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  • Points with position vectors  <img alt="\vec{a}-2\vec{b}+3\vec{c},\; -2\vec{a}+3\vec{b}-\vec{c},\; 4\vec{a}-7\vec{b}+7\vec{c}" src="https://learn.careers360.com/latex-image/?%5Cvec%7Ba%7D-2%5C

Points with position vectors  \vec{a}-2\vec{b}+3\vec{c},\; -2\vec{a}+3\vec{b}-\vec{c},\; 4\vec{a}-7\vec{b}+7\vec{c}  are collinear ?.It is given that vectors are non-coplanar

Option: 1

YES


Option: 2

NO


Option: 3

Can't be said


Option: 4

May be


Answers (1)

best_answer

As we learned

 Two vectors are said to be collinear if they have common line of action.

 Or We can say 

 Two vectors \vec{A} and \vec{B} are collinear if there exists a number n such that \vec{A}=n.\vec{B}

 

 \\*\lambda_{1} (\vec{a}-2\vec{b}+3\vec{c})+\; \lambda_{2} (-2\vec{a}+3\vec{b}-\vec{c})+\; \lambda_{3} (4\vec{a}-7\vec{b}-7\vec{c})=0\\*\\*with\; \; \lambda_{1}+\lambda_{2}+\lambda_{3}=0

equating the coefficient of \vec{a},\vec{b}\, and\, \vec{c}.Separately to Zero ; we get

\\*\lambda _{1}-2\lambda _{2}+4\lambda _{3}=0\\*\\*-2\lambda _{1}+3\lambda _{2}-7\lambda _{3}=0\; \; and\\*\\*3\lambda _{1}-\lambda _{2}+7\lambda _{3}=0\\*\\*\Rightarrow \lambda _{1}=\, -2,\lambda _{2}=\, 1,\, \lambda _{3}=1\\*\\*\Rightarrow \lambda _{1}+\lambda _{2}+\lambda _{3}=0

 

Posted by

Rakesh

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