# Find the vector equation of the plane passing through the intersection of the planes,

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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

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