Exams Joint Entrance Examination Mathematics 3D Geometry Question 63 Question 63 of 521 Medium MCQ
If A ( 3 , 1 , − 1 ) , B ( 5 3 , 7 3 , 1 3 ) , C ( 2 , 2 , 1 ) A(3,1,-1),B(\frac{5}{3},\frac{7}{3},\frac{1}{3}),C(2,2,1) A ( 3 , 1 , − 1 ) , B ( 3 5 , 3 7 , 3 1 ) , C ( 2 , 2 , 1 ) and D ( 10 3 , 2 3 , − 1 3 ) D(\frac{10}{3},\frac{2}{3},\frac{-1}{3}) D ( 3 10 , 3 2 , 3 − 1 ) are the vertices of a quadrilateral A B C D ABCD A B C D , then its area is
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A ( 3 , 1 , − 1 ) , B ( 5 3 , 7 3 , 1 3 ) , C ( 2 , 2 , 1 ) , D ( 10 3 , 2 3 , − 1 3 ) A(3,1,-1),B(\frac{5}{3},\frac{7}{3},\frac{1}{3}),C(2,2,1),D(\frac{10}{3},\frac{2}{3},\frac{-1}{3}) A ( 3 , 1 , − 1 ) , B ( 3 5 , 3 7 , 3 1 ) , C ( 2 , 2 , 1 ) , D ( 3 10 , 3 2 , 3 − 1 ) are vertices of a quadrilateral
A C → ← = ( 2 i ^ + 2 j ^ + k ^ ) − ( 3 i ^ + j ^ − k ^ ) = − i ^ + j ^ + 2 k ^ B D → ← = ( 10 3 i ^ + 2 3 j ^ − 1 3 k ^ ) − ( 5 3 i ^ + 7 3 j ^ + 1 3 k ^ ) B D → ← = 5 3 i ^ − 5 3 j ^ − 2 3 k ^ = 1 2 ∣ A C → ← × B D → ← ∣ = 1 2 ∣ i ^ j ^ k ^ − 1 1 2 5 3 − 5 3 − 2 3 ∣ = 1 2 ( 8 3 ) 2 + ( 8 3 ) 2 [ ∵ A C → ← × B D → ← = 8 3 i ^ + 8 3 j ^ ] = 1 2 × 8 3 × 2 = 4 2 3 \begin{gathered} \overset{\to \leftarrow }{AC}=(2\hat{i}+2\hat{j}+\hat{k})-(3\hat{i}+\hat{j}-\hat{k}) \\ =-\hat{i}+\hat{j}+2\hat{k} \\ \overset{\to \leftarrow }{BD}=(\frac{10}{3}\hat{i}+\frac{2}{3}\hat{j}-\frac{1}{3}\hat{k})-(\frac{5}{3}\hat{i}+\frac{7}{3}\hat{j}+\frac{1}{3}\hat{k}) \\ \overset{\to \leftarrow }{BD}=\frac{5}{3}\hat{i}-\frac{5}{3}\hat{j}-\frac{2}{3}\hat{k} \\ =\frac{1}{2}|\overset{\to \leftarrow }{AC}\times \overset{\to \leftarrow }{BD}| \\ =\frac{1}{2}|\begin{array}{|c|}\hat{i} \quad \hat{j} \quad \hat{k} \\ -1 \quad 1 \quad 2 \\ \frac{5}{3} \quad \frac{-5}{3} \quad \frac{-2}{3}\end{array}| \\ =\frac{1}{2}\sqrt{{(\frac{8}{3})}^{2}+{(\frac{8}{3})}^{2}}[\because \overset{\to \leftarrow }{AC}\times \overset{\to \leftarrow }{BD}=\frac{8}{3}\hat{i}+\frac{8}{3}\hat{j}] \\ =\frac{1}{2}\times \frac{8}{3}\times \sqrt{2}=\frac{4\sqrt{2}}{3} \end{gathered} A C →← = ( 2 i ^ + 2 j ^ + k ^ ) − ( 3 i ^ + j ^ − k ^ ) = − i ^ + j ^ + 2 k ^ B D →← = ( 3 10 i ^ + 3 2 j ^ − 3 1 k ^ ) − ( 3 5 i ^ + 3 7 j ^ + 3 1 k ^ ) B D →← = 3 5 i ^ − 3 5 j ^ − 3 2 k ^ = 2 1 ∣ A C →← × B D →← ∣ = 2 1 ∣ i ^ j ^ k ^ − 1 1 2 3 5 3 − 5 3 − 2 ∣ = 2 1 ( 3 8 ) 2 + ( 3 8 ) 2 [ ∵ A C →← × B D →← = 3 8 i ^ + 3 8 j ^ ] = 2 1 × 3 8 × 2 = 3 4 2 Answer the question to unlock