Exams Joint Entrance Examination Mathematics 3D Geometry Question 140 Question 140 of 521 Medium NUMERICAL
Let the plane x + 3 y − 2 z + 6 = 0 x+3 y-2 z+6=0 x + 3 y − 2 z + 6 = 0 meet the co-ordinate axes at the points A, B, C. If the orthocenter of the triangle \[\mathrm{ABC}\] is \[\left(\alpha, \beta, \frac{6}{7}\right)\], then \[98(\alpha+\beta)^{2}\] is equal to \_\_\_\_\_\_\_\_\_\_\_.
Answer: 288
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A ( − 6 , 0 , 0 ) B ( 0 , − 2 , 0 ) C = ( 0 , 0 , 3 ) A B → ← = 6 i ^ − 2 j ^ , B C → ← = 2 j ^ + 3 k ^ , A C → ← = 6 i ^ + 3 k ^ \begin{gathered} A(-6,0,0)\, B(0,-2,0)C=(0,0,3) \\ \overset{\to \leftarrow }{AB}=6\hat{i}-2\hat{j},\, \overset{\to \leftarrow }{BC}=2\hat{j}+3\hat{k}, \\ \overset{\to \leftarrow }{AC}=6\hat{i}+3\hat{k} \end{gathered} A ( − 6 , 0 , 0 ) B ( 0 , − 2 , 0 ) C = ( 0 , 0 , 3 ) A B →← = 6 i ^ − 2 j ^ , B C →← = 2 j ^ + 3 k ^ , A C →← = 6 i ^ + 3 k ^
A H → ← ⋅ B C → ← = 0 ( α + 6 , β , 6 7 ) ⋅ ( 0 , 2 , 3 ) = 0 β = − 9 7 C H → ← ⋅ A B → ← = 0 ( α , β , − 15 7 ) ⋅ ( 6 , − 2 , 0 ) = 0 6 α − 2 β = 0 α = − 3 7 \begin{gathered} \overset{\to \leftarrow }{AH}\cdot \overset{\to \leftarrow }{BC}=0 \\ (\alpha +6,\beta ,\frac{6}{7})\cdot (0,2,3)=0 \\ \beta =\frac{-9}{7} \\ \overset{\to \leftarrow }{CH}\cdot \overset{\to \leftarrow }{AB}=0 \\ (\alpha ,\beta ,\frac{-15}{7})\cdot (6,-2,0)=0 \\ 6\alpha -2\beta =0 \\ \alpha =\frac{-3}{7} \end{gathered} A H →← ⋅ B C →← = 0 ( α + 6 , β , 7 6 ) ⋅ ( 0 , 2 , 3 ) = 0 β = 7 − 9 C H →← ⋅ A B →← = 0 ( α , β , 7 − 15 ) ⋅ ( 6 , − 2 , 0 ) = 0 6 α − 2 β = 0 α = 7 − 3
98 ( α + β ) 2 = ( 98 ) ( 144 ) 49 = 288 98(\alpha +\beta {)}^{2}=(98)\frac{(144)}{49}=288 98 ( α + β ) 2 = ( 98 ) 49 ( 144 ) = 288 Answer the question to unlock