第 12 课 RD Sharma 解决方案 - 第 29 章飞机 - 练习 29.6

问题 1. 求给定平面之间的角度：

（一世） $\vec{r} \cdot \left( 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) = 1$和$\vec{r} \cdot \left( - \hat{i} + \hat{j} \right) = 4$

解决方案：

Given 

$\vec{r} \cdot \left( 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) = 1$and $\vec{r} \cdot \left( - \hat{i} + \hat{j} \right) = 4$

As we know that the angle between the planes $\vec{r} . \vec{n_1} = d_1 , \vec{r} . \vec{n_2} = d_2$ is given by,

$\cos \theta = \frac{\vec{n_1} . \vec{n_2}}{\left| \vec{n_1} \right| \left| \vec{n_2} \right|}$

Here, $\vec{n_1} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}, \vec{n_2} = - \hat{i} + \hat{j} + 0 \hat{k}$

So, $\cos \theta = \frac{\left( 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right) . \left( - \hat{i} + \hat{j} + 0 \hat{k} \right)}{\left| 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \right| \left| - \hat{i} + \hat{j} + 0 \hat{k} \right|}$

= $\frac{- 2 - 3}{\sqrt{4 + 9 + 16} \sqrt{1 + 1 + 0}}$

= $\frac{- 5}{\sqrt{29} \sqrt{2}}$

= $\frac{- 5}{\sqrt{58}}$

Therefore, $\theta = \cos^{- 1} \left( \frac{- 5}{\sqrt{58}} \right)$.

(二) $\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) = 6$和$\vec{r} \cdot \left( 3 \hat{i} + 6 \hat{j} - 2 \hat{k} \right) = 9$

解决方案：

Given 

$\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) = 6$ and $\vec{r} \cdot \left( 3 \hat{i} + 6 \hat{j} - 2 \hat{k} \right) = 9$

As we know that the angle between the planes $\vec{r} . \vec{n_1} = d_1 , \vec{r} . \vec{n_2} = d_2$ is given by,

$\cos \theta = \frac{\vec{n_1} . \vec{n_2}}{\left| \vec{n_1} \right| \left| \vec{n_2} \right|}$

Here, $\vec{n_1} = 2 \hat{i} - \hat{j} + 2 \hat{k},\vec{n_2} = 3 \hat{i} + 6 \hat{j} - 2 \hat{k}$

So, $\cos \theta = \frac{\left( 2 \hat{i}- \hat{j} + 2 \hat{k} \right) . \left( 3 \hat{i} + 6 \hat{j} - 2 \hat{k} \right)}{\left| 2 \hat{i} - \hat{j} + 2 \hat{k} \right| \left| 3 \hat{i} + 6 \hat{j} - 2 \hat{k} \right|}$

= $\frac{6 - 6 - 4}{\sqrt{4 + 1 + 4} \sqrt{9 + 36 + 4}}$

= $\frac{- 4}{\left( 3 \right) \left( 7 \right)}$

= -4/21

Therefore, θ = cos^-1 (-4/21).

㈢ $\vec{r} \cdot \left( 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \right) = 5$和$\vec{r} \cdot \left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right) = 9$

解决方案：

Given 

$\vec{r} \cdot \left( 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \right) = 5$and $\vec{r} \cdot \left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right) = 9$

As we know that the angle between the planes, $\vec{r} . \vec{n_1} = d_1 , \vec{r} . \vec{n_2} = d_2$ is given by,

$\cos \theta = \frac{\vec{n_1} . \vec{n_2}}{\left| \vec{n_1} \right| \left| \vec{n_2} \right|}$

Here, $\vec{n_1} = 2 \hat{i} + 3 \hat{j} - 6 \hat{k} , \vec{n_2} = \hat{i} - 2 \hat{j} + 2 \hat{k}$

So, $\cos \theta = \frac{\left( 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \right) . \left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right)}{\left| 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \right| \left| \hat{i} - 2 \hat{j} + 2 \hat{k} \right|}$

= $\frac{2 - 6 - 12}{\sqrt{4 + 9 + 36} \sqrt{1 + 4 + 4}}$

= $\frac{- 16}{\left( 7 \right) \left( 3 \right)}$

= -16/21

Therefore, θ = cos^-1 (-16/21).

问题 2. 求平面之间的角度：

(i) 2x - y + z = 4 和 x + y + 2z = 3

解决方案：

Given, 2x − y + z = 4 and x + y + 2z = 3

As we know that the angle between the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 is given by,

$\cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt{{a_2}^2 + {b_2}^2 + {c_2}^2}}$

So, the angle between 2x – y + z = 4 and x + y + 2z = 3 is given by,

$\cos \theta = \frac{\left( 2 \right) \left( 1 \right) + \left( - 1 \right) \left( 1 \right) + \left( 1 \right) \left( 2 \right)}{\sqrt{2^2 + \left( - 1 \right)^2 + 1^2} \sqrt{1^2 + 1^2 + 2^2}}$

= $\frac{2 - 1 + 2}{\sqrt{4 + 1 + 1} \sqrt{1 + 1 + 4}}$

= $\frac{3}{\sqrt{6} \sqrt{6}}$

= 3/6

= 1/2

Therefore, θ = cos^-1 (1/2) = π/3.

(ii) x + y - 2z = 3 和 2x - 2y + z = 5

解决方案：

Given, x + y − 2z = 3 and 2x − 2y + z = 5

As we know that the angle between the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 is given by,

$\cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt{{a_2}^2 + {b_2}^2 + {c_2}^2}}$

So, the angle between x + y – 2z = 3 and 2x – 2y + z = 5 is given by,

$\cos \theta = \frac{\left( 1 \right) \left( 2 \right) + \left( 1 \right) \left( - 2 \right) + \left( - 2 \right) \left( 1 \right)}{\sqrt{1^2 + 1^2 + \left( - 2 \right)^2} \sqrt{2^2 + \left( - 2 \right)^2 + 1^2}}$

= $\frac{2 - 2 - 2}{\sqrt{1 + 1 + 4} \sqrt{4 + 4 + 1}}$

= $\frac{- 2}{\sqrt{6} \sqrt{9}}$

= -2/3√6

Therefore, θ = cos^-1 (-2/3√6).

(iii) x - y + z = 5 和 x + 2y + z = 9

解决方案：

Given, x − y + z = 5 and x + 2y + z = 9

As we know that the angle between the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 is given by,

$\cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt{{a_2}^2 + {b_2}^2 + {c_2}^2}}$

So, the angle between x – y + z = 5 and x + 2y + z = 9 is given by,

$\cos \theta = \frac{\left( 1 \right) \left( 1 \right) + \left( - 1 \right) \left( 2 \right) + \left( 1 \right) \left( 1 \right)}{\sqrt{1^2 + \left( - 1 \right)^2 + 1^2} \sqrt{1^2 + 2^2 + 1^2}}$

= $\frac{1 - 2 + 1}{\sqrt{1 + 1 + 1} \sqrt{1 + 4 + 1}}$

= $\frac{0}{\sqrt{3} \sqrt{6}}$

= 0

Therefore, θ = cos^-1 (0) = π/2.

(iv) 2x - 3y + 4z = 1 和 - x + y = 4

解决方案：

Given, 2x − 3y + 4z = 1 and − x + y = 4

As we know that the angle between the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 is given by,

$\cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt{{a_2}^2 + {b_2}^2 + {c_2}^2}}$

So, the angle between 2x – 3y + 4z = 1 and -x + y + 0z = 4 is given by,

$\cos \theta = \frac{\left( 2 \right) \left( - 1 \right) + \left( - 3 \right) \left( 1 \right) + \left( 4 \right) \left( 0 \right)}{\sqrt{2^2 + \left( - 3 \right)^2 + 4^2} \sqrt{\left( - 1 \right)^2 + 1^2 + 0^2}}$

= $\frac{- 2 - 3 + 0}{\sqrt{4 + 9 + 16} \sqrt{1 + 1 + 0}}$

= $\frac{- 5}{\sqrt{29} \sqrt{2}}$

= $\frac{- 5}{\sqrt{58}}$

Therefore, $\theta = \cos^{- 1} \left( \frac{- 5}{\sqrt{58}} \right)$.

(v) 2x + y - 2z = 5 和 3x - 6y - 2z = 7

解决方案：

Given, 2x + y − 2z = 5 and 3x − 6y − 2z = 7

As we know that the angle between the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 is given by,

$\cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt{{a_2}^2 + {b_2}^2 + {c_2}^2}}$

So, the angle between 2x + y – 2z = 5 and 3x – 6y – 2z = 7 is given by,

$\cos \theta = \frac{\left( 2 \right) \left( 3 \right) + \left( 1 \right) \left( - 6 \right) + \left( - 2 \right) \left( - 2 \right)}{\sqrt{2^2 + 1^2 + \left( - 2 \right)^2} \sqrt{3^2 + \left( - 6 \right)^2 + \left( - 2 \right)^2}}$

= $\frac{6 - 6 + 4}{\sqrt{4 + 1 + 4} \sqrt{9 + 36 + 4}}$

= $\frac{4}{\left( 3 \right) \left( 7 \right)}$

= 4/21

Therefore, θ = cos^-1 (4/21).

问题 3. 证明下列平面成直角。

（一世） $\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + \hat{k} \right) = 5$和$\vec{r} \cdot \left( - \hat{i} - \hat{j} + \hat{k} \right) = 3$

解决方案：

Given, $\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + \hat{k} \right) = 5$and $\vec{r} \cdot \left( - \hat{i} - \hat{j} + \hat{k} \right) = 3$

As we know that the planes $\vec{r} . \vec{n_1} = d_1 , \vec{r} . \vec{n_2} = d_2$ are perpendicular to each other only if $\vec{n_1} . \vec{n_2} =0$.

Here, $\vec{n_1} = 2 \hat{i} - \hat{j} + \hat{k} , \vec{n_2} = - \hat{i} - \hat{j} + \hat{k}$

Now, we have

$\vec{n_1} . \vec{n_2} = \left( 2 \hat{i} - \hat{j} + \hat{k} \right) . \left( - \hat{i} - \hat{j} + \hat{k} \right)$

= -2 + 1 + 1 

= 0

So, the given planes are perpendicular.

Hence proved.

(ii) x - 2y + 4z = 10 和 18x + 17y + 4z = 49

解决方案：

Given, x − 2y + 4z = 10 and 18x + 17y + 4z = 49

As we know that the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 are perpendicular to each other only if,

=> a₁ a₂ + b₁ b₂ + c₁ c₂ = 0

The given planes are x – 2y + 4z = 10 and 18x + 17y + 4z = 49.

Here, a₁ = 1, b₁ = – 2, c₁ = 4, a₂ = 18, b₂ = 17 and c₂ = 4

Now, we have

a₁ a₂ + b₁ b₂ + c₁ c₂ = (1) (18) + (- 2) (17) + (4) (4)

= 18 – 34 + 16 

= 0

So, the given planes are perpendicular.

Hence proved.

问题 4. 确定以下平面相互垂直的 λ 值。

（一世） $\vec{r} \cdot \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) = 7$和$\vec{r} \cdot \left( \lambda \hat{i} + 2 \hat{j} - 7 \hat{k} \right) = 26$

解决方案：

Given, $\vec{r} \cdot \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) = 7$and $\vec{r} \cdot \left( \lambda \hat{i} + 2 \hat{j} - 7 \hat{k} \right) = 26$

As we know that the planes, $\vec{r} . \vec{n_1} = d_1 , \vec{r} . \vec{n_2} = d_2$ are perpendicular to each other only if,  $\vec{n_1} . \vec{n_2} =0$.

Here, $\vec{n_1} = \hat{i}| + 2 \hat{j} + 3 \hat{k} , \vec{n_2} = \lambda \hat{i} + 2 \hat{j} - 7 \hat{k}$

The given planes are perpendicular. So, we have

$\vec{n_1} . \vec{n_2} = 0$

$\left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) . \left( \lambda \hat{i} + 2 \hat{j} - 7 \hat{k} \right) = 0$

λ + 4 – 21 = 0

λ – 17 = 0

λ = 17

Therefore, the value of λ is 17.

(ii) 2x - 4y + 3z = 5 和 x + 2y + λz = 5

解决方案：

Given, 2x − 4y + 3z = 5 and x + 2y + λz = 5

As we know that the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 are perpendicular to each other only if,

=> a₁ a₂ + b₁ b₂ + c₁ c₂ = 0

The given planes are 2x – 4y + 3z = 5 and  x + 2y + λz = 5.

Here, a₁ = 2, b₁ = – 4, c₁ = 3, a₂ = 1, b₂ = 2 and c₂ = λ.

It is given that the given planes are perpendicular. So, we get

a₁ a₂ + b₁ b₂ + c₁ c₂ = 0

(2) (1) + (-4) (2) + (3) (λ) = 0

2 – 8 + 3λ = 0

3λ = 6

λ = 2

Therefore, the value of λ is 2.

(iii) 3x - 6y - 2z = 7 和 2x + y - λz = 5

解决方案：

Given, 3x − 6y − 2z = 7 and 2x + y − λz = 5

As we know that the planes a₁ x + b₁ y + c₁ z + d₁ = 0 and a₂ x + b₂ y + c₂ z + d₂ = 0 are perpendicular to each other only if,

=> a₁ a₂ + b₁ b₂ + c₁ c₂ = 0

The given planes are 3x – 6y – 2z = 7 and 2x + y – λz = 5.

Here, a₁ = 3, b₁ = – 6, c₁ = – 2, a₂ = 2, b₂ = 1 and c₂ = -λ

Here, the given planes are perpendicular.

a₁ a₂ + b₁ b₂ + c₁ c₂ = 0

(3) (2) + (-6) (1) + (-2) (λ) = 0

6 – 6 + 2λ = 0

2λ = 0

λ = 0

问题 5. 求通过点 (-1, -1, 2) 并垂直于平面 3x + 2y - 3z = 1 和 5x - 4y + z = 5 的平面方程。

解决方案：

The equation of any plane passing through  (-1, -1, 2) is,

a (x + 1) + b (y + 1) + c (z – 2) = 0  . . . . (1)

It is given that the above equation is perpendicular to each of the planes 3x + 2y – 3z = 1 and 5x – 4y + z = 5. 

3a + 2b – 3c = 0   . . . . (2)

5a – 4b + c = 0    . . . . (3)

On solving eq (1), (2) and (3), we get,

$\begin{vmatrix}x + 1 & y + 1 & z - 2 \\ 3 & 2 & - 3 \\ 5 & - 4 & 1\end{vmatrix} = 0$

-10 (x + 1) – 18 (y + 1) – 22 (z – 2) = 0

-5 (x + 1) – 9 (y + 1) – 11 (z – 2) = 0

5x + 5 + 9y + 9 + 11z – 22 = 0

5x + 9y + 11z – 8 = 0

Hence, the equation of the plane is 5x + 9y + 11z – 8 = 0

问题 6. 求出通过点 (1, -3, -2) 并垂直于平面 x + 2y + 2z = 5 和 3x + 3y + 2z = 8 的平面方程。

解决方案：

The equation of any plane passing through (1, -3, -2) is,

a (x – 1) + b (y + 3) + c (z + 2) = 0 . . . . (1)

It is given that above equation is perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.

a + 2b + 2c = 0    . . . . (2)

3a + 3b + 2c = 0  . . . . (3)

On solving eq (1), (2) and (3), we get,

$\begin{vmatrix}x - 1 & y + 3 & z + 2 \\ 1 & 2 & 2 \\ 3 & 3 & 2\end{vmatrix} = 0$

-2 (x – 1) + 4 (y + 3) – 3 (z + 2) = 0

-2x + 2 + 4y + 12 – 3z – 6 = 0

2x – 4y + 3z – 8 = 0

Hence, the equation of the plane is 2x – 4y + 3z – 8 = 0

问题 7. 求通过原点并垂直于平面 x + 2y - z = 1 和 3x - 4y + z = 5 的平面的方程。

解决方案：

The equation of any plane passing through the origin (0, 0, 0) is,

a (x – 0) + b (y – 0) + c (z – 0) = 0  . . . . (1)

ax + by + cz = 0 

It is given that above equation is perpendicular to the planes x + 2y – z = 1 and 3x – 4y + z = 5 .

a + 2b – c = 0   . . . . (2)

3a – 4b + c = 0  . . . . (3)

On solving eq(1), (2) and (3), we get,

$\begin{vmatrix}x & y & z \\ 1 & 2 & - 1 \\ 3 & - 4 & 1\end{vmatrix} = 0$

– 2x – 4y – 10z = 0

x + 2y + 5z = 0

Hence, the equation of the plane is x + 2y + 5z = 0

问题 8. 求通过点 (1, -1, 2) 和 (2, -2, 2) 且垂直于平面 6x - 2y + 2z = 9 的平面方程。

解决方案：

The equation of any plane passing through (1, -1, 2) is,

a (x – 1) + b (y + 1) + c (z – 2) = 0  . . . . (1)

It is given that above equation is passing through (2, -2, 2). So,

a (2 – 1) + b (-2 + 1) + c (2 – 2) = 0 

a – b = 0  . . . . (2)

It is given that above equation is perpendicular to the plane 6x – 2y + 2z = 9. So,

6a – 2b + 2c = 0

3a – b + c = 0  . . . . (3)

On solving eq(1), (2) and (3), we get,

$\begin{vmatrix}x - 1 & y + 1 & z - 2 \\ 1 & - 1 & 0 \\ 3 & - 1 & 1\end{vmatrix} = 0$

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

– x + 1 – y – 1 + 2z – 4 = 0

x + y – 2z + 4 = 0

Hence, the equation of the plane is x + y – 2z + 4 = 0

问题 9. 求通过点 (2, 2, 1) 和 (9, 3, 6) 并垂直于平面 2x + 6y + 6z = 1 的平面方程。

解决方案：

The equation of any plane passing through (2, 2, 1) is,

a (x – 2) + b (y – 2) + c (z – 1) = 0  . . . . (1)

It is given that the above equation is passing through (9, 3, 6). So,

a (9 – 2) + b (3 – 2) + c (6 – 1) = 0

7a + b + 5c = 0  . . . . (2)

It is given that the above equation is perpendicular to the plane 2x + 6y + 6z = 1. So,

2a + 6b + 6c = 0

a + 3b + 3c = 0 . . . . (3)

On solving eq(1), (2) and (3), we get

$\begin{vmatrix}x - 2 & y - 2 & z - 1 \\ 7 & 1 & 5 \\ 1 & 3 & 3\end{vmatrix} = 0$

-12 (x – 2) – 16 (y – 2) + 20 (z – 1) = 0

3 (x – 2) + 4 (y – 2) – 5 (z – 1) = 0

3x + 4y – 5z = 9

Hence, the equation of the plane is 3x + 4y – 5z = 9

问题 10. 求通过坐标为 (-1, 1, 1) 和 (1, -1, 1) 的点并垂直于平面 x + 2y + 2z = 5 的平面方程。

解决方案：

The equation of any plane passing through (-1, 1, 1) is,

a (x + 1) + b (y – 1) + c (z – 1) = 0  . . . . (1)

It is given that the above equation passed through (1, -1, 1). So,

a (1 + 1) + b (-1 – 1) + c (1 – 1) = 0 

2a – 2b = 0  . . . . (2)

It is given that the above equation is perpendicular to the plane x + 2y + 2z = 5. So,

a + 2b + 2c = 0  . . . . (3)

On solving eq(1), (2) and (3), we get

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

-4 (x + 1) – 4 (y – 1) + 6 (z – 1) = 0

2 (x + 1) + 2 (y – 1) – 3 (z – 1) = 0

2x + 2y – 3z + 3 = 0

Hence, the equation of the plane is 2x + 2y – 3z + 3 = 0

问题 11. 求 y 轴上截距为 3 且平行于 ZOX 平面的平面方程。

解决方案：

The equation of the plane parallel to the plane ZOX  is,

y = b  . . . . (1)

According to the question, it is given that this plane passes through (0, 3, 0). So,

=> b = 3

On substituting this value in eq(1), we get

y = 3, 

Hence, the equation of the plane is

y = 3 

问题 12. 找到包含点 (1, -1, 2) 并垂直于每个平面 2x + 3y - 2z = 5 和 x + 2y - 3z = 8 的平面方程。

解决方案：

The equation of any plane passing through (1, -1, 2) is,

a (x – 1) + b (y + 1) + c (z – 2) = 0  . . . . (1)

It is given that the above equation is perpendicular to the plane 2x + 3y – 2z = 5. So, 

2a + 3b – 2c = 0  . . . . (2)

It is given that the above equation is perpendicular to the plane x + 2y – 3z = 8. So,

a + 2b – 3c = 0   . . . . (3)

So, on solving eq (1), (2) and (3), we get

$\begin{vmatrix}x - 1 & y + 1 & z - 2 \\ 2 & 3 & - 2 \\ 1 & 2 & - 3\end{vmatrix} = 0$

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

5x – 4y – z = 7

Hence, the equation of the plane is 5x – 4y – z = 7

问题 13. 求通过 (a, b, c) 且平行于平面的平面方程$\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right)$ = 2。

解决方案：

Given equation of the plane is \vec{r} \cdot \left( \hat{i}  + \hat{j}  + \hat{k}  \right)  = 2

So, on substituting $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$ in the given equation of the plane, we get

$\left( x \hat{i} + y \hat{j} + z \hat{k} \right) . \left( \hat{i} + \hat{j} + \hat{k} \right) = 2$

=> x + y + z – 2 = 0   . . . . (1)

The equation of a plane which is parallel to plane (eq 1) is of the form,

x + y + z = k   . . . . (2)

It is given that above equation of plane is passing through the point (a, b, c). So,

a + b + c = k

On substituting this value of  k in eq(2), we get

x + y + z = a + b + c

Hence, the equation of the plane is

x + y + z = a + b + c 

问题 14. 求通过点 (-1, 3, 2) 并垂直于每个平面 x + 2y + 3z = 5 和 3x + 3y + z = 0 的平面方程。

解决方案：

The equation of any plane passing through point (-1, 3, 2) is,

a (x + 1) + b (y – 3) + c (z – 2) = 0  . . . . (1)

It is given that the above equation is perpendicular to the plane x + 2y + 3z = 5. So, 

a + 2b + 3c = 0   . . . . (2)

It is given that the above equation is perpendicular to the plane 3x + 3y + z = 0. So,

3a + 3b + c = 0   . . . . (3)

On solving eq (1), (2) and (3), we get

$\begin{vmatrix}x + 1 & y - 3 & z - 2 \\ 1 & 2 & 3 \\ 3 & 3 & 1\end{vmatrix} = 0$

-7 (x + 1) + 8 (y – 3) – 3 (z – 2) = 0

7x – 8y + 3z + 25 = 0

Hence, the equation of the plane is

7x – 8y + 3z + 25 = 0

问题 15. 求平面通过点 (2, 1, -1) 和 (-1, 3, 4) 并垂直于平面 x - 2y + 4z = 10 的向量方程。

解决方案：

The equation of any plane passing through (2, 1, -1) is,

a (x – 2) + b (y – 1) + c (z + 1) = 0  . . . . (1)

It is given that the above equation passes through (-1, 3, 4). So,

a (-1 – 2) + b (3 – 1) + c (4 + 1) = 0

-3a + 2b + 5c   . . . . (2)

It is given that the above equation is perpendicular to the plane x – 2y + 4z = 10. So,

a – 2b + 4c = 0  . . . . (3)

On solving eq (1), (2) and (3), we get

$\begin{vmatrix}x - 2 & y - 1 & z + 1 \\ - 3 & 2 & 5 \\ 1 & - 2 & 4\end{vmatrix} = 0$

18 (x – 2) + 17 (y – 1) + 4 (z + 1) = 0

18x + 17y + 4z – 49 = 0

Hence, the equation of the plane is

18x + 17y + 4z – 49 = 0