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Mathematics Question on Three Dimensional Geometry

A line ll passing through the origin is perpendicular to the lines L1:(3+t)i^+(1−2t)j^+(4+2t)k^,−∞<t<∞ L_1 : (3+t) \widehat {i} +(1-2t) \widehat {j}+(4+2t)\widehat {k}, - \infty < t < \infty L2:(3+2s)i^+(3+2s)j^+(2+s)k^,−∞<s<∞ L_2 : (3+2s) \widehat {i} +(3+2s) \widehat {j}+(2+s)\widehat {k}, - \infty < s < \infty Then, the coordinate(s) of the point(s) on l2l_2 at a distance of 17\sqrt 17 from the point of intersection of l and l1l_1 is (are)

A

(73,73,53)\bigg(\frac{7}{3},\frac{7}{3},\frac{5}{3}\bigg)

B

(−1,−1,0)(-1,-1,0)

C

(1,1,1)(1, 1, 1)

D

(79,79,89)\bigg(\frac{7}{9},\frac{7}{9},\frac{8}{9}\bigg)

Answer

(79,79,89)\bigg(\frac{7}{9},\frac{7}{9},\frac{8}{9}\bigg)

Explanation

Solution

PLAN Equation of straight line is
\hspace20mm l: \frac{x-x_1}{a} = \frac{y-y_1}{b}=\frac{z-z_1}{c}
Since, l is perpendicular to l1l_1 and c.
So, its DR's are cross - product of l1l_1 and l2l_2.
Now, to find a point on l2l_2 whose distance is given, assume a point
and find its distance to obtain point.
Let l:x−0a=y−0b=z−0c l: \frac{x-0}{a} = \frac{y-0}{b}=\frac{z-0}{c}
which is perpendicular to
   L1:(3j^−j^+4k^)+t(i^+2j^+2k^)\ \ \ L_1 : (3\widehat {j}-\widehat {j} +4\widehat {k}) + t(\widehat {i} +2 \widehat {j}+2\widehat {k})
   L2:(3i^−3j^+2k^)+s(2i^+2j^+k^)\ \ \ L_2 : (3\widehat {i}-3\widehat {j} +2\widehat {k}) + s(2\widehat {i} +2 \widehat {j}+\widehat {k})
∴\therefore DR′s of l is∣i^j^k^ 122 221 ∣=−2i^+3j^+2k^DR's\ of\ l \ is \begin{vmatrix} \widehat {i} &\widehat {j} &\widehat {k} \\\ 1 & 2 & 2 \\\ 2 & 2 & 1 \\\ \end{vmatrix}= -2 \widehat {i}+3\widehat {j} +2\widehat {k}
\hspace30mm l: \frac{x}{-2} = \frac{y}{3}=\frac{z}{-2}=k_1,k_2

Now, A(−2k1,3k1,−2k1)A(-2k_1,3k_1,-2k_1) and B(−2k2,3k2,−2k2).B(-2k_2,3k_2,-2k_2).
Since, A lies on l1l_1
∴\therefore (−2k1)i^+(3k1)j^−(2k1)k^=(3+t)i^+(−1+2t)j^(-2k_1)\widehat {i} + (3k_1)\widehat {j} - (2k_1)\widehat {k}=(3+t)\widehat {i} + (-1+2t) \widehat {j}
\hspace70mm + (4 + 2t) \widehat {k}
⇒    \Rightarrow \ \ \ \ 3+t=−2k1,−1+2t=3k1,4+2t=−2k13+t = -2k_ 1,-1 + 2t = 3k_1, 4 + 2t = -2k_1
∴\therefore     k=−1\ \ \ \ k_ = -1
⇒\Rightarrow A (2, -3, 2)
Let any point on l2(3+2s,3+2s,2+s) l_2 (3+2s,3+2s,2+s)
  (2−3−2s)2(−3−3+2s)2+(2−2−s)2=17\ \ \sqrt{ (2-3-2s)^2 (-3 - 3 +2s)^2+(2-2-s)^2} = \sqrt{17}
⇒\Rightarrow \hspace50mm 9s^2+28s+37 = 17
⇒\Rightarrow \hspace50mm 9s^2+28s+20 = 0
⇒\Rightarrow \hspace20mm 9s^2+18s+10s+20=0
⇒\Rightarrow \hspace30mm (9s+10) (s+2)=0
∴\therefore \hspace50mm s = -2, \frac{-10}{9}
Hence, (-1,-1,0) and (79,79,89)\bigg(\frac{7}{9},\frac{7}{9},\frac{8}{9}\bigg) are required points.