Find the vector and cartesian equations of the plane passing through the points having position vectors \hat{i}+\hat{j}-2\hat{k}, 2\hat{i}-\hat{j}+\hat{k} \:\:and \:\:\hat{i}+2\hat{j}+\hat{k.} Write the equation of a plane passing through a point (2,3,7) and parallel to the plane obtained above. Hence, find the distance between the two parallel planes.

 

 

 

 

 
 
 
 
 

Answers (1)

Let A,B,C be the points with the position vectors 

\hat{i}+\hat{j}-2\hat{k}, 2\hat{i}-\hat{j}+\hat{k} \:\:and \:\:\hat{i}+2\hat{j}+\hat{k.} respectively.

Then ,

 \overrightarrow{AB}=P.V\: of\: B - P.V. \: of\: A \\ = (2\hat{i}-\hat{j}+\hat{k})-(\hat{i}+\hat j -2 \hat k) \\= \hat i - 2 \hat j + 3 \hat k \\ and \: \overrightarrow{BC }= P.V \: of \: C - P.V \: of \: B \\ = (\hat i + 2 \hat j + \hat k ) - ( \hat i + \hat j - 2 \hat k ) \\ = - \hat i + 3 \hat j + 0 \hat k

A vector normal to the plane containing points A, B and C is 

\overrightarrow n = \overrightarrow{AB}\times \overrightarrow{AC} \\ = \begin{bmatrix} \ \hat i & \hat j & \hat k \\ 1& -2 & 3\\ -1 & 3 & 0 \end{bmatrix} \\ = -9\hat i - 3 \hat j + \hat k

The required plane passes through the point having vector \vec{a}= \hat i + \hat j - 2 \hat k and is normal to the vector  -9\hat i - - 3 \hat j + \hat k. So its vector equation is 

(\vec{r}- \vec a)\cdot \vec n = 0 \\ \Rightarrow \vec r \cdot \vec n - \vec a \cdot \vec n = 0 \\ \Rightarrow \vec r \cdot \vec n = \vec a \cdot \vec n \\ \Rightarrow \vec r \cdot (-9 \hat i -3 \hat j + \hat k )= (\hat i + \hat j + 2 \hat k )( -9 \hat i -3 \hat j + \hat k ) \\ \Rightarrow \vec r \cdot (-9 \hat i -3 \hat j + \hat k)= -9-3-2 \\ \Rightarrow \hat r \cdot (-9 \hat i -3 \hat j + \hat k)= -14

This is the required vector equation of the plane the cartesian equation of plane is given by 

(x \hat i + y \hat j + z \hat k) \cdot (-9 \hat i -3 \hat j + \hat k)=-14\\ -9x+3y+z=-14 \\ 9x+3y-z=14

Direction ratios of this plane are (9,3,-1) then the equation of the plane parallel to the above plane and passing through (2,3,7) is 

a(x-x_1)+b(y-y_1)+c(z-z_1) \\ = 9(x-2)+3(y-3)-1(z-7)\\ \Rightarrow 9x+3y-z-20=0

This is the required parallel plane 

Then, Distance between 9x+3y-z+14=0 \: \: and \:\: 9x+3y-z-20=0

 Let P(x_1,y_1,z_1) be any point on 9x+3y-z+14 = 0

Then, 

9x_1+3y_1-z_1-14 = 0

Let d be the distance between planes. Then, 

d = length of perpendicular from P(x_1,y_1,z_1) to 9x+3y-z-20=0

d=\left | \frac{9x_1+3y_1-z_1-20}{\sqrt{(9)^2+(3)^2+(-1)^2}} \right |= \left | +\frac{14-20}{\sqrt{91}} \right | \\ = \frac{6}{\sqrt{91}}units

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