Solveeit Logo
Login

Question

Mathematics Question on Vector Algebra

Let a, b, c are three non-coplanar vectors such that r1=a−b+c,r2=b+c−a,r3=c+a+b,r_{1}=a-b+c, r_{2}=b+c-a, r_{3}=c+a+b, r=2a−3b+4c.If r=λ1r1+λ2r2+λ3r3,r=2a-3b+4c. If \, r=\lambda_{1}r_{1}+\lambda_{2}r_{2}+\lambda_{3}r_{3}, then

A

λ1=7\lambda_{1}=7

B

λ1+λ3=6\lambda_{1}+\lambda_{3}=6

C

λ1+λ2+λ3=4\lambda_{1}+\lambda_{2}+\lambda_{3}=4

D

λ3+λ2=2\lambda_{3}+\lambda_{2}=2

Answer

λ1+λ2+λ3=4\lambda_{1}+\lambda_{2}+\lambda_{3}=4

Explanation

Solution

We have λ1r1+λ2r2+λ3r3\lambda_{1}r_{1}+\lambda_{2}r_{2}+\lambda_{3}r_{3} ⇒2a−3b+4c=(λ1−λ2+λ3)a+(−λ1+λ2+λ3)b+(λ1+λ2+λ3)c\Rightarrow\quad2a-3b+4c=\left(\lambda_{1}-\lambda_{2}+\lambda_{3}\right)a+\left(-\lambda_{1}+\lambda_{2}+\lambda_{3}\right)b+\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)c ⇒ λ1−λ2+λ3=2,−λ1+λ2+λ3=−3,λ1+λ2+λ3=\Rightarrow\, \lambda_{1}-\lambda_{2}+\lambda_{3}=2, -\lambda_{1}+\lambda_{2}+\lambda_{3}=-3, \lambda_{1}+\lambda_{2}+\lambda_{3}= (∵a, b, c are non − coplanar)\left(\because\quad a,\, b, \,c\, are\, non\,-\,coplanar\right) ⇒ λ1=72,λ2=1,λ3=−12\Rightarrow\, \lambda_{1}=\frac{7}{2}, \lambda_{2}=1, \lambda_{3}=-\frac{1}{2} Therefore, λ1+λ3=3 and λ1+λ2+λ3=4.Therefore,\, \lambda_{1}+\lambda_{3}=3 \, and \, \lambda_{1}+\lambda_{2}+\lambda_{3}=4.