logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
0 votes

Let the vectors \( \overrightarrow a, \overrightarrow b, \overrightarrow c\) be given as \(a_1\hat i + a_2\hat j + a_3\hat k, \: b_1\hat i + b_2\hat j + b_3\hat k, \: c_1\hat i + c_2\hat j + c_3\hat k \). Then show that \( \overrightarrow a × (\overrightarrow b + \overrightarrow c) = \overrightarrow a × \overrightarrow b + \overrightarrow a × \overrightarrow c\)

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\hat i\times\hat i=\hat j\times\hat j=\hat k\times\hat k=0$
  • $\hat i\times\hat j=k,\hat j\times\hat k=\hat i,\hat k\times\hat i=\hat j$
  • $\hat j\times\hat i=-\hat k,\hat k\times\hat j=-\hat i,\hat i\times\hat k=-\hat j$
Step 1:
Let $\overrightarrow a=a_1\hat i+a_2\hat j+a_3\hat k,\overrightarrow b=b_1\hat i+b_2\hat j+b_3\hat k$ and $\overrightarrow c=c_1\hat i+c_2\hat j+c_3\hat k$
We are asked to prove that $\overrightarrow a × (\overrightarrow b + \overrightarrow c) = \overrightarrow a × \overrightarrow b + \overrightarrow a × \overrightarrow c$
Step 2:
Consider the LHS:
First let us determine $\overrightarrow b+\overrightarrow c$
$\overrightarrow b+\overrightarrow c=(b_1\hat i+b_2\hat j+b_3\hat k)+(c_1\hat i+c_2\hat j+c_3\hat k)$
$\qquad\quad=(b_1+c_1)\hat i+(b_2+c_2)\hat j+(b_3+c_3)\hat k$
$\overrightarrow a × (\overrightarrow b + \overrightarrow c) =a_1\hat i+a_2\hat j+a_3\hat k\times[(b_1+c_1)\hat i+(b_2+c_2)\hat j+(b_3+c_3)\hat k]$
$\qquad\qquad\quad=a_1\hat i\times (b_1+c_1)\hat i+a_1\hat i\times (b_2+c_2)\hat j+a_1\hat i\times (b_3+c_3)\hat k+a_2\hat j\times (b_1+c_1)\hat i+a_2\hat j\times (b_2+c_2)\hat j+a_2\hat j\times (b_3+c_3)\hat k+a_3\hat k\times (b_1+c_1)\hat i+a_3\hat k\times (b_2+c_2)\hat j+a_3\hat k\times (b_3+c_3)\hat k$
But we know $\hat i\times\hat i=\hat j\times\hat j=\hat k\times\hat k=0$
Also $\hat i\times\hat j=k,\hat j\times\hat k=\hat i,\hat k\times\hat i=\hat j$
$\hat j\times\hat i=-\hat k,\hat k\times\hat j=-\hat i,\hat i\times\hat k=-\hat j$
Hence $LHS=0+(a_1b_2+a_1c_2)\hat k+(-\hat j)(a_1b_3+a_1c_3)+(-\hat k)(a_2b_1+a_2c_1)+0+\hat i(a_2b_3+a_2c_3)+\hat j(a_3b_1+a_3c_1)+(-\hat i)(a_3b_2+a_3c_2)+0$
$\qquad=\hat i(a_2b_3+a_2c_3-a_3b_2-a_3c_2)+\hat j(a_3b_1+a_3c_1-a_1b_3-a_1c_3)+\hat k(a_1b_2+a_1c_2-a_2b_1-a_2c_1)$---(1)
Step 3:
Let us now consider the RHS
First let us find $\overrightarrow a\times \overrightarrow b$
$\overrightarrow a\times \overrightarrow b=\begin{vmatrix}\hat i&\hat j &\hat k\\a_1&a_2 &a_3\\b_1&b_2&b_3\end{vmatrix}=\hat i(a_2b_3-a_3b_2)-\hat j(a_1b_3-a_3b_1)+\hat k(a_1b_2-a_2b_1)$
Similarly $\overrightarrow a\times \overrightarrow c$
$\overrightarrow a\times \overrightarrow c=\begin{vmatrix}\hat i&\hat j &\hat k\\a_1&a_2 &a_3\\c_1&c_2&c_3\end{vmatrix}=\hat i(a_2c_3-a_3c_2)-\hat j(a_1c_3-a_3c_1)+\hat k(a_1c_2-a_2c_1)$
Therefore $(\overrightarrow a\times \overrightarrow b)+(\overrightarrow a\times \overrightarrow c)=\hat i(a_2b_3+a_2c_3-a_3b_2-a_3c_2)+\hat j(a_3b_1+a_3c_1-a_1b_3-a_1c_3)+\hat k(a_1b_2+a_1c_2-a_2b_1-a_2c_1)$---(2)
Step 4:
Since equ(1)=equ(2)
LHS=RHS.
Hence proved.
answered May 22, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...