1.

Find the vector equation of a plane passing through the intersection of the planes `vecr.(hati+hatj+hatk) = 6` and `vecr. (2hati+3hatj+4hatk) - 5 = 0` and through the point `(2,2,1)`.

Answer» Let the equation of the plane through the intersection of given planes is
`[vecr.(hati+hatj+hatk)-6]`
`+lambda[vecr.(2hati+3hatj+4hatk)-5] = 0`
`rArr vecr.[(1+2lambda)hati+(1+3lambda)hatj`
`+(1+4lambda)hatk]=6+5lambda`
It passes through the point `(2,2,1)` i.e., `(2hati+2hatj+hatk)`
`:. (hati+2hatj+hatk),[(1+2lambda)hati+(1+3lambda)hatj`
`+(1+4lambda)hatk] = 6+5lambda`
`rArr 2(1+2lambda)+2(1+3lambda)+(1+4lambda)=6+5lambda`
`rArr 2 + 4 lambda + 2 + 6 lambda+1 + 4lambda= 6 +5lambda`
`rArr 9lambda = 1`
`rArr lambda = 1/9`
Therefore the eqation of the plane is
`vecr.[(1+2/9)hati+(1+3/9)hatj+(1+4/9)hatk] = 6+5/9`
`rArr vecr.(11hati+12hatj+13hatk) = 59`.


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