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April 14th, 2017, 07:19 AM   #1
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find the modulus of vector V.
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 April 14th, 2017, 07:45 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,507 Thanks: 1234 $2\vec{V} + \vec{V} \times (\hat{i}+2\hat{j}) = 2\hat{i}+\hat{k}$ let $\vec{V} = a\hat{i} + b\hat{j} + c\hat{k}$ $\vec{V} \times (\hat{i}+2\hat{j})=\begin{vmatrix} i & j & k\\ a & b & c\\ 1 & 2 & 0 \end{vmatrix} = (-2c)-(c)+(2a-b)=2a\hat{i}-b\hat{j}-3c\hat{k}$ $2(a\hat{i} + b\hat{j} + c\hat{k}) + (2a\hat{i}-b\hat{j}-3c\hat{k}) = 2\hat{i}+\hat{k}$ can you finish ?
 April 14th, 2017, 08:38 AM #3 Newbie   Joined: Apr 2017 From: Bhadohi, U.P., India Posts: 6 Thanks: 0 Yeah. thankyou! it can be done by assuming it. But is there any way to find it without assuming it as ai+bj+ck.
 April 14th, 2017, 08:56 AM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,507 Thanks: 1234 You have two other vectors in the equation given in component form. I used $a,b, \text{ and } c$ as constant multiples of the unit vectors for V in each direction to express V in component form also. You understand you have to solve for the constants $a,b, \text{ and } c$, correct? ... how else would you do it? Thanks from Shariq Faraz Last edited by skeeter; April 14th, 2017 at 09:32 AM.
 April 15th, 2017, 03:44 AM #5 Newbie   Joined: Apr 2017 From: Bhadohi, U.P., India Posts: 6 Thanks: 0 okay. yeah

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