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

Find the sine of the angle between the vectors  a → = 3î + 2ĵ + k̂ and b → = −î + 2ĵ + 4k̂.
0 Follower 7 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 → = 3 i ^ + j ^ + 2 k ^         a n d         b → = 2 i ^ − 2 j ^ + 4 k ^ W e     k n o w     t h a t     | a → * b → | = | a → | | b → | s i n θ ∴           a → * b → = | i ^ j ^ k ^ 3 1 2 2 − 2 4 |                                       = i ^ ( 4 + 4 ) − j ^ ( 1 2 − 4 ) + k ^ ( − 6 − 2 )                                       = 8 i ^ − 8 j ^ − 8 k ^ ∴       | a → * b → | = ( 8 ) 2 + ( − 8 ) 2 + ( − 8 ) 2                                       = 6 4 + 6 4 + 6 4 = 1 9 2 = 6 4 * 3 = 8 3 ⇒                 | a → | = ( 3 ) 2 + ( 1 ) 2 + ( 2 ) 2 = 9 + 1 + 4 = 1 4 ⇒                 | b → | = ( 2 ) 2 + ( − 2 ) 2 + ( 4 ) 2 = 4 + 4 + 1 6 = 2 4 = 2 6 ∴ s i n θ = | a → * b → | | a → | | b → | = 8 3 1 4 . 2 6 = 4 3 8 4 = 4 3 2 2 1 = 2 7 H e n c e ,     s i n θ = 2 7 .

New answer posted

a year ago

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

If a → + b → + c → = 0, show that a → * b → = b → * c = c → * a → . Interpret the result geometrically.
0 Follower 2 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 → + b → + c → = 0 S o ,                                       a → * ( a → + b → + c → ) = a → * 0 ⇒                         a → * a → + a → * b → + a → * c → = 0 ⇒                                         0 → + a → * b → + a → * c → = 0                                     ( ? a → * a → = 0 ) ⇒                                                       a → * b → − c → * a → = 0                                     ( a → * c → = − c → * a → ) ⇒                                                                                     a → * b → = c → * a →                                                                               … ( i ) N o w ,                                                         a → + b → + c → = 0 ⇒                                               b → * ( a → + b → + c → ) = b → * 0 ⇒                             b → * a → + b → * b → + b → * c → = 0 ⇒                                           b → * a → + 0 → + b → * c → = 0                                       ( ? b → * b → = 0 ) ⇒                                         − ( a → * b → ) + b → * c → = 0 ⇒                                                                                       b → * c → = a → * b →                                                                               … ( i i ) F r o m     e q . ( i )     a n d     ( i i )     w e     g e t                         a → * b → = b → * c → = c → * a → .     H e n c e     p r o v e d . G e o m e t r i c a l     I n t e r p r e t a t i o n According  to  figure,  we  have  Area  of  parallelogram  ABCD  is ⇒                   | a → * b → | = | a → | | b → | s i n θ Since,  the  parallelograms  on  the  same  base  and  between  the  same  parallel  lines  are e q u a l     i n     a r e a . ∴                     | 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

Find the angle between the vectors 2î − ĵ + k̂ and 3î + 4ĵ − k̂.
0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

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

Sol:

  L e t     a → = 2 i ^ − j ^ + k ^     a n d     b → = 3 i ^ + 4 j ^ − k ^ a n d     l e t     θ     b e     t h e     a n g l e     b e t w e e n     a →     a n d     b → . ∴                         c o s θ = a → . b → | a → | | b → | = ( 2 i ^ − j ^ + k ^ ) ( 3 i ^ + 4 j ^ − k ^ ) 4 + 1 + 1 . 9 + 1 6 + 1                                                       = 6 − 4 − 1 6 2 6 = 1 2 3 . 1 3 = 1 2 3 9 ∴                                           θ = c o s − 1 1 2 3 9             ⇒ c o s − 1 ( 1 1 5 6 ) H e n c e ,     t h e     r e q u i r e d     v a l u e     o f     θ     i s     c o s − 1 ( 1 1 5 6 ) .

New answer posted

a year ago

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

Find a vector of magnitude 6, which is perpendicular to both the vectors 2î − 2ĵ + k̂ and 4î − 3ĵ + k̂.
0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

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

Sol:

L e t     a → = 2 i ^ − j ^ + 2 k ^     a n d     b → = 4 i ^ − j ^ + 3 k ^ W e     k n o w     t h a t     u n i t     v e c t o r     a r e     p e r p e n d i c u l a r     t o     a →     a n d     b → = ( a → * b → ) | a → * b → |                                 a → * b → = | i ^ j ^ k ^ 2 − 1 2 4 − 1 3 |                                                       = i ^ ( − 3 + 2 ) − j ^ ( 6 − 8 ) + k ^ ( − 2 + 4 ) = − i ^ + 2 j ^ + 2 k ^ ∴                         | a → * b → | = ( − 1 ) 2 + ( 2 ) 2 + ( 2 ) 2 = 1 + 4 + 4 = 9 = 3 S o ,           ( a → * b → ) | a → * b → | = − i ^ + 2 j ^ + 2 k ^ 3 = 1 3 ( − i ^ + 2 j ^ + 2 k ^ ) N o w     t h e     v e c t o r     o f     m a g n i t u d e     6 = 1 3 ( − i ^ + 2 j ^ + 2 k ^ ) . 6                                                         = 2 ( − i ^ + 2 j ^ + 2 k ^ ) = − 2 i ^ + 4 j ^ + 4 k ^ H e n c e ,     t h e     r e q u i r e d     v e c t o r     i s     − 2 i ^ + 4 j ^ + 4 k ^ .

New answer posted

a year ago

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

Find the vector equation of the line which is parallel to the vector 3 i ^ − 2 j ^ + 6 k ^ and which passes through the point (1, –2, 3).
0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

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

W e     k n o w     t h a t     t h e     e q u a t i o n     o f     l i n e     i s                                                           r → = a → + b → λ H e r e ,     a → = i ^ − 2 j ^ + 3 k ^     a n d     b → = 3 i ^ − 2 j ^ + 6 k ^ ∴ E q u a t i o n     o f     l i n e     i s     r → = ( i ^ − 2 j ^ + 3 k ^ ) + λ ( 3 i ^ − 2 j ^ + 6 k ^ ) ⇒ ( x i ^ + y j ^ + z k ^ ) = ( i ^ − 2 j ^ + 3 k ^ ) + λ ( 3 i ^ − 2 j ^ + 6 k ^ ) ⇒ ( x i ^ + y j ^ + z k ^ ) − ( i ^ − 2 j ^ + 3 k ^ ) = λ ( 3 i ^ − 2 j ^ + 6 k ^ ) ⇒ ( x − 1 ) i ^ + ( y + 2 ) j ^ + ( z − 3 ) k ^ = λ ( 3 i ^ − 2 j ^ + 6 k ^ ) H e n c e ,     t h e     r e q u i r e d     e q u a t i o n     i s     ( x − 1 ) i ^ + ( y + 2 ) j ^ + ( z − 3 ) k ^ = λ ( 3 i ^ − 2 j ^ + 6 k ^ ) .

New answer posted

a year ago

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

Find the position vector of a point A in space such that O A → is inclined at 60º to OX and at 45° to OY and | O A → | = 10 units.
0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

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

L e t     α = 6 0 0 ,   β = 4 5 0     a n d     t h e     a n g l e     i n c l i n e d     t o     O Z     a x i s     b e     γ . W e     k n o w     t h a t     c o s 2 α + c o s 2 β + c o s 2 γ = 1 ⇒                                       c o s 2 6 0 0 + c o s 2 4 5 0 + c o s 2 γ = 1 ⇒                                                       ( 1 2 ) 2 + ( 1 2 ) 2 + c o s 2 γ = 1               ⇒ 1 4 + 1 2 + c o s 2 γ = 1 ⇒                 3 4 + c o s 2 γ = 1                     ⇒ c o s 2 γ = 1 − 3 4 = 1 4 ∴                                 c o s γ = ± 1 2                     ⇒ c o s γ = 1 2 ( Rejecting  cosγ=−12,  since  γ<900 ) ∴                     O A → = | O A → | ( 1 2 i ^ + 1 2 j ^ + 1 2 k ^ ) = 1 0 ( 1 2 i ^ + 1 2 j ^ + 1 2 k ^ )                                           = 5 i ^ + 5 2 j ^ + 5 k ^ H e n c e ,     t h e     p o s i t i o n     v e c t o r     o f     A     i s     ( 5 i ^ + 5 2 j ^ + 5 k ^ ) .

New answer posted

a year ago

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

A vector   has magnitude 14 and direction ratios 2, 3, –6. Find the direction cosines and components of r?, given that  makes an acute angle with the x-axis.
0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

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

Sol:

L e t     a → ,     b →     a n d     c →     b e     t h r e e     v e c t o r s     s u c h     t h a t     a → = 2 k ,     b → = 3 k     a n d     c → = − 6 k I f     l ,     m     a n d     n     a r e     t h e     d i r e c t i o n s     c o s i n e s     o f     v e c t o r     r → ,     t h e n                                               l = a → | r → | = 2 k 1 4 = k 7                                             m = b → | r → | = 3 k 1 4             a n d           n = c → | r → | = − 6 k 1 4 = − 3 k 7 W e     k n o w     t h a t     l 2 + m 2 + n 2 = 1         ⇒ k 2 4 9 + 9 k 2 1 9 6 + 9 k 2 4 9 = 1 ⇒                   4 k 2 + 9 k 2 + 3 6 k 2 1 9 6 = 1                   ⇒ 4 9 k 2 = 1 9 6         ⇒ k 2 = 4 ∴                   k = ± 2                 a n d               l = k 7 = 2 7                       m = 3 k 1 4 = 3 * 2 1 4 = 3 7         a n d         n = − 3 k 7 = − 3 * 2 7 = − 6 7 ∴                     r ^ = ± ( 2 7 i ^ + 3 7 j ^ − 6 7 k ^ ) W e     k n o w     t h a t     r → = ( r ^ ) | r → |                                                                       = ± ( 2 7 i ^ + 3 7 j ^ − 6 7 k ^ ) . 1 4 = ± ( 4 i ^ + 6 j ^ − 1 2 k ^ ) H e n c e ,     t h e     r e q u i r e d     d i r e c t i o n     c o s i n e s     a r e 2 7 , 3 7 , − 6 7     a n d     t h e     c o m p o n e n t s     o f     r →     a r e     4 i ^ , 6 j ^     a n d     − 1 2 k ^ .

New answer posted

a year ago

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

A vector r → . is inclined at equal angles to the three axes. If the magnitude of r . is 2 3 units, find r → .
0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

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

Sol:

Since,  the  vector  r→  makes  equal  angles  w ith  the  axes,  their  direction  cosines  should  be  same. ∴                                                                     l = m = n W e     k n o w     t h a t     l 2 + m 2 + n 2 = 1         ⇒ l 2 + l 2 + l 2 = 1 ⇒                   3 l 2 = 1             ⇒ l 2 = 1 3                               ⇒ l = ± 1 3 ∴     r ^ = ± 1 3 i ^ ± 1 3 j ^ ± 1 3 k ^                       ⇒ r ^ = ± 1 3 ( i ^ + j ^ + k ^ ) W e     k n o w     t h a t     r → = ( r ^ ) | r → |                                                                       = ± 1 3 ( i ^ + j ^ + k ^ ) 2 3 = ± 2 ( i ^ + j ^ + k ^ ) H e n c e ,     t h e     r e q u i r e d     v a l u e     o f     r →     i s     ± 2 ( i ^ + j ^ + k ^ ) .

New question posted

a year ago

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

A vector r → . is inclined at equal angles to the three axes. If the magnitude of r . is 2 3 units, find r → .
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New answer posted

a year ago

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

Using vectors, find the value of k such that the points (k, –10, 3), (1, –1, 3), and (3, 5, 3) are collinear.
0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

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

Sol:

Let  the  given  points  are  A(k,−10,3)  ,B(1,−1,3)and  C(3,5,3)                           A B → = ( 1 − k ) i ^ + ( − 1 + 1 0 ) j ^ + ( 3 − 3 ) k ^ = ( 1 − k ) i ^ + 9 j ^ + 0 k ^ ∴                   | A B → | = ( 1 − k ) 2 + ( 9 ) 2 = ( 1 − k ) 2 + 8 1                             B C → = ( 3 − 1 ) i ^ + ( 5 + 1 ) j ^ + ( 3 − 3 ) k ^ = 2 i ^ + 6 j ^ + 0 k ^ ∴                     | B C → | = ( 2 ) 2 + ( 6 ) 2 = 4 + 3 6 = 4 0 = 2 1 0                               A C → = ( 3 − k ) i ^ + ( 5 + 1 0 ) j ^ + ( 3 − 3 ) k ^ = ( 3 − k ) i ^ + 6 j ^ + 0 k ^ ∴                       | A C → | = ( 3 − k ) 2 + ( 1 5 ) 2 = ( 3 − k ) 2 + 2 2 5 I f     A ,     B ,     a n d     C     a r e     c o l l i n e a r ,     t h e n                                                                   | A B → | + | B C → | = | A C → |                         ( 1 − k ) 2 + 8 1 + 2 1 0 = ( 3 − k ) 2 + 2 2 5 S q u a r i n g     b o t h     s i d e s ,     w e     h a v e                                                                                   [ ( 1 − k ) 2 + 8 1 + 4 0 ] 2 = [ ( 3 − k ) 2 + 2 2 5 ] 2 ⇒               ( 1 − k ) 2 + 8 1 + 4 0 + 2 4 0 ( 1 − k ) 2 + 8 1 = ( 3 − k ) 2 + 2 2 5 ⇒ 1 + k 2 − 2 k + 1 2 1 + 2 4 0 1 + k 2 − 2 k + 8 1 = 9 + k 2 − 6 k + 2 2 5 ⇒                                                 1 2 2 − 2 k + 2 4 0 k 2 − 2 k + 8 2 = 2 3 4 − 6 k D i v i d i n g     b y     2 ,     w e     g e t ⇒                                                 6 1 − k + 4 0 k 2 − 2 k + 8 2 = 1 1 7 − 3 k ⇒                                                                                     4 0 k 2 − 2 k + 8 2 = 1 1 7 − 6 1 − 3 k + k ⇒                                                                                     4 0 k 2 − 2 k + 8 2 = 5 6 − 2 k ⇒                                                                                     2 1 0 k 2 − 2 k + 8 2 = 5 6 − 2 k ⇒                                                                                           1 0 k 2 − 2 k + 8 2 = 2 8 − k                   ( D i v i d i n g     b y     2 ) S q u a r i n g     b o t h     s i d e s ,     w e     g e t ⇒                                                                                                 1 0 ( k 2 − 2 k + 8 2 ) = 7 8 4 + k 2 − 5 6 k ⇒                                                                                                 1 0 k 2 − 2 0 k + 8 2 0 = 7 8 4 + k 2 − 5 6 k ⇒                           1 0 k 2 − k 2 − 2 0 k + 5 6 k + 8 2 0 − 7 8 4 = 0 ⇒                       9 k 2 + 3 6 k + 3 6 = 0             ⇒ k 2 + 4 k + 4 = 0 ⇒                       ( k + 2 ) 2 = 0         ⇒ k + 2 = 0           ⇒ k = − 2 H e n c e ,     t h e     r e q u i r e d     v a l u e     i s     k = − 2 .

 

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