What types of vectors are discussed in Class 12 Three Dimensional Geometry?

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    Answered by

    Nishtha Datta

    5 months ago

    The Class 12 Three Dimensional Geometry covers several topics, one of them is type of vectors. Students can chek the below given bullet points for type of vectors;

    • Zero Vector: A vector with zero magnitude and no specific direction.

    • Unit Vector: A vector with a magnitude of 1, representing only direction.

    • Position Vector: A vector representing the position of a point in space from the origin.

    • Direction Vector: A vector indicating the direction of a line in space.

    • Collinear Vectors: All the vectors whch lies in one line, are called Non-Colinear Vectors

    • Non-Collinear Vectors: All the vectors whch are not in one line, are called Non-Coline
    ...more

Similar Questions for you

A
alok kumar singh

π 2 π 2 ( x 2 c o s x 1 + π 2 + 1 + s i n 2 x 1 + e ( s i n x ) 2 0 2 3 ) d x = π 4 ( π + α ) 2

0 π 2 { ( x 2 c o s x 1 + π x + 1 + s i n 2 x 1 + e ( s i n x ) 2 0 2 3 ) + ( x 2 c o s x 1 + π x + 1 + s i n 2 x 1 + e ( s i n x ) 2 0 2 3 ) } d x

= π 4 ( π + α ) 2

0 π 2 ( x 2 c o s x + 1 + s i n 2 x ) d x = π 4 ( π + α ) 2

0 π 2 x 2 c o s x d x + 0 π 2 ( 1 + s i n 2 x ) d x = π 4 ( π + α ) 2 ....(1)

Let I 1 = 0 π 2 ( 1 + s i n 2 x ) d x

I 1 = 0 π 2 1 d x + 0 π 2 ( 1 c o s 2 x 2 ) d x

I 1 = π 2 + 1 2 [ π 2 + 0 ]

I 1 = 3 π 4

Let I 2 = 0 π 2 x 2 c o s x d x

I 2 = [ x 2 ( s i n x ) 2 x c o s x d x ] 0 π 2

I 2 = [ x 2 ( s i n x ) 2 x s i n x ] 0 π 2

I 2 = [ x 2 s i n x 2 ( x ( c o s x ) + c o s x ) ] 0 π 2

I 2 = [ x 2 s i n x 2 ( x c o s x + s i n x ) ] 0 π 2

I 2 = ( π 2 4 2 )

Put l1 and l2 in (1)

π 2 4 2 + 3 π 4

π 2 4 + 3 π 4 2

π 4 ( π + 3 ) 2

α = 3

A
alok kumar singh

Given | a | = 1 , | b | = 4 , a b = 2

c = 2 ( a × b ) 3 b  

Dot product with  a on both sides

c a = 6 ... (1)

Dot product with  b  on both sides

b c = 4 8 ... (2)

c c = 4 | a × b | 2 + 9 | b | 2

| c | 2 = 4 [ | a | 2 | b | 2 ( a b ) 2 ] + 9 | b | 2

| c | 2 = 4 [ ( 1 ) ( 4 ) 2 ( 4 ) ] + 9 ( 1 6 )

| c | 2 = 4 [ 1 2 ] + 1 4 4

| c | 2 = 4 8 + 1 4 4

| c | 2 = 1 9 2

c o s θ = b c | b | | c |

c o s θ = 4 8 1 9 2 4

c o s θ = 4 8 8 3 4

c o s θ = 3 2 3

c o s θ = 3 2 θ = c o s 1 ( 3 2 )

 

A
alok kumar singh

(a – 1) × 2 + (b – 2) × 5 + (g – 3) × 1 = 0

2a + 5b + g – 15 = 0

Also, P lie on line

a + 1 = 2λ

b – 2 = 5λ

g – 4 = λ

2 (2λ – 1) + 5 (5λ + 2) + λ + 4 – 15 = 0

4λ + 25λ + λ – 2 + 10 + 4 – 15 = 0

30λ – 3 = 0

λ = 1 1 0  

a + b + g = (2λ – 1) + (5λ + 2) + (λ + 4)

= 8 λ + 5 = 8 1 0 + 5 = 5 . 8

A
alok kumar singh

Take x 1 2 = y 2 3 = z 3 4 = λ  

x = 2λ + 1, y = 3λ + 2, z = 4λ + 3

  A B  = (α − 2)  i ^ + (β − 3) j ^ + (γ − 4) k ^  

Now,

(α − 2)  2 + (β − 3) 3 + (γ − 4) 4 = 0

2α − 4 + 3β − 9 + 4γ −16 = 0

2α + 3β + 4γ = 29

V
Vishal Baghel

L 1 = x λ 1 = y 1 2 1 2 = z 1 2

S D = | 2 λ + 3 ( 2 λ + 1 2 ) + λ | 1 4 = | 5 λ + 3 2 | 1 4

5 λ + 3 2 = 7 2 5 λ = 5 λ = 1

| λ | = 1

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