    Question

# If the shortest distance between the lines x−1α=y+1−1=z1, (α≠−1) and x+y+z+1=0=2x–y+z+3 is 1√3, then a value of α is :

A
3219
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B
1932
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C
1619
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D
1916
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Solution

## The correct option is A 3219 Family of plane is, x+y+z+1+k(2x–y+z+3)=0(1+2k)x+(1−k)y+(1+k)z+(1+3k)=0 If given family of plane is parallel to the line then (1+2k)α−(1−k)+(1+k)=0⇒α=−2k1+2kPerpendicular distance between x−1α=y+1−1=z1 & (1+2k)x+(1−k)y+(1+k)z+(1+3k)=0 is|(1+2k)(1)+(1−k)(−1)+(1+k)(0)+(1+3k)|√(1+2k)2+(1−k)2+(1+k)2=1√3⇒k=0,−32102⇒α=0,3219  Suggest Corrections  0      Similar questions
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