第12类RD Sharma解决方案–第29章飞机–练习29.10

问题1.找到平行平面2x – y + 3z – 4 = 0和6x – 3y + 9z + 13 = 0之间的距离

解决方案：

Let P(x₁, y₁, z₁) be any point on plane 2x – y + 3z – 4 = 0.

⟹ 2x₁ – y₁ + 3z₁ = 4 (equation-1)

Distance between (x₁, y₁, z₁) and the plane

6x – 3y + 9z + 13 = 0:

As we know that, the distance of point (x₁, y₁, z₁) from the plane π: ax + by + cz + d = 0 is given by:

p = $\left|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}\right|$

Now, substitute the values, we get

p = $\left|\frac{(6)(x_1)+(-3)(y_1)+(9)(z_1)+13}{\sqrt{6^2+(-3)^2+9^2}}\right|$

= $\left|\frac{3(2x_1-y_1+3z_1)+13}{\sqrt{6^2+(-3)^2+9^2}}\right|$

= $\left|\frac{3(4)+13}{3\sqrt{14}}\right|$  [by using equation 1]

= $\frac{25}{3\sqrt{14}}$

Therefore, the distance between the parallel planes 2x – y + 3z – 4 = 0 and 6x – 3y + 9z + 13 = 0 is $\frac{25}{3\sqrt{14}}$ units.

问题2。找到经过点(3，4，-1)并与2x – 3y + 5z + 7 = 0平面平行的平面的方程。还要找到两个平面之间的距离。

解决方案：

Since the plane is parallel to 2x – 3y + 5z + 7 = 0, it must be of the form:

2x – 3y + 5z + θ = 0

It is given that,

The plane passes through (3, 4, –1)

⟹ 2(3) – 3(4) +5(–1) + θ = 0

θ = -11

Thus, 

The equation of the plane is as follows:

2x – 3y + 5z – 11 = 0

Distance of the plane 2x – 3y + 5z + 7 = 0 from (3, 4, –1):

As we know that, the distance of point (x₁, y₁, z₁) from the plane π: ax + by + cz + d = 0 is given by:

p = $\left|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}\right|$

Now, after substituting the values, we will get

= $\left|\frac{(2)(3)+(-3)(4)+(5)(-1)+7}{\sqrt{2^2+(-3)^2+5^2}}\right|$

= $\frac{4}{\sqrt{38}}$

Therefore, the distance of the plane 2x – 3y + 5z + 7 = 0 from (3, 4, -1) is $\frac{4}{\sqrt{38}}$

问题3.找到与平面2x – 2y + z + 3 = 0和2x – 2y + z + 9 = 0平行的平面的方程

解决方案：

Given:

Equation of planes: 

π₁= 2x – 2y + z + 3 = 0

π₂= 2x – 2y + z + 9 = 0

Let the equation of the plane mid–parallel to these planes be:

π₃: 2x – 2y + z + θ = 0

Now,

Let P(x₁, y₁, z₁) be any point on this plane,

⟹ 2(x₁) – 2(y₁) + (z₁) + θ = 0 —(equation-1)

As we know that, the distance of point (x₁, y₁, z₁) from the plane π: ax + by + cz + d = 0 is given by:

p = $\left|\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}}\right|$

Distance of P from π₁:

p = $\left|\frac{(2x_1-2y_2+z_1)+3}{\sqrt{2^2+(-2)^2+1^2}}\right|$

=  $\left|\frac{(-θ)+3}{3}\right|$  (By using equation 1)

Similarly,

DIstance of q from π₂:

q = $\left|\frac{(2x_1-2y_2+z_1)+9}{\sqrt{2^2+(-2)^2+1^2}}\right|$

= $\left|\frac{(-θ)+9}{3}\right|$   (By using equation 1)

As π₃ is mid-parallel is π₁ and π₂:

p = q 

So,

$\left|\frac{(-θ)+3}{3}\right|=\left|\frac{-θ +9}{3}\right|$

Now square on both sides, we get

$\left(\frac{(-θ)+3}{3}\right)^2=\left(\frac{-θ+9}{3}\right)^2$

(3 – θ)² = (9 – θ)²

9 – 2×3×θ + θ² = 81 – 2×9×θ + θ²

θ = 6

Now, substitute the value of θ = 6 in equation 2x – 2y + z + θ = 0, we get

Hence, the equation of the mid-parallel plane is 2x – 2y + z + 6 = 0

问题4.找出平面之间的距离$\vec{r}.(\hat{i}+2\hat{j}+3\hat{k})+7=0$和$\vec{r}.(2\hat{i}+4\hat{j}+6\hat{k})+7=0$

解决方案：

Let $\vec{a}$ be the position vector of any pont P on the plane

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

So, 

$\vec{a}(\hat{i}+2\hat{j}+3\hat{k})+7=0$ —(equation 1)

As we know that, the distance of from $\vec{a}$ the plane $\vec{r}.\vec{n}-d=0$ is given by:

p = $\left|\frac{\vec{a}.\vec{n}-d}{|n|}\right|$

Length of perpendicular from$P(\vec{a}) to plane \vec{r}.(2\hat{i}+4\hat{j}+6\hat{k})+7=0$ is given by substituting the values of$\vec{a}\ and\ \vec{n}$, we get

p = $\left|\frac{2\vec{a}.(2\hat{i}+4\hat{j}+6\hat{k})+7}{|2\hat{i}+4\hat{j}+6\hat{k}|}\right|$

= $\left|\frac{2\vec{a}.(\hat{i}+2\hat{j}+3\hat{k})+7}{\sqrt{2^2+4^2+6^2}}\right|$

= $\left|\frac{2(-7)+7}{\sqrt{56}}\right|$

p = $\frac{7}{\sqrt{56}}$

Therefore, the distance between the planes

$\vec{r}.(\hat{i}+2\hat{j}+3\hat{k})+7=0$ and $\vec{r}.(2\hat{i}+4\hat{j}+6\hat{k})+7=0$ is $\frac{7}{\sqrt{56}}$