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  • Find the vector and cartesian equations of the plane passing through the points having position vectors 

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

Posted by

Ravindra Pindel

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