Question

Find the vector equation of the plane passing through the intersection of planes r. (2i + 2j - 3k) = 7, r.(2i + 5j + 3k) = 9 and through the point (2, 1, 3).

Answer 1

Answer:

Equation of plane

$\vec r.[38\hat i+68\hat j+3\hat k]=153$

Step-by-step explanation:

To find the vector equation of the plane passing through the intersection of planes r. (2i + 2j - 3k) = 7, r.(2i + 5j + 3k) = 9 and through the point (2, 1, 3).

these plane equations are in normal form

$\vec r.\vec n_{1}=d_{1}\\\\and\\\\\vec r.\vec n_{2}=d_{2}\\\\where\\\\\vec n_{1}=(2\hat i + 2\hat j - 3\hat k)\\\\\\d_{1}=7\\\\\\\vec n_{2}=(2\hat i + 5\hat j + 3\hat k)\\\\d_{2}=9$

The equation of the plane passes throught the  intersection of the plane

$\vec r.(\vec n_{1}+\lambda \vec n_{2})=(d_{1}+\lambda d_{2})\\\\\vec r.((2\hat i + 2\hat j - 3\hat k)+\lambda( 2\hat i + 5\hat j + 3\hat k)=(7+\lambda 9)\\\\\\=> \vec r.[(2+2\lambda)\hat i+(2+5\lambda)\hat j+(-3+3\lambda)\hat k]=(7+\lambda 9)$....eq1

Since the plane in eq 1 passes through the point (2,1,3),means it satisfies the eq 1

$\vec r= 2\hat i+\hat j+3\hat k\\\\\\2\hat i+\hat j+3\hat k.[(2+2\lambda)\hat i+(2+5\lambda)\hat j+(-3+3\lambda)\hat k]=(7+\lambda 9)\\\\\\=[(2+2\lambda)2+(2+5\lambda)+(-3+3\lambda)3]=(7+\lambda 9)\\\\\\=> 9\lambda =10\\\\\lambda=\frac{10}{9}$

Now put the value of lambda in eq 1 to get the equation of the plane

$\vec r.[(2+2(\frac{10}{9}))\hat i+(2+5(\frac{10}{9}))\hat j+(-3+3(\frac{10}{9}))\hat k]=(7+(\frac{10}{9}) 9)\\\\\\=> \vec r.[(2+\frac{20}{9})\hat i+(2+\frac{50}{9})\hat j+(-3+\frac{10}{3})\hat k]=(7+10)\\\\\\=>\vec r.[38\hat i+68\hat j+3\hat k]=153$

is the required palne in vector form.