Maths-
General
Easy

Question

Assertion : For a equals fraction numerator 1 over denominator square root of 3 end fraction the volume of the parallel piped formed by vectors i with ˆ on top plus a j with ˆ on top comma a i with ˆ on top plus j with ˆ on top plus k with ˆ on top and j with ˆ on top plus a k with ˆ on top is maximum (The vectors form a right-handed system)
Reason: The volume of the parallel piped having three coterminous edges stack a with ‾ on top comma stack b with ‾ on top and stack c with ‾ on top

  1. Statement negative 1 is true, statement negative 2 is true; statement negative 2 is a correct explanation for statement negative 1    
  2. Statement negative 1 is true, statement negative 2 is true statement negative 2 is not a correct explanation for statement negative 1    
  3. Statement negative 1 is true, statement negative 2 is false    
  4. statement negative 1 is false, statement negative 2 is true    

hintHint:

We are given two statements. We have to tell which of the statement is true. For statement 1, we will check if for the given value of "a" the volume is maximum. We will use the formula of volume of parallelepiped. For statement 2, we have to check if parallelepiped have coterminous edges.

The correct answer is: Statement negative 1 is true, statement negative 2 is true; statement negative 2 is a correct explanation for statement negative 1


    We are given two statements we have to check which of them are true.
    Statment 1:
    The first statement is about the maximum volume of parallelepiped. We will find the maximum volume of parallelepiped.
    T h e space g i v e n space v e c t o r s space a r e
A with rightwards arrow on top space equals space i with hat on top space plus space a j with hat on top
B space equals a i with hat on top space plus space j with hat on top space plus k with hat on top
C equals j with hat on top space plus space a k with hat on top
    T h e space v o l u m e space o f space a space p a r a l l e l space p i p e d space i s space g i v e n space b y space a space s c a l a r space p r o d u c t.
V space equals left square bracket stack A space with rightwards arrow on top space B with rightwards arrow on top space stack C right square bracket with rightwards arrow on top
space space space space equals open vertical bar table row 1 a 0 row a 1 1 row 0 1 a end table close vertical bar
space space space space space equals 1 left parenthesis a space minus space 1 right parenthesis space minus a left parenthesis a squared space minus space 0 right parenthesis space plus space 0
V space space equals a space minus space 1 space minus a cubed
    We  will find the maximum value of parallelepiped.
    V space equals space a space minus space 1 space minus space a cubed
fraction numerator d V over denominator d a end fraction space equals space 1 space minus space 0 space minus space 3 a squared
W e space h a v e space m a x i m a space w h e n space d e r i v a t e space o f space t h e space f u n c t i o n
i s space z e r o.
W e space w i l l space s u b s t i t u t e space t h e space v a l u e space o f space d e r i v a t i v e space t o space b e space z e r o
t o space f i n d space t h e space v a l u e space o f space a space f o r space w h i c h space w e space a r e space g e t t i n g space z e r o.
0 space equals space 1 space minus space 3 a squared
1 space equals space 3 a squared
3 a squared space equals space 1
a squared equals space 1 third
T a k i n g space s q u a r e space r o o t s space w e space g e t comma
a space equals space fraction numerator 1 over denominator square root of 3 end fraction
    So, the first statement is true.
    Statement 2:
    We are getting the maximum value of parallelepiped at this value of a.
    Now, we will see statment 2.
    Coterminous edges means the vectors or sides end at single common point. So, yes parallelepiped has coterminous edges.
    Three edges meet at one point.
    As the edges end at one point, we can consider two of them as base. We can take their cross product and take them as base. And, the projection of the third vector with the cross product will give is the height.
    Hence the statement 2 is true and a correct explanation of statement 1.

    For such questions, we should know the formula of a parallelepiped. We should also know how to take a scalar product.

    Related Questions to study

    General
    Maths-

    Assertion (A): Let a with rightwards arrow on top equals 3 i with ˆ on top minus j with ˆ on top comma b with rightwards arrow on top equals 2 i with ˆ on top plus j with ˆ on top minus 3 k with ˆ on top. If stack b with rightwards arrow on top equals stack b with rightwards arrow on top subscript 1 end subscript plus stack b with rightwards arrow on top subscript 2 end subscript such that stack b with rightwards arrow on top subscript 1 end subscript is collinear with stack a with rightwards arrow on top and stack b with rightwards arrow on top subscript 2 end subscript is perpendicular to stack a with rightwards arrow on top is possible, then b with rightwards arrow on top subscript 2 equals i with ˆ on top plus 3 j with ˆ on top minus 3 k with ˆ on top.
    Reason (R): If stack a with rightwards arrow on top and stack b with rightwards arrow on top are non-zero, non-collinear vectors, then stack b with rightwards arrow on top can be expressed as stack b with rightwards arrow on top equals stack b with rightwards arrow on top subscript 1 end subscript plus stack b with rightwards arrow on top subscript 2 end subscript where stack b with rightwards arrow on top subscript 1 end subscript is collinear with stack a with rightwards arrow on top and stack b with rightwards arrow on top subscript 2 end subscript is perpendicular to stack a with rightwards arrow on top

    Assertion (A): Let a with rightwards arrow on top equals 3 i with ˆ on top minus j with ˆ on top comma b with rightwards arrow on top equals 2 i with ˆ on top plus j with ˆ on top minus 3 k with ˆ on top. If stack b with rightwards arrow on top equals stack b with rightwards arrow on top subscript 1 end subscript plus stack b with rightwards arrow on top subscript 2 end subscript such that stack b with rightwards arrow on top subscript 1 end subscript is collinear with stack a with rightwards arrow on top and stack b with rightwards arrow on top subscript 2 end subscript is perpendicular to stack a with rightwards arrow on top is possible, then b with rightwards arrow on top subscript 2 equals i with ˆ on top plus 3 j with ˆ on top minus 3 k with ˆ on top.
    Reason (R): If stack a with rightwards arrow on top and stack b with rightwards arrow on top are non-zero, non-collinear vectors, then stack b with rightwards arrow on top can be expressed as stack b with rightwards arrow on top equals stack b with rightwards arrow on top subscript 1 end subscript plus stack b with rightwards arrow on top subscript 2 end subscript where stack b with rightwards arrow on top subscript 1 end subscript is collinear with stack a with rightwards arrow on top and stack b with rightwards arrow on top subscript 2 end subscript is perpendicular to stack a with rightwards arrow on top

    Maths-General
    General
    Maths-

    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.

    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.

    Maths-General
    General
    Maths-

    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

    Maths-General
    parallel
    General
    Maths-

    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

    Maths-General
    General
    Maths-

    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
    Statement- 2 open parentheses S subscript 2 end subscript close parentheses:The condition for coplanarity of four points A left parenthesis stack a with ‾ on top right parenthesis comma B left parenthesis stack b with ‾ on top right parenthesis comma C left parenthesis stack c with ‾ on top right parenthesis comma D left parenthesis stack d with ‾ on top right parenthesis is that there exists scalars 1 comma m comma n comma p not all zeros such that  l a with ‾ on top plus m b with ‾ on top plus n c with ‾ on top plus p d with ‾ on top equals 0 with minus on top where l plus m plus n plus p equals 0.

    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
    Statement- 2 open parentheses S subscript 2 end subscript close parentheses:The condition for coplanarity of four points A left parenthesis stack a with ‾ on top right parenthesis comma B left parenthesis stack b with ‾ on top right parenthesis comma C left parenthesis stack c with ‾ on top right parenthesis comma D left parenthesis stack d with ‾ on top right parenthesis is that there exists scalars 1 comma m comma n comma p not all zeros such that  l a with ‾ on top plus m b with ‾ on top plus n c with ‾ on top plus p d with ‾ on top equals 0 with minus on top where l plus m plus n plus p equals 0.

    Maths-General
    parallel
    General
    Maths-

    Assertion (A): The number of vectors of unit length and perpendicular to both the vectors. i with ˆ on top plus j with ˆ on top and j with ˆ on top plus k with ˆ on top is zero Reason
    (R): stack a with ‾ on top and stack b with ‾ on top are two non-zero and non-parallel vectors it is true that stack a with ‾ on top cross times stack b with ‾ on top is perpendicular to the plane containing stack a with ‾ on top and stack b with ‾ on top

    Assertion (A): The number of vectors of unit length and perpendicular to both the vectors. i with ˆ on top plus j with ˆ on top and j with ˆ on top plus k with ˆ on top is zero Reason
    (R): stack a with ‾ on top and stack b with ‾ on top are two non-zero and non-parallel vectors it is true that stack a with ‾ on top cross times stack b with ‾ on top is perpendicular to the plane containing stack a with ‾ on top and stack b with ‾ on top

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    The value of p for which the straight lines r with rightwards arrow on top equals left parenthesis 2 i with ˆ on top plus 9 j with ˆ on top plus 13 k with ˆ on top right parenthesis plus t left parenthesis i with ˆ on top plus 2 j with ˆ on top plus 3 k with ˆ on top right parenthesis and r with rightwards arrow on top equals left parenthesis negative 3 i with ˆ on top plus 7 j with ˆ on top plus p k with ˆ on top right parenthesis plus s left parenthesis negative i with ˆ on top plus 2 j with ˆ on top minus 3 k with ˆ on top right parenthesis are coplanar is

    For such questions, we should know the condition for two lines to be coplanar. We should also know about the scalar triple product.

    The value of p for which the straight lines r with rightwards arrow on top equals left parenthesis 2 i with ˆ on top plus 9 j with ˆ on top plus 13 k with ˆ on top right parenthesis plus t left parenthesis i with ˆ on top plus 2 j with ˆ on top plus 3 k with ˆ on top right parenthesis and r with rightwards arrow on top equals left parenthesis negative 3 i with ˆ on top plus 7 j with ˆ on top plus p k with ˆ on top right parenthesis plus s left parenthesis negative i with ˆ on top plus 2 j with ˆ on top minus 3 k with ˆ on top right parenthesis are coplanar is

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    The position vector of the centre of the circle vertical line r with rightwards arrow on top vertical line equals 5 comma r with rightwards arrow on top times left parenthesis i with ˆ on top plus j with ˆ on top plus k with ˆ on top right parenthesis equals 3 square root of 3

    The position vector of the centre of the circle vertical line r with rightwards arrow on top vertical line equals 5 comma r with rightwards arrow on top times left parenthesis i with ˆ on top plus j with ˆ on top plus k with ˆ on top right parenthesis equals 3 square root of 3

    Maths-General
    parallel
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    Let stack r with rightwards arrow on top equals left parenthesis stack a with rightwards arrow on top cross times stack b with rightwards arrow on top right parenthesis s i n invisible function application x plus left parenthesis stack b with rightwards arrow on top cross times stack c with rightwards arrow on top right parenthesis c o s invisible function application y plus 2 left parenthesis stack c with rightwards arrow on top cross times stack a with rightwards arrow on top right parenthesis where stack a with rightwards arrow on top stack c with rightwards arrow on top are three noncoplanar vectors. If stack r with rightwards arrow on top is perpendicular to stack a with rightwards arrow on top plus stack b with rightwards arrow on top plus stack v with rightwards arrow on top, then minimum value of x squared plus y squared

    Let stack r with rightwards arrow on top equals left parenthesis stack a with rightwards arrow on top cross times stack b with rightwards arrow on top right parenthesis s i n invisible function application x plus left parenthesis stack b with rightwards arrow on top cross times stack c with rightwards arrow on top right parenthesis c o s invisible function application y plus 2 left parenthesis stack c with rightwards arrow on top cross times stack a with rightwards arrow on top right parenthesis where stack a with rightwards arrow on top stack c with rightwards arrow on top are three noncoplanar vectors. If stack r with rightwards arrow on top is perpendicular to stack a with rightwards arrow on top plus stack b with rightwards arrow on top plus stack v with rightwards arrow on top, then minimum value of x squared plus y squared

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