Vector Algebra

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New answer posted

a year ago

0 Follower 63 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image 

 

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

|a|=|b|=|c|=1
|a-b|²+|a-c|²=8
⇒|a|²+|b|²-2a.b+|a|²+|c|²-2a.c=8
⇒4-2 (a.b+a.c)=8
⇒a.b+a.c=-2
|a+2b|²+|a+2c|²
=|a|²+4|b|²+4a.b+|a|²+4|c|²+4a.c
=10+4 (a.b+a.c)
=10-8=2

New answer posted

a year ago

0 Follower 16 Views

R
Raj Pandey

Contributor-Level 9

p = 2i + 3j + k, q = I + 2j + k
r = αi + βj + γk is ⊥ to p+q and p-q
∴ r is collinear with (p+q) * (p-q) = -2 (p*q)
p*q = |i, j, k; 2,3,1; 1,2,1| = I - j + k
∴ r = λ (i - j + k)
|r| = √3 ⇒ λ = 1
∴ r = I - j + k = αi + βj + γk
⇒ α=1, β=-1, γ=1
|α|+|β|+|γ| = 3

New answer posted

a year ago

0 Follower 14 Views

A
alok kumar singh

Contributor-Level 10

f (x+y)=f (x)f (y). f (n)=f (1)?
Σf (x)=f (1)/ (1-f (1)=2 ⇒ f (1)=2/3.
f (4)/f (2) = (f (1)? )/ (f (1)²) = f (1)² = 4/9.

New answer posted

a year ago

0 Follower 32 Views

R
Raj Pandey

Contributor-Level 9

Vectors are coplanar. Determinant is zero. Row operations.
This leads to 2a=b+c.

New answer posted

a year ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

= 1 56 * 277.85 = 496

± 1 = 1 1 λ 1 1 3 2 1 1 ⇒ = - λ + 3 = ± 1 or ⇒ λ = 2

For λ = 4

λ = 4

c o s ? θ = 2 + 1 + 4 6 18 = 7 6 3

 

 

New answer posted

a year ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

= 128 3

3 + 2 ( a ? ⋅ b ? + b ? ⋅ c ? + c ? ⋅ a ? ) = 0

a ? ⋅ b ? + b ? ⋅ c ? + c ? ⋅ a ? = - 3 2 d ? = a ? * b ? + b ? * c ? + c ? * a ? d ? = 3 ( a ? * b ? )

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

Cross product will represent a vector perpendicular to both the vector.

Dir. Of Motion of P : = ( i ^ + j ^ ) * ( j ^ * k ^ )

= k ^ − j ^ + i ^

Dir. Of Motion of Q = ( i ^ + j ^ ) * ( − i ^ + j ^ )

= 2 k ^

c o s θ = a → . b → a b = 2 2 3 = 1 3 ⇒ θ = c o s − 1 ( 1 3 )

New answer posted

a year ago

0 Follower 20 Views

A
alok kumar singh

Contributor-Level 10

 a→⋅c^=α+6+21+4+4

103=8+α3⇒α=2

b→*c→=i^ (2β−8)+j^ (10)+k^ (6+β)

=−6i^+10j^+7k^

So = 1

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