Exams Joint Entrance Examination Mathematics 3D Geometry Question 126 Question 126 of 521 Medium NUMERICAL
If the shortest distance between the lines x − λ 3 = y − 2 − 1 = z − 1 1 \frac{x-\lambda }{3}=\frac{y-2}{-1}=\frac{z-1}{1} 3 x − λ = − 1 y − 2 = 1 z − 1 and x + 2 − 3 = y + 5 2 = z − 4 4 \frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4} − 3 x + 2 = 2 y + 5 = 4 z − 4 is 44 30 \frac{44}{\sqrt{30}} 30 44 , then the largest possible value of ∣ λ ∣ |\lambda | ∣ λ ∣ is equal to \_\_\_\_\_\_\_\_\_.
Answer: 43
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L 1 : x − λ 3 = y − 2 − 1 = z − 1 1 L 2 : x + 2 − 3 = y + 5 2 = z − 4 4 \begin{gathered} {L}_{1}:\frac{x-\lambda }{3}=\frac{y-2}{-1}=\frac{z-1}{1} \\ {L}_{2}:\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4} \end{gathered} L 1 : 3 x − λ = − 1 y − 2 = 1 z − 1 L 2 : − 3 x + 2 = 2 y + 5 = 4 z − 4
n 1 × n 2 = ∣ i ^ j ^ k ^ 3 − 1 1 − 3 2 4 ∣ = − 6 i ^ − 15 j ^ + 3 k ^ d = ∣ [ ( λ + 2 ) i ^ + 7 j ^ − 3 k ^ ] [ − 6 i ^ − 15 j ^ + 3 k ^ ] ∣ − 6 i ^ − 15 j ^ + 3 k ^ ∣ ∣ = 44 30 ∣ − 6 λ − 12 − 105 − 9 270 ∣ = 44 30 ∣ 6 λ + 126 ∣ = 132 ∣ λ + 21 ∣ = 22 λ + 21 = ± 22 ∣ λ ∣ m a x = 43 \begin{gathered} {n}_{1}\times {n}_{2}=|\begin{array}{|c|}\hat{i} \quad \hat{j} \quad \hat{k} \\ 3 \quad -1 \quad 1 \\ -3 \quad 2 \quad 4\end{array}| \\ =-6\hat{i}-15\hat{j}+3\hat{k} \\ d=|\frac{[(\lambda +2)\hat{i}+7\hat{j}-3\hat{k}][-6\hat{i}-15\hat{j}+3\hat{k}]}{|-6\hat{i}-15\hat{j}+3\hat{k}|}|=\frac{44}{\sqrt{30}} \\ |\frac{-6\lambda -12-105-9}{\sqrt{270}}|=\frac{44}{\sqrt{30}} \\ |6\lambda +126|=132 \\ |\lambda +21|=22 \\ \lambda +21=\pm 22 \\ |\lambda {|}_{max}=43 \end{gathered} n 1 × n 2 = ∣ i ^ j ^ k ^ 3 − 1 1 − 3 2 4 ∣ = − 6 i ^ − 15 j ^ + 3 k ^ d = ∣ ∣ − 6 i ^ − 15 j ^ + 3 k ^ ∣ [( λ + 2 ) i ^ + 7 j ^ − 3 k ^ ] [ − 6 i ^ − 15 j ^ + 3 k ^ ] ∣ = 30 44 ∣ 270 − 6 λ − 12 − 105 − 9 ∣ = 30 44 ∣6 λ + 126∣ = 132 ∣ λ + 21∣ = 22 λ + 21 = ± 22 ∣ λ ∣ ma x = 43 Answer the question to unlock