Maths-
General
Easy

Question

Statement negative 1 colon If a comma b comma c are distinct non-negative numbers and the vectors â plus a j with ˆ on top plus c k with ˆ on top comma i with ˆ on top plus k with ˆ on top and c i with ˆ on top plus c j with ˆ on top plus b k with ˆ on top are coplanar then c is arithmetic mean of a and b.
Statement -2: Parallel vectors have proportional direction ratios.

  1. Statement negative 1 is True, Statement negative 2 is True; Statement negative 2 is a correct explanation for Statement - 1    
  2. Statement negative 1 is True, Statement negative 2 is True; Statement negative 2 is NOT a correct explanation for Statement - 1    
  3. Statement negative 1 is True, Statement negative 2 is False    
  4. Statement negative 1 is False, Statement negative 2 is True    

The correct answer is: Statement negative 1 is True, Statement negative 2 is True; Statement negative 2 is NOT a correct explanation for Statement - 1


    Here open vertical bar table row a a c row 1 0 1 row c c b end table close vertical bar equals 0 rightwards double arrow c to the power of 2 end exponent equals a b

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    If a with ‾ on top equals i plus j minus k comma b with ‾ on top equals 2 i plus j minus 3 k and stack r with ‾ on top is a vector satisfying 2 stack r with ‾ on top plus stack r with ‾ on top cross times stack a with ‾ on top equals stack b with ‾ on top.
    Assertion left parenthesis A right parenthesis colon stack r with ‾ on top can be expressed in terms of stack a with ‾ on top comma stack b with ‾ on top and stack a with ‾ on top cross times stack b with ‾ on top.
    Reason left parenthesis R right parenthesis colon r with ‾ on top equals 1 over 7 left parenthesis 7 i plus 5 j minus 9 k plus a with ‾ on top cross times b with ‾ on top right parenthesis

    If a with ‾ on top equals i plus j minus k comma b with ‾ on top equals 2 i plus j minus 3 k and stack r with ‾ on top is a vector satisfying 2 stack r with ‾ on top plus stack r with ‾ on top cross times stack a with ‾ on top equals stack b with ‾ on top.
    Assertion left parenthesis A right parenthesis colon stack r with ‾ on top can be expressed in terms of stack a with ‾ on top comma stack b with ‾ on top and stack a with ‾ on top cross times stack b with ‾ on top.
    Reason left parenthesis R right parenthesis colon r with ‾ on top equals 1 over 7 left parenthesis 7 i plus 5 j minus 9 k plus a with ‾ on top cross times b with ‾ on top right parenthesis

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    If stack a with minus on top comma stack b with minus on top are non-zero vectors such that vertical line stack a with minus on top plus stack b with minus on top vertical line equals vertical line stack a with minus on top minus 2 stack b with minus on top vertical line then
    Assertion left parenthesis A right parenthesis : Least value of stack a with ‾ on top times stack b with ‾ on top plus fraction numerator 4 over denominator vertical line stack b with ‾ on top vertical line to the power of 2 end exponent plus 2 end fraction is 2 square root of 2 minus 1
    Reason (R): The expression stack a with minus on top times stack b with minus on top plus fraction numerator 4 over denominator vertical line stack b with minus on top vertical line to the power of 2 end exponent plus 2 end fraction is least when magnitude of stack b with minus on top is square root of 2 t a n invisible function application open parentheses fraction numerator pi over denominator 8 end fraction close parentheses end root

    If stack a with minus on top comma stack b with minus on top are non-zero vectors such that vertical line stack a with minus on top plus stack b with minus on top vertical line equals vertical line stack a with minus on top minus 2 stack b with minus on top vertical line then
    Assertion left parenthesis A right parenthesis : Least value of stack a with ‾ on top times stack b with ‾ on top plus fraction numerator 4 over denominator vertical line stack b with ‾ on top vertical line to the power of 2 end exponent plus 2 end fraction is 2 square root of 2 minus 1
    Reason (R): The expression stack a with minus on top times stack b with minus on top plus fraction numerator 4 over denominator vertical line stack b with minus on top vertical line to the power of 2 end exponent plus 2 end fraction is least when magnitude of stack b with minus on top is square root of 2 t a n invisible function application open parentheses fraction numerator pi over denominator 8 end fraction close parentheses end root

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    Statement- 1: If a with rightwards arrow on top equals 3 i with ˆ on top minus 3 j with ˆ on top plus k with ˆ on top comma b with rightwards arrow on top equals negative i with ˆ on top plus 2 j with ˆ on top plus k with ˆ on top and c with rightwards arrow on top equals i with ˆ on top plus j with ˆ on top plus k with ˆ on top and d with rightwards arrow on top equals 2 i with ˆ on top minus j with ˆ on top, then there exist real numbers alpha comma beta, gamma such that stack a with rightwards arrow on top equals alpha stack b with rightwards arrow on top plus beta stack c with rightwards arrow on top plus gamma d
    Statement- 2: stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top comma stack d with rightwards arrow on top are four vectors in a 3 - dimensional space. If stack b with rightwards arrow on top comma stack c with rightwards arrow on top comma stack d with rightwards arrow on top are non-coplanar, then there exist real numbers alpha comma beta comma gamma such that stack a with rightwards arrow on top equals alpha stack b with rightwards arrow on top plus beta stack c with rightwards arrow on top plus gamma stack d with rightwards arrow on top

    Statement- 1: If a with rightwards arrow on top equals 3 i with ˆ on top minus 3 j with ˆ on top plus k with ˆ on top comma b with rightwards arrow on top equals negative i with ˆ on top plus 2 j with ˆ on top plus k with ˆ on top and c with rightwards arrow on top equals i with ˆ on top plus j with ˆ on top plus k with ˆ on top and d with rightwards arrow on top equals 2 i with ˆ on top minus j with ˆ on top, then there exist real numbers alpha comma beta, gamma such that stack a with rightwards arrow on top equals alpha stack b with rightwards arrow on top plus beta stack c with rightwards arrow on top plus gamma d
    Statement- 2: stack a with rightwards arrow on top comma stack b with rightwards arrow on top comma stack c with rightwards arrow on top comma stack d with rightwards arrow on top are four vectors in a 3 - dimensional space. If stack b with rightwards arrow on top comma stack c with rightwards arrow on top comma stack d with rightwards arrow on top are non-coplanar, then there exist real numbers alpha comma beta comma gamma such that stack a with rightwards arrow on top equals alpha stack b with rightwards arrow on top plus beta stack c with rightwards arrow on top plus gamma stack d with rightwards arrow on top

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    Statement- 1 open parentheses S subscript 1 end subscript close parentheses:If A open parentheses x subscript 1 end subscript comma y subscript 1 end subscript close parentheses comma B open parentheses x subscript 2 end subscript comma y subscript 2 end subscript close parentheses comma C open parentheses x subscript 3 end subscript comma y subscript 3 end subscript close parentheses are non-collinear points. Then every point left parenthesis x comma y right parenthesis in the plane of capital delta to the power of text le  end text end exponent A B C, can be expressed in the form open parentheses fraction numerator k x subscript 1 end subscript plus l x subscript 2 end subscript plus m x subscript 3 end subscript over denominator k plus l plus m end fraction comma fraction numerator k y subscript 1 end subscript plus l y subscript 2 end subscript plus m y subscript 3 end subscript over denominator k plus l plus m end fraction close parentheses
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