Question
Download Solution PDFयदि 2 रेखाएं \(\rm {x -1\over 2}={y+1\over3}={z\over-1}\) और \(\rm {x+2\over 1}={y-1\overλ}={z-2\over2}\) एक-दूसरे के समतलीय हैं, तो λ का मान क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
यदि 2 रेखाएं \(\rm {x -x_1\over a}={y-y_1\over b}={z-z_1\over c}\) और \(\rm {x -x_2\over p}={y-y_2\over q}={z-z_2\over r}\) एक-दूसरे के समतलीय हैं, तो
\(\begin{vmatrix} x_2-x_1& y_2-y_1 & z_2-z_1\\ a& b &c \\ p& q &r \end{vmatrix} = 0\) है।
गणना:
समतलीय रेखाओं के दिए गए समीकरण
\(\rm {x -1\over 2}={y+1\over3}={z\over-1}\) और \(\rm {x+2\over 1}={y-1\overλ}={z-2\over2}\) हैं।
x1 = 1, y1 = - 1, z1 = 0 और x2 = - 2, y2 = 1, z2 = 2
∵ रेखाएं एक-दूसरे के समतलीय हैं इसलिए,
\(\begin{vmatrix} x_2-x_1&y_2-y_1& z_2-z_1\\ 2& 3 &-1 \\ 1&λ &2 \end{vmatrix} = 0\)
\(\begin{vmatrix} -2-1& 1-(-1)& 2-0\\ 2& 3 &-1 \\ 1&λ &2 \end{vmatrix} = 0\)
\(\begin{vmatrix} -2-1& 1+1& 2-0\\ 2& 3 &-1 \\ 1&λ &2 \end{vmatrix} = 0\)
\(\begin{vmatrix} -3& 2& 2\\ 2& 3 &-1 \\ 1&λ &2 \end{vmatrix} = 0\)
-3(6 + λ) - 2(4 + 1) + 2(2λ - 3) = 0
λ - 34 = 0
λ = 34
Last updated on May 31, 2025
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