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Mathematics Question on Three Dimensional Geometry

The shortest distance between the straight lines through the points A1=(6,2,2)A_1 = (6, 2, 2) and A2=(−4,0,−1)A_2 = (-4, 0, -1), in the directions of (1,−2,2)(1, -2, 2) and (3,−2,−2)(3, -2, -2) is

A

6

B

8

C

12

D

9

Answer

9

Explanation

Solution

Equation of first line,
x−61=y−22=z−22=k(say)\frac{x-6}{1}=\frac{y-2}{2}=\frac{z-2}{2}=k\left(say\right)
∴x=k+6,y=−2k+2,z=2k+2\therefore x = k + 6, y = -2k + 2, z = 2k + 2
Hence, general point on the first line,
P≡(k+6,−2k+2,2k+2)P \equiv \left(k + 6, - 2k + 2, 2k + 2\right)
Equation of second line,
x+43=y−2=Z+1−2=l(say)\frac{x+4}{3}=\frac{y}{-2}=\frac{Z+1}{-2}=l\quad\left(say\right)
∴x=3l−4,y=−2l,z=−2l−1\therefore x = 3l - 4, y = -2l, z = -2l -1
Hence, general point on the second line,
Q≡(3l−4,−2l,−2l−1)Q \equiv \left(3l - 4, - 2l, -2l - 1\right)
Direction ratios of PQ are
3l−4−k−6,−2l+2k−2,−2l−1−2k−23l - 4 - k -6, -2l + 2k -2, - 2l -1 -2k - 2
i.e. 3l−k−10,−2l+2k−2,−2l−2k−33l - k - 10, -2l + 2k- 2, - 2l -2k - 3
Now |PQ| will be the shortest distance
between the two lines if PQ is perpendicular
to both the lines. Hence,
1(3l−k−10)+(−2)1\left(3l - k-10\right)+ \left(-2\right)
(−2l+2k−2)+2(−2l−2k−3)=0\left(-2l + 2k - 2\right) + 2\left(-2l - 2k - 3\right)= 0
and 3(3l−k−10)+(−2)(−2l+2k−2)+(−2)(−21−2k−3)=03 \left(3l - k - 10\right) + \left(-2\right)\left(-2l+2k-2\right)+\left(-2\right)\left(-21-2k-3\right)=0
i.e.3l−9k=12orl−3k=4...(i)i.e. 3l - 9k = 12 or l - 3k = 4\quad\quad\quad ...\left(i\right)
and 17l−3k=20...(ii)17l - 3k = 20 \quad\quad\quad\quad\quad\quad\quad\quad... \left(ii\right)
Subtracting equation (i)\left(i\right) from(ii) \left(ii\right), we get
161=16∴l=1161 = 16 \quad\quad\quad\quad\therefore l = 1
Putting this value of l in equation \left(i\right), we get
−3k=3,∴k=−1- 3k = 3, \therefore k = - 1
∴P≡(−1+6,−2(−1)+2,2(−1)+2)\therefore P \equiv \left(-1 + 6, - 2 \left(- 1\right) + 2, 2 \left(- 1\right) + 2\right)
≡(5,4,0)\quad\quad\equiv \left(5, 4, 0\right)
Similarly, Q=(−1,−2,−3)Q = \left(-1, -2, -3\right)
Hence, shortest distance, PQ,
=(−1−5)2+(−2−4)2+(−3−0)2\sqrt{\left(-1-5\right)^{2}+\left(-2-4\right)^{2}+\left(-3-0\right)^{2}}
=(−6)2+(−6)2+(−3)2=36+36+9=\sqrt{\left(-6\right)^{2}+\left(-6\right)^{2}+\left(-3\right)^{2}}=\sqrt{36+36+9}
=9 units