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Question 124 of 162

Q.The point of intersection of the lines r⃗=(−i⃗+2j⃗+3k⃗)+t(−2i⃗+j⃗+k⃗)\vec{r} = (-\vec{i} + 2\vec{j} + 3\vec{k}) + t(-2\vec{i} + \vec{j} + \vec{k}) and r⃗=(2i⃗+3j⃗+5k⃗)+s(i⃗+2j⃗+3k⃗)\vec{r} = (2\vec{i} + 3\vec{j} + 5\vec{k}) + s(\vec{i} + 2\vec{j} + 3\vec{k}) is :

(a) (1,1,2)(1, 1, 2)
(b) (2,1,1)(2, 1, 1)
(c) (1,1,1)(1, 1, 1)
(d) (1,2,1)(1, 2, 1)
Puducherry TnboardTamil Nadu HSC (DGE) Board 2018MCQ· 1mImportance★★★★★
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Solving the parametric equations of the two given lines simultaneously gives the intersection point (1,1,2)(1,1,2).

  1. Line 1: r⃗=(−i⃗+2j⃗+3k⃗)+t(−2i⃗+j⃗+k⃗)\vec r=(-\vec i+2\vec j+3\vec k)+t(-2\vec i+\vec j+\vec k), i.e. (x,y,z)=(−1−2t, 2+t, 3+t)(x,y,z)=(-1-2t,\ 2+t,\ 3+t).
  2. Line 2: r⃗=(2i⃗+3j⃗+5k⃗)+s(i⃗+2j⃗+3k⃗)\vec r=(2\vec i+3\vec j+5\vec k)+s(\vec i+2\vec j+3\vec k), i.e. (x,y,z)=(2+s, 3+2s, 5+3s)(x,y,z)=(2+s,\ 3+2s,\ 5+3s).
  3. Equate the yy-components: 2+t=3+2s⇒t=1+2s2+t=3+2s \Rightarrow t=1+2s.
  4. Equate the zz-components: 3+t=5+3s⇒t=2+3s3+t=5+3s \Rightarrow t=2+3s.
  5. Set the two expressions for tt equal: 1+2s=2+3s⇒−1=s⇒s=−11+2s=2+3s \Rightarrow -1=s \Rightarrow s=-1.
  6. Then t=1+2(−1)=−1t=1+2(-1)=-1. …

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