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JEE Main & Advanced Sample Paper JEE Main Sample Paper-40

  • question_answer
    Let coordinates of a point 'p' with respect to the system of non-coplanar vectors \[\vec{a},\vec{b}\]and \[\vec{c}\] is (3, 2, 1). Then coordinates of "p' with respect to the system of vectors \[\vec{a}+\vec{b}+\vec{c},\]\[\vec{a}-\vec{b}+\vec{c}\]and \[\vec{a}+\vec{b}-\vec{c}\]is

    A) \[\left( \frac{3}{2},\frac{1}{2},1 \right)\]                   

    B) \[\left( \frac{3}{2},1,\frac{1}{2} \right)\]

    C) \[\left( \frac{1}{2},\frac{3}{2},1 \right)\]                   

    D) None

    Correct Answer: A

    Solution :

    Any vector (x, y, z) can be represented as a linear combination of three non-coplanar vectors. \[\overrightarrow{OP}=x\vec{a}+y\vec{b}+z\vec{c}\] But \[\overrightarrow{P}\] is (3,2,1) \[\Rightarrow \overrightarrow{OP}=3\vec{a}+2\vec{b}+\vec{c}\] Now if coordinate axis changes then the vector is represented by \[x(\vec{a}+\vec{b}+\vec{c})+y(\vec{a}-\vec{b}-\vec{c})+2(\vec{a}+\vec{b}-\vec{c})\] \[=3\vec{a}+2\vec{b}+\vec{c}\] Comparing coeff of \[\vec{a},\vec{b}\] & \[\vec{c},\] we get \[x+y+z=3,x-y+z=2,x+y-z=1\] On solving, we get \[x=\frac{3}{2},y=\frac{1}{2},z=1\] Hence the new co-ordinates of point P are \[\left( \frac{3}{2},\frac{1}{2},1 \right)\]


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