1.

Find the equation of the plane passing through the line of intersection of the planes `vecr.(hati+hatj+hatk)=1 and vecr.(2hati+3hatj-hatk)+4=0` and parallel to x-axis.

Answer» Equation of a plane through the intersection of planes is
`[vecr.(hati+hatj+hatk)-1)]+lambda(vecr.(2hati+3hatj-hatk)+4]=0`
`rArr vecr.[(2lambda+1)hati+(3lambda+1)hatj + (1-lambda)hatk]+(4lambda-1)=0"……"(1)`
The direction ratio of the normal to the plane are `(2lambda+1),(3lambda+1)` and `(1-lambda)`.
Required plane is parallel to X-axis, so the normal to the plane will be ratios of X-axis are `1,0` and `0`.
`:. 1(2lamda+1)+0(3lambda+1)+0(1-lambda) = 0`
`rArr 2 lambda +1=0 rArr lambda = - 1/2`
put the value of `lambda` in equation (1),
`vecr.(-(1)/(2)hatj+3/2hatk)+(-3)=0 rArr vecr.(hatj-3hatk) + 6 =0`
Therefore, cartessian equation of plane is
`y - 3z + 6 = 0`
Which is the equation of the required plane.


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