Exams Joint Entrance Examination Mathematics 3D Geometry Question 75 Question 75 of 521 Medium MCQ
The shortest distance, between lines L 1 L_1 L 1 and L 2 L_2 L 2 , where \[L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}\] and L 2 L_2 L 2 is the line, passing through the points \[\mathrm{A}(-4,4,3), \mathrm{B}(-1,6,3)\] and perpendicular to the line...
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L 2 = x + 4 3 = y − 4 2 = z − 3 0 ∴ S . D = ∣ x 2 − x 1 y 2 − y 1 z 2 − z 1 2 − 3 2 3 2 0 ∣ ∣ n 1 → × n 2 → ∣ = ∣ 5 − 5 − 7 2 − 3 2 3 2 0 ∣ ∣ n 1 → × n 2 → ∣ = 141 ∣ − 4 i ^ + 6 j ^ + 13 k ^ ∣ = 141 16 + 36 + 169 = 141 221 \begin{gathered} {L}_{2}=\frac{x+4}{3}=\frac{y-4}{2}=\frac{z-3}{0} \\ \therefore S.D=\frac{|\begin{array}{|c|}{x}_{2}-{x}_{1} \quad {y}_{2}-{y}_{1} \quad {z}_{2}-{z}_{1} \\ 2 \quad -3 \quad 2 \\ 3 \quad 2 \quad 0\end{array}|}{|\overset{\to }{{n}_{1}}\times \overset{\to }{{n}_{2}}|} \\ =\frac{|\begin{array}{|c|}5 \quad -5 \quad -7 \\ 2 \quad -3 \quad 2 \\ 3 \quad 2 \quad 0\end{array}|}{|\overset{\to }{{n}_{1}}\times \overset{\to }{{n}_{2}}|} \\ =\frac{141}{|-4\hat{i}+6\hat{j}+13\hat{k}|} \\ =\frac{141}{\sqrt{16+36+169}} \\ =\frac{141}{\sqrt{221}} \end{gathered} L 2 = 3 x + 4 = 2 y − 4 = 0 z − 3 ∴ S . D = ∣ n 1 → × n 2 → ∣ ∣ x 2 − x 1 y 2 − y 1 z 2 − z 1 2 − 3 2 3 2 0 ∣ = ∣ n 1 → × n 2 → ∣ ∣ 5 − 5 − 7 2 − 3 2 3 2 0 ∣ = ∣ − 4 i ^ + 6 j ^ + 13 k ^ ∣ 141 = 16 + 36 + 169 141 = 221 141 Answer the question to unlock