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Find the vector equation of the plane passing through the intersection of the planes, r • (2i + j + k ) = -4 and r •(3i + 4j + k ) = 6 and the point (1, 3, 1).

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The equation of plane passing through given plenes is

\(\vec r((2\hat i + \hat j + \hat k) + \lambda(3\hat i+4\hat j+\hat k))=-4+6\lambda\)

⇒ \(\vec r((2+3\lambda)\hat i+(1+4\lambda)\hat j+(1+\lambda)\hat k)= -4+6\lambda\) 

⇒ \((x\hat i+y\hat j+z\hat k).((2+3\lambda)\hat i+(1+4\lambda)\hat j+(1+\lambda)\hat k)=-4+6\lambda\) 

(\(\because\hat r = x\hat i+y\hat j+z\hat k\))

⇒ (2 + 3\(\lambda\))x + (1 + 4\(\lambda\))y + (1 + \(\lambda\))z = -4 + 6\(\lambda\)----(1)

It passes through (1, 3, 1)

\(\therefore\) (2 + 3\(\lambda\)).1 + (1 + 4\(\lambda\)).3 + (1 +\(\lambda\)).1 = -4 + 6\(\lambda\)

⇒ 2 + 3\(\lambda\) + 3 + 12\(\lambda\) + 1 + \(\lambda\) = -4 + 6\(\lambda\)

⇒ 10\(\lambda\) + 10 = 0

\(\lambda\) = -10/10 = -1

\(\therefore\) From(1),

(2 - 3)x + (1 - 4)y + (1 - 1) z = -4 - 6

⇒ -x - 3y = -10

⇒ x + 3y =10 

\(\therefore(x\hat i+y\hat j+z\hat k).(\hat i+3\hat j+0\hat k)\) = 10

Hence, vector equation of required plane is

\((x\hat i+y\hat j+z\hat k).(\hat i + 3\hat j+0\hat k)=10\)

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