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Find the vector equation of the plane which contains the line of intersection of the planes vec("r").(hat"i"+2hat"j"+3hat"k"),-4=0, vec("r").(2hat"i"+hat"j"-hat"k")+5=0 - Mathematics

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Question

Find the vector equation of the plane which contains the line of intersection of the planes `vec("r").(hat"i"+2hat"j"+3hat"k"),-4=0, vec("r").(2hat"i"+hat"j"-hat"k")+5=0`and which is perpendicular to the plane`vec("r").(5hat"i"+3hat"j"-6hat"k"),+8=0`

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Solution

The equations of the given planes are

`vec("r").(hat"i"+2hat"j"+3hat"k")-4=0` .............(1)
`vec("r").(2hat"i"+hat"j"-hat"k")+5=0`  ..............(2)

The equation of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is

`[vec("r").(hat"i"+2hat"j"+3hat"k")-4]+lambda[vec("r").(2hat"i"+hat"j"-hat"k")+5]=0`

`vec("r").[(2lambda+1)hat"i"+(lambda+2)hat"j"+(3-lambda)hat"k"]+(5lambda-4)=0` ....(3)

The plane in equation (3) is perpendicular to the plane, `vec("r").(5hat"i"+3hat"j"-6hat"k"),+8=0`

∴5(2λ+1) + 3(λ+2) - 6(3-λ) = 0

⇒19λ - 7 = 0

`⇒ lambda =7/19`

Substituting `lambda =7/19` in equation (3), we obtain

`⇒ vec("r").[33/19hat"i" +45/19hat"j"+50/19hat"k"](-41)/19=0`

`⇒ vec("r").(33hat"i" +45hat"j"+50hat"k")-41=0` .............(4)

This is the vector equation of the required plane.

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2018-2019 (March) 65/4/3

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