Vector Algebra

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New answer posted

6 days ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

  a + 5 b  is collinear with c  

  a + 5 b = c           …(1)

b + 6 c is collinear with a  

⇒   b + 6 c = μ a               …(2)

From (1) and (2)

  b + 6 c = μ ( λ c 5 b )          

-> ( 1 + 5 μ ) b + ( 6 λ μ ) c = 0

? b and c  are non-collinear

-> 1 + 5m = 0 μ = 1 5  and 6 – lm = 0 Þ lm = 6

-> l = – 30

Now,

b = 6 c = 1 5 a

5 b + 3 0 c = a

a + 5 b + 3 0 c = 0 a + α b + β c = 0 ]

On comparing

α = 5, β = 30  α + β = 35

New answer posted

2 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

a = i ^ + 2 j ^ k ^ , b = i ^ j ^ , c = i ^ j ^ k ^

r * a = c * a

r = c + λ a

Now, 0 = b . c + λ a . b a s r . b = 0

λ = b . c a . b = 2

r . a = a . c + 2 a 2 = 1 2

New answer posted

2 weeks ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

a 1 = x i ^ j ^ + k ^ & a 2 = i ^ + y j ^ + z k ^           

given  a 1 & a 2 are collinear then a 1 = λ a 2  

( x i ^ j ^ + k ^ ) = λ ( i ^ + y j ^ + z k ^ )         

Since i ^ , j ^ & k ^ are not collinear so

  S o x i ^ + y j ^ + z k ^ = λ i ^ 1 λ j ^ + 1 λ k ^         

Hence possible unit vector parallel to it be  1 3 ( i ^ j ^ + k ^ ) for λ =

New answer posted

2 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Data contradiction.

a * ( b * c ) = ( a c ) b ( a b ) c

New answer posted

2 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Mid point of BC is 1 2 ( 5 i ^ + ( α 2 ) j ^ + 9 k ^ )

A B ¯ = i ^ + ( α 4 ) j ^ + k ^

A C ¯ = i ^ + ( 2 α ) j ^ + k ^

For = 1, A B ¯  and A C ¯  will be collinear. So for non collinearity

= 2

New answer posted

2 weeks ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Mid point of BC is 1 2 ( 5 i ^ + ( α 2 ) j ^ + 9 k ^ )  

A B ¯ = i ^ + ( α 4 ) j ^ + k ^  

A C ¯ = i ^ + ( 2 α ) j ^ + k ^

For a = 1,   A B ¯ and A C ¯  will be collinear. So for non collinearity

a = 2

New answer posted

2 weeks ago

0 Follower 3 Views

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 answer posted

2 weeks ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Data contradiction.

a * ( b * c ) = ( a c ) b ( a b ) c

New answer posted

2 weeks ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Given : a = ( α , 1 , 1 ) a n d b = ( 2 , 1 , α ) c = a * b = | i ^ j ^ k ^ α 1 1 2 1 α |  

= ( α + 1 ) i ^ + ( α 2 2 ) j ^ + ( α 2 ) k ^ Projection of c on d = i ^ + 2 j ^ 2 k ^ = | c . d | d | | = 3 0 { G i v e n }

| α 1 4 + 2 α 2 2 α + 4 1 + 4 + 4 | = 3 0  

On solving Þ α = 1 3 2  (Rejected as a > 0) and a = 7

New answer posted

3 weeks ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

x 2 ( y + z ) y 2 ( z + x ) z 2 ( x + y ) = a 3 b 3 c 3 = x 3 y 3 z 3

( x + y ) ( y + z ) ( z + x ) = x y z

x 2 ( y + z ) + y 2 ( z + y ) + z 2 ( x + y ) + x y z = 0 a 3 + b 3 + c 3 + a b c = 0

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