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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Choose the correct answer from the given four options in each of the Exercises 19 to 33:

The vector in the direction of the vector î − 2? + 2k? that has magnitude 9 is:

(A)  î − 2ĵ + 2k̂

(B) ( i ^ −   2 j ^   +   2 k ^ ) 3  

(C) 9(î − 2ĵ + 2k̂)

(D) (î − 2ĵ + 2k̂)/3
0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

This is a Objective type Questions as classified in NCERT Exemplar

Sol: 

  L e t     a → = i ^ − 2 j ^ + 2 k ^ U n i t     v e c t o r     i n     t h e     d i r e c t i o n     o f     a → = a → | a → |                             = i ^ − 2 j ^ + 2 k ^ ( 1 ) 2 + ( − 2 ) 2 + ( 2 ) 2 = i ^ − 2 j ^ + 2 k ^ 1 + 4 + 4 = i ^ − 2 j ^ + 2 k ^ 3 ∴         v e c t o r     o f     m a g n i t u d e     9 = 9 ( i ^ − 2 j ^ + 2 k ^ ) 3 = 3 ( i ^ − 2 j ^ + 2 k ^ ) H e n c e ,     t h e     c o r r e c t     o p t i o n     i s     ( c ) .

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Find the angle between the lines r → = ( 3 i ^ + 2 j ^ + 6 k ^ ) + λ ( 2 i ^ + j ^ + 6 k ^ )   and r → = ( 2 i ^ + 5 k ^ ) + μ ( 6 i ^ + 3 j ^ + 2 k ^ ) .

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

H e r e ,                   b 1 → = 2 i ^ + j ^ + 2 k ^     a n d     b 2 → = 6 i ^ + 3 j ^ + 2 k ^ ∴                   c o s θ = b 1 → b 2 → | b 1 → | | b 2 → | = ( 2 i ^ + j ^ + 2 k ^ ) . ( 6 i ^ + 3 j ^ + 2 k ^ ) ( 2 ) 2 + ( 1 ) 2 + ( 2 ) 2 . ( 6 ) 2 + ( 3 ) 2 + ( 2 ) 2                                               = 1 2 + 3 + 4 4 + 1 + 4 3 6 + 9 + 4 = 1 9 9 4 9 = 1 9 3 . 7 = 1 9 2 1 ∴                                   θ = c o s − 1 ( 1 9 2 1 ) H e n c e ,     t h e     r e q u i r e d     a n g l e     i s     c o s − 1 ( 1 9 2 1 ) .

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Show that the lines x − 1 2 = y − 2 3 = z − 3 4   and x − 4 5 = y − 1 2 = z   intersect. Also, find their point of intersection.
0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

T h e     g i v e n     e q u a t i o n s     a r e         x − 1 2 = y − 2 3 = z − 3 4     a n d     x − 4 5 = y − 1 2 = z L e t                               x − 1 2 = y − 2 3 = z − 3 4 = λ ∴     x = 2 λ + 1 ,   y = 3 λ + 2     a n d     z = 4 λ + 3 a n d                             x − 4 5 = y − 1 2 = z 1 = μ ∴     x = 5 μ + 4 ,   y = 2 μ + 1     a n d     z = μ If  the  two  lines  intersect  each  other  at  one  point, t h e n         2 λ + 1 = 5 μ + 4                 ⇒ 2 λ − 5 μ = 3                                                                                     … ( i )                           3 λ + 2 = 2 μ + 1                 ⇒ 3 λ − 2 μ = − 1                                                                               … ( i i ) a n d           4 λ + 3 = μ                                     ⇒ 4 λ − μ = − 3                                                                                 … ( i i i ) S o l v i n g     e q n . ( i )     a n d ( i i )     w e     g e t                                           2 λ − 5 μ = 3                                       ( M u l t i p l y     b y     3 )                                         3 λ − 2 μ = − 1                                   ( M u l t i p l y     b y     2 ) ⇒                   6 λ           − 1 5 μ =                 9                             6 λ               − 4 μ =             − 2                   ( − )                   ( + )                             ( + )       _                                               − 1 1 μ         =         1 1                                   ∴ μ = − 1 P u t t i n g     t h e     v a l u e     o f     μ     i n     e q n . ( i )     w e     g e t ,                                               2 λ − 5 ( − 1 ) = 3 ⇒                                                         2 λ + 5 = 3 ⇒                                                                       2 λ = − 2                     ∴ λ = − 1 N o w ,     p u t t i n g     t h e     v a l u e     o f     λ     μ     i n     e q n . ( i i i )     t h e n ,                                                       4 ( − 1 ) − ( − 1 ) = − 3                                                                                 − 4 + 1 = − 3                         − 3 = − 3 ( s a t i s f i e d ) ∴  Coordinates  of  the  point  of  intersection  are                                                     x = 5 ( − 1 ) + 4 = − 5 + 4 = − 1                                                     y = 2 ( − 1 ) + 1 = − 2 + 1 = − 1                                                       z = − 1 Hence,  the  given  lines  intersect  each  other  at  (−1,−1,−1).

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If a →   = î + ĵ + k̂ and b̂ = ĵ − k̂, find a vector c such that a →   *   c →   =   b →   and a →   .   c →   = 3.
0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer type Questions as classified in NCERT Exemplar

Sol:

  L e t     c → = c 1 i ^ + c 2 j ^ + c 3 k ^ A l s o     g i v e n     t h a t         a → = i ^ + j ^ + k ^     a n d     b → = j ^ − k ^ S i n c e ,                     a → * c → = b → ∴             | i ^ j ^ k ^ 1 1 1 c 1 c 2 c 3 | = j ^ − k ^         i ^ ( c 3 − c 2 ) − j ^ ( c 3 − c 1 ) + k ^ ( c 2 − c 1 ) = j ^ − k ^ O n     c o m p a r i n g     t h e     l i k e     t e r m s ,     w e     g e t                                     c 3 − c 2 = 0                                                                                                                               … ( i )                                     c 1 − c 3 = 1                                                                                                                                 … ( i i )                                     c 2 − c 1 = − 1                                                                                                                         … ( i i i ) N o w                     f o r     a → . c → = 3               ( i ^ + j ^ + k ^ ) . ( c 1 i ^ + c 2 j ^ + c 3 k ^ ) = 3 ∴                                                                                 c 1 + c 2 + c 3 = 3                                                                   … ( i v ) A d d i n g     e q . ( i i )     a n d     e q . ( i i i )     w e     g e t ,                                                                                                         c 2 − c 3 = 0                                                                   … ( v ) F r o m     ( i v )     a n d     ( v )     w e     g e t                                                                                                       c 1 + 2 c 2 = 3                                                                   … ( v i ) F r o m     ( i i i )     a n d     ( v i )     w e     g e t                                           c 1 + 2 c 2 = 3                                           − c 1 + c 2 = − 1 _ A d d i n g                               3 c 2 = 2 ∴                                                               c 2 = 2 3                                                 c 3 − c 2 = 0               ⇒ c 3 − 2 3 = 0             ∴ c 3 = 2 3 N o w ,                           c 2 − c 1 = − 1           ⇒ 2 3 − c 1 = − 1 ⇒                           c 1 = 1 + 2 3 = 5 3 ∴                                                   c → = 5 3 i ^ + 2 3 j ^ + 2 3 k ^ = 1 3 ( 5 i ^ + 2 j ^ + 2 k ^ ) H e n c e ,     c → = 1 3 ( 5 i ^ + 2 j ^ + 2 k ^ ) .

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Show that the area of the parallelogram whose diagonals are given by a →   and b →   is 1 2 ∣ a →   *   b →   2 ∣ Also, find the area of the parallelogram whose diagonals are 2î − ĵ + k̂ and  î + 3ĵ − k̂.
0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer type Questions as classified in NCERT Exemplar
Let  ABCD  be  a  parallelogram  such  that,                     A B → = p → ,     A D → = q → = B C → ∴     b y     l a w     o f     t r i a n g l e ,     w e     g e t                     A C → = a → = p → + q →                                                                     … ( i ) a n d     B D → = b → = − p → + q →                                                             … ( i i ) A d d i n g     e q . ( i )     a n d     ( i i )     w e     g e t ,               a → + b → = 2 q →               ⇒ q → = ( a → + b → 2 ) S u b t r a c t i n g     e q . ( i i )     f r o m     ( i )     w e     g e t ,               a → − b → = 2 p →               ⇒ p → = ( a → − b → 2 ) ∴         p → * q → = 1 4 ( a → + b → ) ( a → − b → ) = 1 4 ( a → * a → − a → * b → + b → * a → − b → * b → )                                     = 1 4 ( − a → * b → + b → * a → )                                 [ ? a → * a → = 0           b → * b → = 0 ]                                     = 1 4 ( a → * b → + b → * a → ) = 1 4 . 2 ( a → * b → ) = | a → * b → | 2 So,  the  area  of  parallelogram  ABCD=|p→*q→|=12|a→*b→| Now  area  of  parallelogram  whose  diagonals  are  2i^−j^+k^  and  i^+3j^−k^                                                   = 1 2 | ( 2 i ^ − j ^ + k ^ ) * ( i ^ + 3 j ^ − k ^ ) |                                                     = 1 2 | i ^ j ^ k ^ 2 − 1 1 1 3 − 1 |                                                     = 1 2 [ i ^ ( 1 − 3 ) − j ^ ( − 2 − 1 ) + k ^ ( 6 + 1 ) ] = 1 2 | − 2 i ^ + 3 j ^ + 7 k ^ |                                                     = 1 2 ( − 2 ) 2 + ( 3 ) 2 + ( 7 ) 2 = 1 2 4 + 9 + 4 9 = 1 2 6 2 s q . u n i t s . H e n c e ,     t h e     r e q u i r e d     a r e a     i s     1 2 6 2     s q . u n i t s .

 

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If    a →   ,   b →   ,   c → determine the vertices of a triangle, show that 1 2 (   b →   *     c → +   c →   *   a → + a →   *   b →   )   gives the a →   ,   b →   ,   c → vector area of the triangle. Hence deduce the condition that the three points are collinear. Also, find the unit vector normal to the plane of the triangle.
0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer type Questions as classified in NCERT Exemplar

Since,  a→,b→  and  c→  are  the  vertices  of  ΔABC ∴                   A B → = b → − a → ,     B C → = c → − b →     a n d     A c → = c → − a → ∴                 A r e a     o f     Δ A B C = 1 2 | A B → * A C → |                   = 1 2 | ( b → − a → ) * ( c → − b → ) | = 1 2 | b → * c → − b → * a → − a → * c → + a → * a → |                   = 1 2 | b → * c → + a → * b → + c → * a → |                                             [ ? a → * b → = − b → * a →           c → * a → = − a → * c →           a → * a → = 0 → ] F o r     t h r e e     v e c t o r s     a r e     c o l l i n e a r ,     a r e a     o f     Δ A B C = 0 ∴                     1 2 | b → * c → + a → * b → + c → * a → | = 0                                   | a → * b → + b → * c → + c → * a → | = 0 w h i c h     i s     t h e     c o n d i t i o n     o f     c o l l i n e a r i t y     o f     a → , b →     a n d     c → . L e t     n ^     b e     t h e     u n i t     v e c t o r     n o r m a l     t o     t h e     p l a n e     o f     t h e     Δ A B C ∴                                             n ^ = A B → * A C → | A B → * A C → | = a → * b → + b → * c → + c → * a → | a → * b → + b → * c → + c → * a → |

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Prove that in any triangle ABC, c o s   A = b 2 + c 2 − a 2 2 b c   where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively.
0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer type Questions as classified in NCERT Exemplar

Sol:

H e r e ,     i n     t h e     g i v e n     f i g u r e ,     t h e     c o m p o n e n t s     o f     c     a r e     c   c o s A     a n d     a n d     c   s i n A . ∴                                   C D → = b − c   c o s A I n     Δ B D C ,                                           a 2 = C D 2 + B D 2 ⇒                                 a 2 = ( b − c   c o s A ) 2 + ( c   s i n A ) 2 ⇒                                 a 2 = b 2 + c 2   c o s 2 A − 2 b c c o s A + c 2   s i n 2 A ⇒                                 a 2 = b 2 + c 2 ( c o s 2 A + s i n 2 A ) − 2 b c c o s A ⇒                                 a 2 = b 2 + c 2 − 2 b c c o s A ⇒ 2 b c c o s A = b 2 + c 2 − a 2 ∴                         c o s A = b 2 + c 2 − a 2 2 b c H e n c e     p r o v e d .

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Using vectors, prove that parallelograms on the same base and between the same parallels are equal in area.
0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a Short Answer type Questions as classified in NCERT Exemplar

  Let  ABCD  and  ABFE  be  two  parallelograms  on  the  same  base  AB  and  between  same p a r a l l e l     l i n e s     A B     a n d     D F . L e t                   A B → = a →         a n d         A D → = b → ∴        Area  of  parallelogram  ABCD=|a→*b→| Now  Area  of  parallelogram  ABFE=|AB→*AE→|                                                       = | a → * ( A D → + D E → ) | = | a → * ( b → + K a → ) |                                                       = | ( a → * b → ) + K ( a → * a → ) | = | a → * b → | + 0                     [ ? a → * a → = 0 ]                                                       = | a → * b → | H e n c e     p r o v e d .

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, –1, 4), C(4, 5, –1).
0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

This is a Short Answer type Questions as classified in NCERT Exemplar

Sol:

  G i v e n     t h a t     A ( 1 , 2 , 3 ) ,     B ( 2 , − 1 , 4 )     a n d     C ( 4 , 5 , − 1 )                   A B → = ( 2 − 1 ) i ^ + ( − 1 − 2 ) j ^ + ( 4 − 3 ) k ^                   A B → = i ^ − 3 j ^ + k ^                   A C → = ( 4 − 1 ) i ^ + ( 5 − 2 ) j ^ + ( − 1 − 3 ) k ^ = 3 i ^ + 3 j ^ − 4 k ^ A r e a     o f     Δ A B C = 1 2 | A B → * A C → | = 1 2 | i ^ j ^ k ^ 1 − 3 1 3 3 − 4 |                                                       = 1 2 [ i ^ ( 1 2 − 3 ) − j ^ ( − 4 − 3 ) + k ^ ( 3 + 9 ) ] = 1 2 | 9 i ^ + 7 j ^ + 1 2 k ^ |                                                     = 1 2 ( 9 ) 2 + ( 7 ) 2 + ( 1 2 ) 2 = 1 2 8 1 + 4 9 + 1 4 4 = 1 2 2 7 4 H e n c e ,     t h e     r e q u i r e d     a r e a     i s     2 7 4 2 .

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If A, B, C, D are the points with position vectors î + ĵ − k̂, 2î − 3ĵ + k̂, 2î − 3k̂, and 3î + 2ĵ − k̂, respectively, find the projection of A B ¯  along   C D . ¯

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

This is a Short Answer type Questions as classified in NCERT Exemplar

  H e r e ,     P o s i t i o n     v e c t o r     o f     A = i ^ + j ^ − k ^                           P o s i t i o n     v e c t o r     o f     B = 2 i ^ − j ^ + 3 k ^                           P o s i t i o n     v e c t o r     o f     C = 2 i ^ − 3 k ^                           P o s i t i o n     v e c t o r     o f     D = 3 i ^ − 2 j ^ + k ^                   A B → = P . V     o f     B − P . V     o f     A                                     = ( 2 i ^ − j ^ + 3 k ^ ) − ( i ^ + j ^ − k ^ ) = i ^ − 2 j ^ + 4 k ^                   C D → = P . V     o f     D − P . V     o f     C                                     = ( 3 i ^ − 2 j ^ + k ^ ) − ( 2 i ^ − 3 k ^ ) = i ^ − 2 j ^ + 4 k ^ Projection  of  AB→  on  CD→=AB¯.CD¯|CD¯|=(i^−2j^+4k^)(i^−2j^+4k^)(1)2+(−2)2+(4)2                                                                                                                 = 1 + 4 + 1 6 1 + 4 + 1 6 = 2 1 2 1 = 2 1 Hence,  the  required  projection=21.

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