13. Find the equation of the plane passing through the point (– 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Answers (1)

Given 

two planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

the normal vectors  of these plane are 

n_1=\hat i+2\hat j+ 3\hat k

n_2=3\hat i+3\hat j+ \hat k

Since the normal vector of the required plane is perpendicular to the normal vector of given planes, the required plane's normal vector will be :

\vec n = \vec n_1\times\vec n_2

\vec n = (\hat i+2\hat j +3\hat k )\times (3\hat i + 3\hat j +\hat k)

\vec n =\begin{vmatrix} \hat i &\hat j &\hat k \\ 1 &2 &3 \\ 3& 3& 1 \end{vmatrix}=\hat i(2-9)-\hat j(1-9)+\hat k (3-6)

\vec n =-7\hat i+8\hat j-3\hat k

Now, as we know 

the equation of a plane in vector form is :

\vec r\cdot\vec n=d

\vec r\cdot(-7\hat i+8\hat j-3\hat k)=d

Now Since this plane passes through the point (-1,3,2)

(-\hat i+3\hat j+2\hat k)\cdot(-7\hat i+8\hat j-3\hat k)=d

7+24-6=d

d=25

Hence the equation of the plane is 

\vec r\cdot(-7\hat i+8\hat j-3\hat k)=25

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