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New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

R ( 1 0 + α 3 , 8 3 )

| m A Q | = | m A P |

| 4 5 α | = | 3 2 α |

⇒a = -7 not possible α = 2 3 7 . 7 α + 3 β = 2 3 + 8 = 3 1

 

New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

a = 2 i ^ + j ^ + 3 k ^

b = 3 i ^ + 3 j ^ + k ^ c = c 1 i ^ + c 2 j ^ + c 3 k ^

C o p l n a n a r | 2 1 3 3 3 1 c 1 c 2 c 3 | = 0

8 c 1 + 7 c 2 + 1 2 c 3 = 0 . . . . . . . . ( i ) a . c = 5 2 c 1 + c 2 + 3 c 3 = 5 . . . . . . . . ( i i ) b . c = 0 3 c 1 + 3 c 2 + c 3 = 0 . . . . . . . . ( i i i )

Solving (i), (ii), (iii)

C 1 = 1 0 1 2 2 , c 2 = 8 5 1 2 2 , c 3 = 2 2 5 1 2 2

1 2 2 ( c 1 + c 2 + c 3 ) = 1 5 0

New question posted

2 months ago

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New answer posted

2 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

e 4 x + 4 e 3 x 5 8 e 2 x + 4 e x + 1 = 0

( e 2 x + 1 e 2 x ) + 4 ( e x + 1 e x ) 5 8 = 0

( e x + 1 e x + 2 ) 2 = 6 4

e x = 6 ± 3 2 2 = 3 ± 2 2

New answer posted

2 months ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

A = {1, 2, 3, ….50}

R1 = (2, 1), (2, 2), (2, 4)…. (2, 32)

(3, 1) (3, 3) (3, 9) (3, 27)

(13, 1) (13, 13)   etc.

R2 = { (2, 1), (2, 2), (3, 1), (3, 3), (5, 1), (5, 5), (7, 1), (7, 7), (1, 1), (11, 11)…}

R 1 R 2  contains only 9 elements.

New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

kindly consider the following Image 

 

New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

t a n π 8 = 2 h 8 0 . . . . . . . . . ( i )

t a n θ = h 8 0 . . . . . . . . . . ( i i )

t a n π 8 = 2 t a n θ

t a n 2 θ = 3 2 2 4

New question posted

2 months ago

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New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

P : ( x + 3 y z 6 ) = λ ( 6 x + 5 y z 7 ) = 0

passes ( 2 , 3 , 1 2 )

( 2 + 9 1 2 6 ) = λ ( 1 2 + 1 5 1 2 7 ) = 0

λ = 1

| 1 3 a | 2 d 2 = ( 1 3 ) 2 ( 9 3 ) ( 1 3 ) 2 = 9 3

New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

( 5 x + 8 y + 1 3 z 2 9 ) + λ ( 8 x 7 y + z 2 0 ) = 0  

P1 passing (2, 1, 3)

(10 + 8 + 39 – 29) + λ ( 1 6 7 + 3 2 0 ) = 0  

2 8 8 λ = 0 λ = 7 / 2  

 2X – Y + Z – 6 = 0      ….(i)

For P2 passes (0, 1, 2)

( 8 + 2 6 2 9 ) + λ ( 7 + 2 2 0 ) = 0

5 2 5 λ = 0 λ = 1 5  

P 2 : x + y + 2 z 5 = 0 . . . . . . . . ( i i )  

Acute angle between the planes

  c o s θ = 2 1 + 2 4 + 1 + 1 1 + 1 + 4 = 1 2  

θ = π 3  

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