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Maths-

General

Easy

Question

If the matrix $open square brackets table row 0 cell 2 beta end cell gamma row alpha beta cell negative gamma end cell row alpha cell negative beta end cell gamma end table close square brackets$ is orthogonal, then -

$α$ = ± $\frac{1}{\sqrt{2}}$
$β$ = ± $\frac{1}{\sqrt{6}}$
$γ$ = ± $\frac{1}{\sqrt{3}}$
all of these

The correct answer is: all of these

Let A = $open square brackets table row 0 cell 2 beta end cell gamma row alpha beta cell negative gamma end cell row alpha cell negative beta end cell gamma end table close square brackets$ , A' = $open square brackets table row 0 alpha alpha row cell 2 beta end cell beta cell negative beta end cell row gamma cell negative gamma end cell gamma end table close square brackets$
Since A is orthogonal, $therefore$ AA' = I
$rightwards double arrow open square brackets table row 0 cell 2 beta end cell gamma row alpha beta cell negative gamma end cell row alpha cell negative beta end cell gamma end table close square brackets open square brackets table row 0 alpha alpha row cell 2 beta end cell beta cell negative beta end cell row gamma cell negative gamma end cell gamma end table close square brackets$ = $open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets$
$rightwards double arrow open square brackets table row cell 4 beta squared plus gamma squared end cell cell 2 beta squared minus gamma squared end cell cell negative 2 beta squared plus gamma squared end cell row cell 2 beta squared minus gamma squared end cell cell alpha squared plus beta squared plus gamma squared end cell cell alpha squared minus beta squared minus gamma squared end cell row cell negative 2 beta squared plus gamma squared end cell cell alpha squared minus beta squared minus gamma squared end cell cell alpha squared plus beta squared plus gamma squared end cell end table close square brackets$
= $open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets$
Equation the corresponding elements, we have
$open table row cell 4 beta to the power of 2 end exponent plus gamma to the power of 2 end exponent equals 1 end cell row cell 2 beta to the power of 2 end exponent minus gamma to the power of 2 end exponent equals 0 end cell end table close curly brackets$ $rightwards double arrow$ $beta equals plus-or-minus fraction numerator 1 over denominator square root of 6 end fraction comma gamma equals plus-or-minus fraction numerator 1 over denominator square root of 3 end fraction$
$alpha to the power of 2 end exponent plus beta to the power of 2 end exponent plus gamma to the power of 2 end exponent equals 1 rightwards double arrow alpha to the power of 2 end exponent plus fraction numerator 1 over denominator 6 end fraction plus fraction numerator 1 over denominator 3 end fraction equals 1 rightwards double arrow a equals plus-or-minus fraction numerator 1 over denominator square root of 2 end fraction$
Hence (D) is correct answer.

Related Questions to study

For each real x such that $negative 1 less than x less than 1$ . Let A (x) denote the matrix (1 – x)^–1/2 $open parentheses table row 1 cell negative x end cell row cell negative x end cell 1 end table close parentheses$ and $A open parentheses x close parentheses A open parentheses y close parentheses equals k A open parentheses z close parentheses blank$ where $x comma y element of R comma negative 1 less than x comma y less than 1$ and z = $fraction numerator x plus y over denominator 1 plus x y end fraction$ then k is -

For each real x such that $negative 1 less than x less than 1$ . Let A (x) denote the matrix (1 – x)^–1/2 $open parentheses table row 1 cell negative x end cell row cell negative x end cell 1 end table close parentheses$ and $A open parentheses x close parentheses A open parentheses y close parentheses equals k A open parentheses z close parentheses blank$ where $x comma y element of R comma negative 1 less than x comma y less than 1$ and z = $fraction numerator x plus y over denominator 1 plus x y end fraction$ then k is -

If the system of equations x + 2y – 3z = 2, (k + 3)z = 3, (2k + 1)y + z = 2 is inconsistent, then value of k can be -

If the system of equations x + 2y – 3z = 2, (k + 3)z = 3, (2k + 1)y + z = 2 is inconsistent, then value of k can be -

If A is a 3 × 3 skew-symmetric matrix, then |A| is given by -

If A is a 3 × 3 skew-symmetric matrix, then |A| is given by -

parallel

The solubility of iodine in water is greatly increased by:

The solubility of iodine in water is greatly increased by:

chemistry-General

Let A = $open square brackets table row 2 3 row cell negative 1 end cell 5 end table close square brackets$ if A^–1 = xA + yI₂ then x and y are respectively -

Let A = $open square brackets table row 2 3 row cell negative 1 end cell 5 end table close square brackets$ if A^–1 = xA + yI₂ then x and y are respectively -

The value of a for which the system of equationsx + y + z = 0ax + (a + 1)y + (a + 2)z = 0 a³x + (a + 1)³y + (a + 2)³z = 0has a non-zero solution is -

The value of a for which the system of equationsx + y + z = 0ax + (a + 1)y + (a + 2)z = 0 a³x + (a + 1)³y + (a + 2)³z = 0has a non-zero solution is -

parallel

If A is non-singular matrix of order 3 × 3, then adj (adj A) is equal to -

If A is non-singular matrix of order 3 × 3, then adj (adj A) is equal to -

If A = $open square brackets table row a b c row x y z row p q r end table close square brackets$ , B = $open square brackets table row q cell negative b end cell y row cell negative p end cell a cell negative x end cell row r cell negative c end cell z end table close square brackets$ then -

If A = $open square brackets table row a b c row x y z row p q r end table close square brackets$ , B = $open square brackets table row q cell negative b end cell y row cell negative p end cell a cell negative x end cell row r cell negative c end cell z end table close square brackets$ then -

Let A, B, C be three square matrices of the same order, such that whenever AB = AC then B = C, if A is -

Let A, B, C be three square matrices of the same order, such that whenever AB = AC then B = C, if A is -

parallel

The inverse of a skew symmetric matrix (if it exists) is -

The inverse of a skew symmetric matrix (if it exists) is -

Let E( $alpha$ ) = $open square brackets table row cell cos to the power of 2 end exponent invisible function application alpha end cell cell cos invisible function application alpha sin invisible function application alpha end cell row cell cos invisible function application alpha sin invisible function application alpha end cell cell sin to the power of 2 end exponent invisible function application alpha end cell end table close square brackets$ . If $alpha$ and $beta$ differs by an odd multiple of $pi$ /2, then E( $alpha$ ) E( $beta$ ) is a -

Let E( $alpha$ ) = $open square brackets table row cell cos to the power of 2 end exponent invisible function application alpha end cell cell cos invisible function application alpha sin invisible function application alpha end cell row cell cos invisible function application alpha sin invisible function application alpha end cell cell sin to the power of 2 end exponent invisible function application alpha end cell end table close square brackets$ . If $alpha$ and $beta$ differs by an odd multiple of $pi$ /2, then E( $alpha$ ) E( $beta$ ) is a -

Let A and B be two 2 × 2 matrices. Consider the statements
i) AB = O $rightwards double arrow$ A = O or B = O
ii) AB = I₂ $rightwards double arrow$ A = B^–1
iii) (A + B)² = A² + 2AB + B²
Then

Let A and B be two 2 × 2 matrices. Consider the statements
i) AB = O $rightwards double arrow$ A = O or B = O
ii) AB = I₂ $rightwards double arrow$ A = B^–1
iii) (A + B)² = A² + 2AB + B²
Then

parallel

Statement-I :If $x equals cos space theta left parenthesis 1 plus sin space theta right parenthesis plus sin space theta left parenthesis 1 minus cos space theta right parenthesis$ and $y equals cos space theta left parenthesis 1 plus cos space theta right parenthesis minus sin space theta left parenthesis 1 minus sin space theta right parenthesis$ then $x plus y less or equal than 3$
Statement-II : $sum from straight r equals 1 to straight n minus 1 of cos squared space open parentheses fraction numerator straight r pi over denominator straight n end fraction close parentheses equals straight n over 2 minus 1$
Statement-III : $sin space open parentheses pi over n close parentheses plus sin space open parentheses fraction numerator 3 pi over denominator n end fraction close parentheses plus sin space open parentheses fraction numerator 5 pi over denominator n end fraction close parentheses plus midline horizontal ellipsis plus n text terms end text equals 0$

Statement-I :If $x equals cos space theta left parenthesis 1 plus sin space theta right parenthesis plus sin space theta left parenthesis 1 minus cos space theta right parenthesis$ and $y equals cos space theta left parenthesis 1 plus cos space theta right parenthesis minus sin space theta left parenthesis 1 minus sin space theta right parenthesis$ then $x plus y less or equal than 3$
Statement-II : $sum from straight r equals 1 to straight n minus 1 of cos squared space open parentheses fraction numerator straight r pi over denominator straight n end fraction close parentheses equals straight n over 2 minus 1$
Statement-III : $sin space open parentheses pi over n close parentheses plus sin space open parentheses fraction numerator 3 pi over denominator n end fraction close parentheses plus sin space open parentheses fraction numerator 5 pi over denominator n end fraction close parentheses plus midline horizontal ellipsis plus n text terms end text equals 0$

$C l subscript 2 space i s space u s e d space i n space t h e space e x t r a c t i o n space o f colon$

$C l subscript 2 space i s space u s e d space i n space t h e space e x t r a c t i o n space o f colon$

chemistry-General

Barium adipate $not stretchy ⟶ with text Dry distillation end text on top left parenthesis straight A right parenthesis not stretchy ⟶ with MeCO subscript 3 straight H on top left parenthesis straight B right parenthesis$ The compound (A) and (B) , respectively, are:

Barium adipate $not stretchy ⟶ with text Dry distillation end text on top left parenthesis straight A right parenthesis not stretchy ⟶ with MeCO subscript 3 straight H on top left parenthesis straight B right parenthesis$ The compound (A) and (B) , respectively, are:

chemistry-General

parallel

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