English

If a, b, c are the lengths of sides, BC, CA and AB of a triangle ABC, prove that → B C + → C A + → A B = → 0 and deduce that a sin A = b sin B = c sin C . - Mathematics

Advertisements
Advertisements

Question

If abc are the lengths of sides, BCCA and AB of a triangle ABC, prove that \[\vec{BC} + \vec{CA} + \vec{AB} = \vec{0}\]  and deduce that \[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} .\]

 
 
Sum
Advertisements

Solution

\[\text{ We have } \]
\[ \vec{BC} = \vec{a} \]
\[ \vec{CA} = \vec{b} \]
\[ \vec{AB} = \vec{c} \]
\[\left| \vec{a} \right|=a\]
\[\left| \vec{b} \right| =b ( \because \text{ Length is always positive} )\]
\[ \vec{c} =c \]
\[ \text{ Now } , \]
\[ \vec{BC} + \vec{CA} + \vec{AB} = \vec{0} ( \text{ Given } )\]
\[ \Rightarrow \vec{a} + \vec{b} + \vec{c} = \vec{0} \]
\[ \Rightarrow \vec{a} \times \left( \vec{a} + \vec{b} + \vec{c} \right) = \vec{a} \times \vec{0} \]
\[ \Rightarrow \vec{a} \times \vec{a} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = \vec{0} \]
\[ \Rightarrow \vec{0} + \vec{a} \times \vec{b} - \vec{c} \times \vec{a} = \vec{0} \]
\[ \Rightarrow \vec{a} \times \vec{b} = \vec{c} \times \vec{a} \]
\[ \Rightarrow \left| \vec{a} \right| \left| \vec{b} \right|\sin C = \left| \vec{c} \right| \left| \vec{a} \right| \sin B\]
\[ \Rightarrow ab \sin C = ca \sin B\]
\[\text{ Dividing both sides by abc,we get } \]
\[ \Rightarrow \frac{\sin C}{c} = \frac{\sin B}{b} . . . (1)\]
\[\text{ Again } ,\]
\[ \vec{BC} + \vec{CA} + \vec{AB} = \vec{0} \]
\[ \Rightarrow \vec{a} + \vec{b} + \vec{c} = \vec{0} \]
\[ \Rightarrow \vec{b} \times \left( \vec{a} + \vec{b} + \vec{c} \right) = \vec{b} \times \vec{0} \]
\[ \Rightarrow \vec{b} \times \vec{a} + \vec{b} \times \vec{b} + \vec{b} \times \vec{c} = \vec{0} \]
\[ \Rightarrow - \vec{a} \times \vec{b} + \vec{0} + \vec{b} \times \vec{c} = \vec{0} \]
\[ \Rightarrow \vec{a} \times \vec{b} = \vec{b} \times \vec{c} \]
\[ \Rightarrow \left| \vec{a} \right| \left| \vec{b} \right| \sin C = \left| \vec{b} \right| \left| \vec{c} \right| \sin A\]
\[ \Rightarrow ab \sin C = bc \sin A\]
\[\text{ Dividing both sides by abc,we get } \]
\[ \Rightarrow \frac{\sin C}{c} = \frac{\sin A}{a} . . . (2)\]
\[\text{ From (1) and (2), we get } \]
\[\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\]
\[ \Rightarrow \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 25: Vector or Cross Product - Exercise 25.1 [Page 30]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 25 Vector or Cross Product
Exercise 25.1 | Q 20 | Page 30

RELATED QUESTIONS

Find `|veca × vecb|`, if `veca = hati - 7hatj + 7hatk` and `vecb = 3hati - 2hatj + 2hatk`.


Show that `(veca - vecb) xx (veca + vecb) = 2(veca xx vecb)`.


Given that `veca.vecb = 0` and `veca xx vecb = 0` What can you conclude about the vectors `veca and vecb`?


Let the vectors `veca` and `vecb` be such that `|veca| = 3` and `|vecb| = sqrt2/3`, then `veca xx vecb` is a unit vector, if the angle between `veca` and `vecb` is ______.


If θ is the angle between two vectors `hati - 2hatj + 3hatk and 3hati - 2hatj + hatk` find `sin theta`


Find the area of the parallelogram determined by the vector \[\hat{ i }  - 3 \hat{ j } + \hat{ k }  \text{ and }  \hat{ i }  + \hat{ j } + \hat{ k }  .\]

 


Find the area of the parallelogram whose diagonals are \[2 \hat{ i }  + 3 \hat{ j } + 6 \hat{ k } \text{ and }  3 \hat{ i }  - 6 \hat{ j }  + 2 \hat{ k } \]

 


Find the angle between two vectors \[\vec{a} \text{ and }  \vec{b}\] , if \[\left| \vec{a} \times \vec{b} \right| = \vec{a} \cdot \vec{b} .\]

 

if \[\vec{a} \times \vec{b} = \vec{b} \times \vec{c} \neq 0,\]  then  show that \[\vec{a} + \vec{c} = m \vec{b} ,\]  where m is any scalar.

 
 

 


What inference can you draw if \[\vec{a} \times \vec{b} = \vec{0} \text{ and }  \vec{a} \cdot \vec{b} = 0 .\]

 

Let \[\vec{a} = \hat{ i } + 4 \hat{ j }  + 2 \hat{ k } , \vec{b} = 3 \hat{ i }- 2 \hat{ j } + 7 \hat{ k }  \text{ and } \vec{c} = 2 \hat{ i } - \hat{ j }  + 4 \hat{ k }  .\]  Find a vector \[\vec{d}\] which is perpendicular to both \[\vec{a} \text{ and } \vec{d}\] \[\text{ and }  \vec{c} \cdot \vec{d} = 15 .\]

 
 

 


Using vectors find the area of the triangle with vertices, A (2, 3, 5), B (3, 5, 8) and C (2, 7, 8).


Using vectors, find the area of the triangle with vertice A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5) .


Find all vectors of magnitude \[10\sqrt{3}\] that are perpendicular to the plane of \[\hat{ i }  + 2 \hat{ j }  + \hat{ k } \] and \[- \hat { i }  + 3 \hat{ j }  + 4 \hat{ k } \] .

 

Define vector product of two vectors.

 

Write the value  \[\left( \hat{ i }  \times \hat{ j }  \right) \cdot \hat{ k }  + \hat{ i }  \cdot \hat{ j }  .\]

 


Write the value of  \[\hat{ i } . \left( \hat{ j } \times \hat{ k }  \right) + \hat{ j }  . \left( \hat{ k } \times \hat{ i }  \right) + \hat{ k }  . \left( \hat{ j }  \times \hat{ i }  \right) .\]

 


For any two vectors  \[\vec{a} \text{ and }  \vec{b}\] write the value of \[\left( \vec{a} . \vec{b} \right)^2 + \left| \vec{a} \times \vec{b} \right|^2\] in terms of their magnitudes.

 
 

\[\text{ If }  \left| \vec{a} \right| = 10, \left| \vec{b} \right| = 2 \text{ and }  \left| \vec{a} \times \vec{b} \right| = 16, \text{ find }  \vec{a} . \vec{b} .\]

 


Write the value of \[\hat{ i }  \times \left(\hat{  j }  \times \hat{ k }  \right) .\]

 

If \[\vec{a} \text{ and }  \vec{b}\] are unit vectors such that \[\vec{a} \times \vec{b}\] is also a unit vector, find the angle between \[\vec{a} \text{ and } \vec{b}\] .

 
 

 


If \[\vec{a} \text{ and } \vec{b}\] are two vectors such that \[\left| \vec{a} . \vec{b} \right| = \left| \vec{a} \times \vec{b} \right|,\]  write the angle between \[\vec{a} \text{ and } \vec{b} .\]

 
 

 


If \[\vec{a}\] is a unit vector such that \[\vec{a} \times \hat{ i }  = \hat{ j }  , \text{ find }  \vec{a} . \hat{ i } \] .

 

If \[\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}\] and \[\vec{a} \times \vec{b} = \vec{a} \times \vec{c,} \vec{a} \neq 0,\] then


If \[\vec{a} = \hat{ i }  + \hat{ j }  - \hat{ k }  , \vec{b} = - \hat{ i }  + 2\hat{ j }  + 2 \hat{ k }  \text{ and }  \vec{c} = - \hat{ i } + 2 \hat{ j }  - \hat{ k }  ,\]  then a unit vector normal to the vectors \[\vec{a} + \vec{b} \text{ and }  \vec{b} - \vec{c}\]  is

 

If \[\hat{ i }  , \hat{ j }  , \hat{ k } \] are unit vectors, then


If θ is the angle between the vectors \[2 \hat{ i }  - 2 \hat{ j}  + 4 \hat{ k }  \text{ and } 3 \hat{ i }  + \hat { j }  + 2 \hat{ k }  ,\]  then sin θ =

 

If θ is the angle between any two vectors `bara` and `barb` and `|bara · barb| = |bara xx barb|` then θ is equal to ______.


Find a unit vector perpendicular to both the vectors `veca and vecb` , where `veca = hat i - 7 hatj +7hatk`  and  `vecb = 3hati - 2hatj + 2hatk` . 


Find the area of the triangle with vertices A(1, l, 2), (2, 3, 5) and (1, 5, 5).


Let `veca = hati + hatj, vecb = hati - hatj` and `vecc = hati + hatj + hatk`. If `hatn` is a unit vector such that `veca.hatn` = 0 and `vecb.hatn` = 0, then find `|vecc.hatn|`.


If the vector `vecb = 3hatj + 4hatk` is written as the sum of a vector `vec(b_1)`, parallel to `veca = hati + hatj` and a vector `vec(b_2)`, perpendicular to `veca`, then `vec(b_1) xx vec(b_2)` is equal to ______.


Let `veca = 2hati + hatj - 2hatk, vecb = hati + hatj`. If `vecc` is a vector such that `veca . vecc = \|vecc|, |vecc - veca| = 2sqrt(2)` and the angle between `veca xx vecb` and `vecc` is 30°, then `|(veca xx vecb) xx vecc|` equals ______.


If the angle between `veca` and `vecb` is `π/3` and `|veca xx vecb| = 3sqrt(3)`, then the value of `veca.vecb` is ______.


Find the area of a parallelogram whose adjacent sides are determined by the vectors `veca = hati - hatj + 3hatk` and `vecb = 2hati - 7hatj + hatk`.


Find the area of the parallelogram whose diagonals are `hati - 3hatj + hatk` and `hati + hatj + hatk`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×