第 12 类 RD Sharma 解决方案 - 第 29 章飞机 - 练习 29.13

问题 1. 证明$\vec{r} = (2\hat{j}-3\hat{k})+λ(\hat{i}+2\hat{j}+3\hat{k})$和$\vec{r}=(2\hat{i}+6\hat{j}+3\hat{k})+μ(2\hat{i}+3\hat{j}+4\hat{k})$是共面的。此外，找到包含它们的平面方程。

解决方案：

We know, the lines $\vec{r} = \vec{a_1}+λ\vec{b_1}$ and $\vec{r} = \vec{a_2}+λ\vec{b_2}$ are coplanar if:

$\vec{a_1}.(\vec{b_1}×\vec{b_2}) = \vec{a_2}.(\vec{b_1}×\vec{b_2})$

$⇒ \vec{b_1}×\vec{b_2}=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&2&3\\2&3&4\\\end{array}\right|$

$⇒ \vec{b_1}×\vec{b_2}=-\hat{i}+2\hat{j}-\hat{k}$

$⇒ \vec{a_1}.(\vec{b_1}×\vec{b_2})=(2\hat{j}-3\hat{k}).(-\hat{i}+2\hat{j}-\hat{k})$

$⇒ \vec{a_1}.(\vec{b_1}×\vec{b_2}) = 0 + 4 + 3 = 7$

$⇒ \vec{a_2}.(\vec{b_1}×\vec{b_2})=(2\hat{i}+6\hat{j}+3\hat{k}).(-\hat{i}+2\hat{j}-\hat{k})$

$⇒ \vec{a_2}.(\vec{b_1}×\vec{b_2})=-2+12-3=7$

Since, $\vec{a_1}.(\vec{b_1}×\vec{b_2}) = \vec{a_2}.(\vec{b_1}×\vec{b_2})$, the lines are coplanar.

Equation of the plane containing them: $\vec{r}.(\vec{b_1}×\vec{b_2})=\vec{a_1}.(\vec{b_1}×\vec{b_2})$

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

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

$⇒ \vec{r}.(\hat{i}-2\hat{j}+\hat{k}) + 7 = 0.$

问题 2. 显示线条$\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$和$\frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}$是共面的。此外，找到包含它们的平面方程。

解决方案：

We know the lines $\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$ and $\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}$ are coplanar if,

$\left|\begin{array}{cc}x_2-x_1&y_2-y_1&z_2-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\\\end{array}\right|=0$

So, $=\left|\begin{array}{cc}0+1&7-3&-7+2\\-3&2&1\\1&-3&2\\\end{array}\right|$

$= \left|\begin{array}{cc}1&4&-5\\-3&2&1\\1&-3&2\\\end{array}\right|$

= 1(4 + 3) − 4(−6 − 1) − 5(9 − 2)

= 7 + 28 − 35 

= 0.

So the lines are coplanar.

Equation of the plane: $= \left|\begin{array}{cc}x-x_1&y-y_1&z-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\\\end{array}\right|=0$

 $= \left|\begin{array}{cc}x+1&y-3&z+2\\-3&2&1\\1&-3&2\\\end{array}\right|=0$

⇒ 7x + 7y + 7z = 0.

问题 3. 求包含直线的平面方程$\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$和点 (0,7,-7) 并显示该线$\frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}$也位于同一平面。

解决方案：

We know the equation of a plane passing through a point (x₁,y₁,z₁) is given by

a(x−x₁) + b(y−y₁) + c(z−z₁) = 0            ……..(1)

Since the required plane passes through (0,7,-7), the equation becomes

ax + b(y − 7) + c(z + 7) = 0                …….(2)

It also contains $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and point is (−1,3,−2).

a(−1) + b(3 − 7) + c(−2 + 7) = 0

⇒ −a − 4b + 5c = 0

Also, −3a + 2b + c = 0

Solving the equations, we get x + y + z = 0

So, $\frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}$ lies on the plane x + y + z = 0.

问题 4. 找出包含两条平行线的平面方程$\frac{x-4}{1}=\frac{y-3}{-4}=\frac{z-2}{5}$和$\frac{x-3}{1}=\frac{y+2}{-4}=\frac{z}{5}.$

解决方案：

We know the equation of a plane passing through a point (x₁,y₁,z₁) is given by

a(x−x₁) + b(y−y₁) + c(z−z₁) = 0            ……..(1)

The required plane passes through (4,3,2). Hence,

a(x − 4) + b(y − 3) + c(z − 2) = 0

It also passes through (3,-2,0). Hence,

a(3 − 4) + b(−2 − 3) + c(0 − 2) = 0

⇒ a + 5b + 2c = 0                …….(2)

Also, a − 4b + 5c = 0             ……..(3)

Solving (2) and (3)  by cross multiplication, we get the equation of the plane as:

⇒ 11x − y − 3z − 35 = 0.

问题 5. 显示线条$\frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}$和 3x - 2y + 5 = 0 = 2x + 3y + 4z - 4 相交。求平面方程。

解决方案：

Using a₁a₂ + b₁b₂ + c₁c₂ = 0, we get

3a − 2b + c = 0                ….(1)

Also, 2a + 3b + 4c = 0.     ….(2)

Solving (1) and (2) by cross multiplication, we have

$\frac{a}{(-2)(4)-(3)(1)}=\frac{b}{(2)(1)-(3)(4)}=\frac{c}{(3)(3)-(-2)(2)}$

$⇒ \frac{a}{-11}=\frac{b}{-10}=\frac{c}{13}$

Hence, the equation of the plane is 45x − 17y + 25z + 53 = 0

and the point of intersection is (2,4,−3).

问题 6. 证明向量方程为的平面$\vec{r}.(\hat{i}+2\hat{j}-\hat{k})=3$包含其矢量方程为的线$\vec{r}=\hat{i}+\hat{j}+λ(2\hat{i}+\hat{j}+4\hat{k}).$  

解决方案：

Here, $\vec{b}.\vec{n}=(2\hat{i}+\hat{j}+4\hat{k}).(\hat{i}+2\hat{j}-\hat{k})$

= 2(1) +1(2) + 4(−1)

$⇒ \vec{b}.\vec{n} = 0$

Now, $\vec{a}.\vec{n}=(\hat{i}+\hat{j}).(\hat{i}+2\hat{j}-\hat{k})$

= 1(1) + 1(2) + 0(−1)

= 3 

$⇒ \vec{a}.\vec{n}=d$

Hence, the given line lies on the plane.

问题 7. 求直线的交点确定的平面方程$\frac{x+3}{3}=\frac{y}{-2}=\frac{z-7}{6}$和$\frac{x+6}{1}=\frac{y+5}{-3}=\frac{z-1}{2}.$

解决方案：

Let the plane be ax + by + cz + d = 0

Since the plane passes through the intersection of the given lines, normal of the plane is perpendicular to the two lines.

⇒ 3a − 2b + 6c = 0

and, a − 3b + 2c = 0

Using cross multiplication, we have

$\frac{a}{-4+18}=\frac{b}{6-6}=\frac{c}{-9+2}$

⇒ $\frac{a}{2}=\frac{b}{0}=\frac{c}{-1}$

问题 8. 求通过点 (3,4,2) 和 (7,0,6) 并垂直于平面 2x - 5y - 15 = 0 的平面的矢量方程。同时，证明由此得到的平面包含该行$\vec{r}=\hat{i}+3\hat{j}-2\hat{k}+λ(\hat{i}-\hat{j}+\hat{k})$

解决方案：

Let the equation of the plane be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

Since the plane passes through (3,4,2) and (7,0,6), we have

$\frac{3}{a}+\frac{4}{b}+\frac{2}{c}=1$ and $\frac{7}{a}+\frac{0}{b}+\frac{6}{c}=1$

Since the required plane is perpendicular to 2x − 5y − 15 = 0, we have,

$\frac{2}{a}+\frac{(-5)}{b}+\frac{0}{c}=1$

⇒ b = 2.5a

Substituting the value of b in the above equations we have, 

$\frac{3}{a}+\frac{4}{2.5a}+\frac{2}{c}=1$ and $\frac{7}{a}+\frac{6}{c}=1$

Solving the above equations, we have

a = 17/5, b = 17/2 and c = −17/3.

Substituting the values in the equation of plane, we obtain

5x + 2y − 3z = 17.

Vector equation of the plane becomes: $\vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=17$.

问题 9. 如果线条$\frac{x-1}{-3}=\frac{y-2}{-2k}=\frac{z-3}{2}$和$\frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5}$是垂直的，找到 k 的值以及包含这些线的平面的方程。

解决方案：

The direction ratios of the two lines are r₁ = (−3,−2k,2) and r₂ = (k,1,5).

Since the lines are perpendicular, we have

 (−3,−2k,2).(k,1,5) = 0

⇒ 3k + 2k − 10 = 0

⇒ 5k = 10

⇒ k = 2

Now, equation of the plane containing the lines is: $\left|\begin{array}{cc}x&y&z\\-3&-4&2\\2&1&5\\\end{array}\right|=0$

⇒ −22x + 19y + 5z + 31 = 0.

问题10.找到线所在点的坐标$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}$与平面 x - y + z - 5 = 0 相交。另外，求直线与平面之间的夹角。

解决方案：

Any point on the given line is of the form (3k + 2, 4k − 1, 2k + 2).

We have, (3k + 2) − (4k − 1) + (2k + 2) − 5 = 0

⇒ k = 0.

Thus, the coordinates of the point become (2,−1,2).

Let v be the angle between the line and the plane. Then, 

$sinv=\frac{al+bm+cn}{\sqrt{a^2+b^2+c^2}\sqrt{l^2+m^2+n^2}}$

Here, l = 3, m = 4, n =2, a =1, b = −1, c = 1.

Hence, $sinv=\frac{1(3)+4(-1)+1(2)}{\sqrt{1^2+(-1)^2+1^2}\sqrt{3^2+4^2+2^2}}$

⇒ $sinv=\frac{1}{\sqrt{3}\sqrt{29}}$

⇒ $v=sin^{-1}[\frac{1}{\sqrt3\sqrt29}]$

问题 11. 用位置向量求过三点平面的向量方程$\hat{i}+\hat{j}-2\hat{k},2\hat{i}-\hat{j}+\hat{k},\hat{i}+2\hat{j}+\hat{k}.$

解决方案：

Let  A, B and C be the three given vectors respectively.

$\vec{AB}=(2\hat{i}-\hat{j}+\hat{k})-(\hat{i}+\hat{j}-2\hat{k})$

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

and, $\vec{AC}=\hat{j}+3\hat{k}$

Now, $\vec{AB}.\vec{AC} = \left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&-2&3\\0&1&3\\\end{array}\right|$

⇒ $\vec{n}=-9\hat{i}-3\hat{j}+\hat{k}$

Equation of the plane is: $\vec{r}.(9\hat{i}+3\hat{j}-\hat{k})=14.$

Coordinates of the points are (1,1,−2).

问题 12. 显示线条$\frac{5-x}{-4}=\frac{y-7}{4}=\frac{z+3}{-5}$和$\frac{x-8}{7}=\frac{2y-8}{2}=\frac{z-5}{3}$是共面的。

解决方案：

We know the lines $\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$ and $\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}$ are coplanar if,

$\left|\begin{array}{cc}x_2-x_1&y_2-y_1&z_2-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\\\end{array}\right|=0$

or, $\left|\begin{array}{cc}8-5&4-7&5+3\\4&4&-5\\7&1&3\\\end{array}\right|$

= $\left|\begin{array}{cc}3&-3&8\\4&4&-5\\7&1&3\\\end{array}\right|$

= 3(12 + 5) + 3(12 + 35) + 8(4 − 28)

= 0.

Hence the lines are coplanar.

问题 13. 求通过点 (3,2,0) 并包含直线的平面方程$\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}.$

解决方案：

Required equation of the plane passing through (3,2,0) is:

a(x − 3) + b(y − 2) + cz = 0            ……(1)

Since the plane also passes through the given line, we have

4b + 4c = 0               ……(2)

Also, the plane will be parallel so,

a + 5b + 4c = 0        ……(3)

Solving (2) and (3), we have

$\frac{a}{16-20}=\frac{b}{4-0}=\frac{c}{0-4}=z$

⇒ $-\frac{a}{4}=\frac{b}{4}=-\frac{c}{4}=z$

⇒ a = −z, b = z and c = −z

Putting the values in (1), we have

x − y + z − 1 = 0.