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UPSC 2014 Maths Optional Paper 1 Q1a — Step-by-Step Solution

10 marks · Section A

Subspaces · Linear Algebra · asked 6× in 13 yrs · Read the full method →

Question

Find one vector in R3\mathbb R^3 which generates the intersection of VV and WW, where VV is the xyxy plane and WW is the space generated by the vectors (1,2,3)(1,2,3) and (1,1,1)(1,-1,1).

Technique

Parametric description of WW + linear constraint from VV + back-substitution.

Solution

Strategy. Write a general element of WW, impose the VV condition (z=0z=0), solve the resulting linear constraint.

Step 1 — Parametrise WW

W={a(1,2,3)+b(1,1,1):a,bR}={(a+b,  2ab,  3a+b)}W=\{a(1,2,3)+b(1,-1,1):a,b\in\mathbb R\}=\{(a+b,\;2a-b,\;3a+b)\}.

Step 2 — Intersect with V={(x,y,0)}V=\{(x,y,0)\}

Require zz-component =0=0:

3a+b=0    b=3a.3a+b=0\;\Longrightarrow\;b=-3a.

Step 3 — Substitute

For any scalar aa:

(a+b,  2ab,  3a+b)=(a3a,  2a(3a),  0)=(2a,  5a,  0)=a(2,5,0).(a+b,\;2a-b,\;3a+b)=(a-3a,\;2a-(-3a),\;0)=(-2a,\;5a,\;0)=a(-2,5,0).

Hence VW=span{(2,5,0)}V\cap W=\operatorname{span}\{(-2,5,0)\} — a one-dimensional subspace.

Answer

  VW is generated by (2,5,0).  \boxed{\;V\cap W\text{ is generated by }(-2,\,5,\,0).\;}
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