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NCERT Class XII Mathematics Chapter - - Solutions

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Question : 22 of 78
Marks: +1, -0
Find the shortest distance between the lines x+17=y+1−6=z+11\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} and x−31=y−5−2=z−71\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}
Solution:  
The shortest distance between the lines l1:x+17=y+1−6=z+11l_1:\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} and l2:x−31=y−5−2=z−71l_2:\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1} is
∣∣x2−x1y2−y1z2−z1a1b1c1a2b2c2∣(b1c2−b2c1)2+(c1a2−a1c2)2+(a1b2−a2b1)2∣\left|\frac{\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}}{\sqrt{(b_1c_2-b_2c_1)^2+(c_1a_2-a_1c_2)^2+(a_1b_2-a_2b_1)^2}}\right|
Here, x1=−1,y1=−1,z1=−1x_1 = -1, y_1 = -1, z_1 = -1; x2=3,y2=5,z2=7x_2 = 3, y_2 = 5, z_2 = 7 and a1=7,b1=−6,c1=1a_1 = 7, b_1 = -6, c_1 = 1; a2=1,b2=−2,c2=1a_2 = 1, b_2 = -2, c_2 = 1
Required shortest distance
=∣∣4687−611−21∣[−6−(−2)]2+(1−7)2+(−14−(−6))2∣=\left|\frac{\begin{vmatrix} 4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix}}{\sqrt{[-6-(-2)]^2+(1-7)^2+(-14-(-6))^2}}\right|
=116116=116=\frac{116}{\sqrt{116}}=\sqrt{116} =229=2\sqrt{29}units
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