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Maths-

General

Easy

Question

$stack a with minus on top equals stack i with minus on top plus stack j with minus on top plus stack k with minus on top comma stack b with minus on top equals 2 stack i with minus on top plus 3 stack j with minus on top plus 5 stack k with minus on top$ and $stack c with minus on top$ is a unit vector. The maximum value of the scalar triple product $left square bracket stack a b c with bar on top right square bracket$ is

14
$\frac{1}{\sqrt{14}}$
$\sqrt{14}$
$\frac{1}{14}$

The correct answer is: $square root of 14$

The maximum value of $left square bracket a b stack c with minus on top right square bracket$ _,when $stack a with minus on top$ and $stack b with minus on top$ are fixed and $stack c with minus on top$ is variable is obtained when $stack c with minus on top$ is in the direction of $stack a with minus on top cross times stack b with minus on top$
Now $stack a with minus on top cross times stack b with minus on top equals open vertical bar table row cell i blank stack j with minus on top blank stack k with minus on top end cell row cell 1 blank 1 blank 1 end cell row cell 2 blank 3 blank 5 end cell end table close vertical bar equals 2 stack i with minus on top minus 3 stack j with minus on top plus stack k with minus on top$
Maximum value of $stack a with minus on top cross times stack b with minus on top times stack c with minus on top equals vertical line stack a with minus on top cross times stack b with minus on top vertical line vertical line stack c with minus on top vertical line equals square root of 14$

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In metals the time of relaxation of electrons

In metals the time of relaxation of electrons

Physics-General

$P left parenthesis stack p with minus on top right parenthesis text end text$ and $Q left parenthesis stack q with minus on top right parenthesis$ are the position vectors of two fixed points and $R left parenthesis stack r with minus on top right parenthesis$ is the position vector of a variable point. If R moves such that $left parenthesis stack r with minus on top minus stack p with minus on top right parenthesis cross times left parenthesis stack r with minus on top minus stack q with minus on top right parenthesis$ =0 then the locus of R is

$P left parenthesis stack p with minus on top right parenthesis text end text$ and $Q left parenthesis stack q with minus on top right parenthesis$ are the position vectors of two fixed points and $R left parenthesis stack r with minus on top right parenthesis$ is the position vector of a variable point. If R moves such that $left parenthesis stack r with minus on top minus stack p with minus on top right parenthesis cross times left parenthesis stack r with minus on top minus stack q with minus on top right parenthesis$ =0 then the locus of R is

The functions $s i n to the power of 1 end exponent invisible function application x comma c o s to the power of 1 end exponent invisible function application x comma t a n to the power of 1 end exponent invisible function application x comma c o t to the power of 1 end exponent invisible function application x comma c o s e c to the power of 1 end exponent invisible function application x$ and $s e c to the power of 1 end exponent invisible function application x$ are called inverse circular or inverse trigonometric functions which are defined as follows
$s i n to the power of 1 end exponent invisible function application x text 1 end text x 1 pi divided by 2 s i n to the power of 1 end exponent invisible function application x pi divided by 2$
$c o s to the power of 1 end exponent cross times 1 cross times 10 c o s to the power of 1 end exponent cross times pi$
$t a n to the power of 1 end exponent invisible function application x cross times R pi divided by 2 less than t a n to the power of 1 end exponent invisible function application x less than pi divided by 2$
$c o s e c to the power of 1 end exponent invisible function application x vertical line x vertical line 1 pi divided by 2 c o s e c to the power of 1 end exponent invisible function application x 0$
$s e c to the power of 1 end exponent invisible function application x vertical line x vertical line 10 s e c to the power of 1 end exponent invisible function application x pi pi divided by 2$
$c o t to the power of 1 end exponent invisible function application x x R 0 less than c o t to the power of 1 end exponent invisible function application x less than pi$
Number of solutions of $t a n to the power of 1 end exponent invisible function application vertical line x vertical line c o s to the power of 1 end exponent invisible function application x equals 0$ is/are

The functions $s i n to the power of 1 end exponent invisible function application x comma c o s to the power of 1 end exponent invisible function application x comma t a n to the power of 1 end exponent invisible function application x comma c o t to the power of 1 end exponent invisible function application x comma c o s e c to the power of 1 end exponent invisible function application x$ and $s e c to the power of 1 end exponent invisible function application x$ are called inverse circular or inverse trigonometric functions which are defined as follows
$s i n to the power of 1 end exponent invisible function application x text 1 end text x 1 pi divided by 2 s i n to the power of 1 end exponent invisible function application x pi divided by 2$
$c o s to the power of 1 end exponent cross times 1 cross times 10 c o s to the power of 1 end exponent cross times pi$
$t a n to the power of 1 end exponent invisible function application x cross times R pi divided by 2 less than t a n to the power of 1 end exponent invisible function application x less than pi divided by 2$
$c o s e c to the power of 1 end exponent invisible function application x vertical line x vertical line 1 pi divided by 2 c o s e c to the power of 1 end exponent invisible function application x 0$
$s e c to the power of 1 end exponent invisible function application x vertical line x vertical line 10 s e c to the power of 1 end exponent invisible function application x pi pi divided by 2$
$c o t to the power of 1 end exponent invisible function application x x R 0 less than c o t to the power of 1 end exponent invisible function application x less than pi$
Number of solutions of $t a n to the power of 1 end exponent invisible function application vertical line x vertical line c o s to the power of 1 end exponent invisible function application x equals 0$ is/are

parallel

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Sum of all solutions is

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Sum of all solutions is

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