Exams Joint Entrance Examination Mathematics Vector Algebra Question 86 Question 86 of 379 Medium MCQ
The sum of all values of α \alpha α , for which the points whose position vectors are i ^ − 2 j ^ + 3 k ^ , 2 i ^ − 3 j ^ + 4 k ^ , ( α + 1 ) i ^ + 2 k ^ \hat{i}-2\hat{j}+3\hat{k},2\hat{i}-3\hat{j}+4\hat{k},(\alpha +1)\hat{i}+2\hat{k} i ^ − 2 j ^ + 3 k ^ , 2 i ^ − 3 j ^ + 4 k ^ , ( α + 1 ) i ^ + 2 k ^ and 9 i ^ + ( α − 8 ) j ^ + 6 k ^ 9\hat{i}+(\alpha -8)\hat{j}+6\hat{k} 9 i ^ + ( α − 8 ) j ^ + 6 k ^ are coplanar, is equal to :
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Let O A → ← = i ^ − 2 j ^ + 3 k ^ \overset{\to \leftarrow }{OA}=\hat{i}-2\hat{j}+3\hat{k} O A →← = i ^ − 2 j ^ + 3 k ^
O B → ← = 2 i ^ − 3 j ^ + 4 k ^ O C → ← = ( a + 1 ) i ^ + 2 k ^ \begin{gathered} \overset{\to \leftarrow }{OB}=2\hat{i}-3\hat{j}+4\hat{k} \\ \overset{\to \leftarrow }{OC}=(a+1)\hat{i}+2\hat{k} \end{gathered} O B →← = 2 i ^ − 3 j ^ + 4 k ^ O C →← = ( a + 1 ) i ^ + 2 k ^
and O D → ← = 9 i ^ + ( a − 8 ) j ^ + 6 k ^ \overset{\to \leftarrow }{OD}=9\hat{i}+(a-8)\hat{j}+6\hat{k} O D →← = 9 i ^ + ( a − 8 ) j ^ + 6 k ^
∴ A B → ← = i ^ − j ^ + k ^ A C → ← = a i ^ + 2 j ^ − k ^ A D → ← = 8 i ^ + ( a − 6 ) j ^ + 3 k ^ \begin{gathered} \therefore \overset{\to \leftarrow }{AB}=\hat{i}-\hat{j}+\hat{k} \\ \overset{\to \leftarrow }{AC}=a\hat{i}+2\hat{j}-\hat{k} \\ \overset{\to \leftarrow }{AD}=8\hat{i}+(a-6)\hat{j}+3\hat{k} \end{gathered} ∴ A B →← = i ^ − j ^ + k ^ A C →← = a i ^ + 2 j ^ − k ^ A D →← = 8 i ^ + ( a − 6 ) j ^ + 3 k ^
Since, given point are coplanar
∴ [ A B → ← , A C → ← , A D → ← ] = 0 ⇒ ∣ 1 − 1 1 a 2 − 1 8 a − 6 3 ∣ = 0 ⇒ 1 ( 6 + a − 6 ) + 1 ( 3 a + 8 ) + 1 ( a 2 − 6 a − 16 ) = 0 ⇒ a + 3 a + 8 + a 2 − 6 a − 16 = 0 ⇒ a 2 − 2 a − 8 = 0 ⇒ a 2 − 4 a + 2 a − 8 = 0 ⇒ a ( a − 4 ) + 2 ( a − 4 ) = 0 ⇒ ( a − 4 ) ( a + 2 ) = 0 ⇒ a = 4 , − 2 ∴ a = 4 − 2 = 2 \begin{gathered} \therefore \, [\overset{\to \leftarrow }{AB},\overset{\to \leftarrow }{AC},\overset{\to \leftarrow }{AD}]=0 \\ \Rightarrow |\begin{array}{|c|}1 \quad -1 \quad 1 \\ a \quad 2 \quad -1 \\ 8 \quad a-6 \quad 3\end{array}|=0 \\ \Rightarrow 1(6+a-6)+1(3a+8)+1({a}^{2}-6a-16)=0 \\ \Rightarrow a+3a+8+{a}^{2}-6a-16=0\Rightarrow {a}^{2}-2a-8=0 \\ \Rightarrow {a}^{2}-4a+2a-8=0\Rightarrow a(a-4)+2(a-4)=0 \\ \Rightarrow (a-4)(a+2)=0\Rightarrow a=4,-2 \\ \therefore a=4-2=2 \end{gathered} ∴ [ A B →← , A C →← , A D →← ] = 0 ⇒ ∣ 1 − 1 1 a 2 − 1 8 a − 6 3 ∣ = 0 ⇒ 1 ( 6 + a − 6 ) + 1 ( 3 a + 8 ) + 1 ( a 2 − 6 a − 16 ) = 0 ⇒ a + 3 a + 8 + a 2 − 6 a − 16 = 0 ⇒ a 2 − 2 a − 8 = 0 ⇒ a 2 − 4 a + 2 a − 8 = 0 ⇒ a ( a − 4 ) + 2 ( a − 4 ) = 0 ⇒ ( a − 4 ) ( a + 2 ) = 0 ⇒ a = 4 , − 2 ∴ a = 4 − 2 = 2 Answer the question to unlock