Exams Joint Entrance Examination Mathematics Vector Algebra Question 37 Question 37 of 379 Hard MCQ
Let the point A divide the line segment joining the points P ( − 1 , − 1 , 2 ) P(-1,-1,2) P ( − 1 , − 1 , 2 ) and Q ( 5 , 5 , 10 ) Q(5,5,10) Q ( 5 , 5 , 10 ) internally in the ratio r : 1 ( r > 0 ) r:1(r>0) r : 1 ( r > 0 ) . If O is the origin and ( O Q → ← ⋅ O A → ← ) − 1 5 ∣ O P → ← × O A → ← ∣ 2 = 10 (\overset{\to \leftarrow }{OQ}\cdot \overset{\to \leftarrow }{OA})-\frac{1}{5}|\overset{\to \leftarrow }{OP}\times \overset{\to \leftarrow }{OA}{|}^{2}=10 ( O Q →← ⋅ O A →← ) − 5 1 ∣ O P →← × O A →← ∣ 2 = 10 , then the value of r is :
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A ≡ ( 5 r − 1 r + 1 , 5 r − 1 r + 1 , 10 r + 2 r + 1 ) ( O Q → ← ⋅ O A → ← ) − ∣ O P → ← × O A → ← ∣ 2 5 = 10 O Q → ← ⋅ O A → ← = 10 r + 1 ( 15 r + 1 ) ∣ O P → ← × O A → ← ∣ 2 = r 2 ( r + 1 ) 2 ( 800 ) 10 r + 1 ( 15 r + 1 ) − 1 5 r 2 ( 800 ) ( r + 1 ) 2 = 10 2 r 2 − 14 r = 0 r = 7 , r ≠ 0 \begin{gathered} A\equiv (\frac{5r-1}{r+1},\frac{5r-1}{r+1},\frac{10r+2}{r+1}) \\ (\overset{\to \leftarrow }{OQ}\cdot \overset{\to \leftarrow }{OA})-\frac{|\overset{\to \leftarrow }{OP}\times \overset{\to \leftarrow }{OA}{|}^{2}}{5}=10\, \\ \overset{\to \leftarrow }{OQ}\cdot \overset{\to \leftarrow }{OA}=\frac{10}{r+1}(15r+1) \\ |\overset{\to \leftarrow }{OP}\times \overset{\to \leftarrow }{OA}{|}^{2}=\frac{{r}^{2}}{(r+1{)}^{2}}(800) \\ \\ \frac{10}{r+1}(15r+1)-\frac{1}{5}\frac{{r}^{2}(800)}{(r+1{)}^{2}}=10 \\ 2{r}^{2}-14r=0 \\ r=7,r\ne 0 \end{gathered} A ≡ ( r + 1 5 r − 1 , r + 1 5 r − 1 , r + 1 10 r + 2 ) ( O Q →← ⋅ O A →← ) − 5 ∣ O P →← × O A →← ∣ 2 = 10 O Q →← ⋅ O A →← = r + 1 10 ( 15 r + 1 ) ∣ O P →← × O A →← ∣ 2 = ( r + 1 ) 2 r 2 ( 800 ) r + 1 10 ( 15 r + 1 ) − 5 1 ( r + 1 ) 2 r 2 ( 800 ) = 10 2 r 2 − 14 r = 0 r = 7 , r = 0 Answer the question to unlock