Exams Joint Entrance Examination Mathematics 3D Geometry Question 250 Question 250 of 521 Medium MCQ
If the lines r ⃗ = ( i ^ − j ^ + k ^ ) + λ ( 3 j ^ − k ^ ) \vec r=(\hat i-\hat j+\hat k)+\lambda(3\hat j-\hat k) r = ( i ^ − j ^ + k ^ ) + λ ( 3 j ^ − k ^ ) and r ⃗ = ( α i ^ − j ^ ) + μ ( 2 i ^ − 3 k ^ ) \vec r=(\alpha\hat i-\hat j)+\mu(2\hat i-3\hat k) r = ( α i ^ − j ^ ) + μ ( 2 i ^ − 3 k ^ ) are coplanar, then the distance of the plane containing these two lines from the point ( α , 0 , 0 ) (\alpha,0,0) ( α , 0 , 0 ) is:
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Points ( 1 , − 1 , 1 ) (1,-1,1) ( 1 , − 1 , 1 ) , ( α , − 1 , 0 ) (\alpha,-1,0) ( α , − 1 , 0 ) ; directions ( 0 , 3 , − 1 ) (0,3,-1) ( 0 , 3 , − 1 ) , ( 2 , 0 , − 3 ) (2,0,-3) ( 2 , 0 , − 3 ) .
\[(0,3,-1)\times(2,0,-3)=(-9,-2,-6).\]
Coplanarity: ( α − 1 , 0 , − 1 ) ⋅ ( − 9 , − 2 , − 6 ) = 0 ⟹ − 9 α + 15 = 0 ⟹ α = 5 3 . (\alpha-1,0,-1)\cdot(-9,-2,-6)=0 \implies -9\alpha+15=0 \implies \alpha=\frac53. ( α − 1 , 0 , − 1 ) ⋅ ( − 9 , − 2 , − 6 ) = 0 ⟹ − 9 α + 15 = 0 ⟹ α = 3 5 .
Plane (normal ( 9 , 2 , 6 ) (9,2,6) ( 9 , 2 , 6 ) ) through ( 1 , − 1 , 1 ) (1,-1,1) ( 1 , − 1 , 1 ) : 9 x + 2 y + 6 z − 13 = 0 9x+2y+6z-13=0 9 x + 2 y + 6 z − 13 = 0 .
\[d=\frac{|9(\frac53)-13|}{\sqrt{81+4+36}}=\frac{2}{11}.\] Answer the question to unlock