λ + 1, -λ + 3, λ + 4)          -Equation(1)

Here, Q is the image of P, so R is the mid point of PQ

Coordinates of R = $\left(\frac{2λ+1+1}{2},\frac{-λ+3+3}{2},\frac{λ+4+4}{2}\right)\\ =\left(\frac{2λ+2}{2},\frac{-λ+6}{2},\frac{λ+8}{2}\right)\\ =\left(λ+1,\frac{-λ+6}{2},\frac{λ+8}{2}\right)$

Point R is lies on the plane 2x – y + z + 3 = 0

= 2(λ + 1) – $\left(\frac{-λ+6}{2}+\frac{λ+8}{2}\right) = 0$

4λ + 4 + λ – 6 + λ + 8 + 6 = 0

6λ = -12

λ = -2

Now, put the value of λ in equation(1), we get

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

= (-3, 5, 2)

Hence, the image of point P(1, 3, 4) is (-3, 5, 2)

问题9.用位置向量求点的距离$-\hat{i}-5\hat{j}-10\hat{k}$从线的交点$\vec{r}=(2\hat{i}-\hat{j}+2\hat{k})+λ(3\hat{i}+4\hat{j}+12\hat{k})$与飞机$\vec{r}.(\hat{i}-\hat{j}+\hat{k})=5$ .

解决方案：

According to the question we have to find distance of a point A with position 

vector $(-\hat{i}-5\hat{j}-10\hat{k})$from the point of intersection of 

line $\vec{r}=(2\hat{i}-\hat{j}+2\hat{k})+λ(3\hat{i}+4\hat{j}+12\hat{k})$

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

Let the point of intersection of line and plan be $B(\vec{b})$

The line and the plane will intersect when,

$[(2\hat{i}-\hat{j}+2\hat{k})+λ(3\hat{i}+4\hat{j}+12\hat{k})](\hat{i}-\hat{j}+\hat{k})=5\\ [(2+3λ)\hat{i}+(-1+4λ)\hat{j}+(2+12λ)\hat{k}](\hat{i}-\hat{j}+\hat{k})=5\\$

(2 + 3λ)(1) + (-1 + 4λ)(-1) + (2 + 12λ)(1) = 5

2 + 3λ + 1 – 4λ + 2 + 12λ = 5

11λ = 5 – 5

λ = 0

So, the point B is given by

$\vec{b}=(2\hat{i}-\hat{j}+2\hat{k})+(0)(3\hat{i}+4\hat{j}+12\hat{k})\\ \vec{b}=(2\hat{i}-\hat{j}+2\hat{k})$

$\vec{AB}=\vec{b}-\vec{a}$

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

$=(2\hat{i}-\hat{j}+2\hat{k}+\hat{i}+5\hat{j}+10\hat{k})=(3\hat{i}+4\hat{j}+12\hat{k})\\ |\vec{AB}|=\sqrt{(3)^2+(4)^2+(12)^2}=\sqrt{9+16+144}=\sqrt{169}=13$

The required distance is 13 units.

问题 10. 求点 (1, 1, 2) 到平面的垂线的长度和底$\vec{r}.(\hat{i}-2\hat{j}+4\hat{k})+5=0$ .

解决方案：

Plane = x – 2y + 4z + 5 = 0          -Equation(1)

Point = (1, 1, 2)

D = $\left|\frac{1-2+8+5}{\sqrt{1+4+16}}\right|$

= 12/√21

The length of the perpendicular from the given point to the plane = 12/√21

Let us assume that the foot of perpendicular be (x, y, z). 

So DR’s are in proportional

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

x = k + 1

y = -2k + 1

z = 4k + 2

Substitute (x, y, z) = (k + 1, -2k + 1, 4k + 2) in the plane equation(1)

k + 1 + 4k – 2 + 16k + 8 + 5 = 0

21k = -12

k = -12/21 = -4/7 

Hence, the coordinate of the foot of the perpendicular  = (3/7, 15/7, -2/7) 

问题 11. 求垂线脚的坐标和点 P(3, 2, 1) 到平面 2x – y + z + 1 = 0 的垂线距离。同时求该点在平面上的像.

解决方案：

Given:

Plane = 2x – y + z + 1 = 0          -Equation(1)

Point P = (3, 2, 1)

D =$\left|\frac{6-2+1+1}{\sqrt{4+1+1}}\right|=\frac{6}{\sqrt{6}}=\sqrt{6}$

The perpendicular distance of the point P from the plane(D) = √6

Let us assume that the foot of perpendicular be (x, y, z). 

So DR’s are in proportional

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

x = 2k + 3

y = -k + 2

z = k + 1

Substitute (x, y, z) = (2k + 3, -k + 2, k + 1) in the plane equation(1)

4k + 6 + k – 2 + k + 1 + 1 = 0

6k = -6

k = -6/6 = -1

The coordinate of the foot of the perpendicular = (1, 3, 0) 

问题12.求垂直于平面的单位向量的方向余弦$\vec{r}.(6\hat{i}-3\hat{j}-2\hat{k})+1=0$穿越原点。

解决方案：

Given:

Equation of the plane $\vec{r}.(6\hat{i}-3\hat{j}-2\hat{k})+1=0$

Thus, the direction ratios normal to the plane are 6, -3 and -2

Hence, the direction cosines to the normal to the plane are

= $\frac{6}{\sqrt{6^2+(-3)^2+(-2)^2}},\frac{-3}{\sqrt{6^2+(-3)^2+(-2)^2}},\frac{-2}{\sqrt{6^2+(-3)^2+(-2)^2}}\\$

= 6/7, -3/7, -2/7

= -6/7, 3/7, 2/7

The direction cosines of the unit vector perpendicular to the plane 

are same as the direction cosines of the unit vector perpendicular

to the plane are: -6/7, 3/7, 2/7 

问题 13. 求从原点到平面 2x – 3y + 4z – 6 = 0 的垂线的底坐标。

解决方案：

According to the question,

Plane = 2x – 3y + 4z – 6 = 0

The direction ratios of the normal to the plane are 2, -3 and 4.

Thus, the direction ratios of the line perpendicular to the plane are 2, -3 and 4.

The equation of the line passing (x₁, y₁, z₁) having direction ratios a, b and c is

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

Thus, the equation of the line passing through the origin 

with direction ratios 2, -3 and 4 is

$\frac{x-0}{2}=\frac{y-0}{-3}=\frac{z-0}{4}\\ \frac{x}{2}=\frac{y}{-3}=\frac{z}{4}=r$ 

Here, r is same constant.

Any point on the line is of the form 2r, -3r, and 4r, 

if the point P(2r, -3r, 4r) lies on the plane 2x – 3y + 4z – 6 = 0.

Thus, we have,

2(2r) – 3(-3r) + 4(4r) – 6 = 0

4r + 9r + 16r – 6 = 0

29r = 6

r = 6/29

Thus, the coordinates of the point of intersection of the perpendicular 

from the origin and the plane are:

P(2×6/29, -3×629, 4×6/29) = P(12/29, -18/29, 24/29) 

问题 14. 求点 (1, 3/2, 2) 到平面 2x – 2y + 4z +5 = 0 的垂线的长度和底。

解决方案：

Given: 

Point = (1, 3/2, 2) 

Plane = 2x – 2y + 4z + 5 = 0

D = $\left|\frac{2-3+8+5}{\sqrt{4+4+16}}\right|=\frac{12}{2\sqrt{6}}$

= √6

So, the length of the perpendicular from the point to the plane(D) = √6

Let the foot of perpendicular be (x, y, z). So, DR’s are in proportional 

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

x = 2k + 1

y = -2k + 3/2

z = 4k + 2

So, using the values of x, y, z in equation of the plane we have,

2(2k + 1) – 2(-2k + 2/3) +4(4k + 2) + 5 = 0

4k + 2 + 4k – 3 + 16k + 8 + 5 = 0

24k = -12

k = -1/2

So, the coordinate of the foot of the perpendicular = (0, 5/2, 0)