Chap 6
Question 7)
ok what have u tried so far ?
I don't know how to begin...
Can you guide me..
maybe for writing sake, say those \(\lambda \)'s are x,y,z...
\(\large s=-3i-j+6k=\lambda_1p+\lambda_2q+\lambda_3r \\=\lambda_1(-i+2j+3k)+\lambda_2(3i+j-2k)+\lambda_3(i+5j+2k)\)
\(\lambda 's\) are values , so if u have a vector \(v=v_1i+v_2j+v_3k\) then \(\lambda v =\lambda v_1i+\lambda v_2j+ \lambda v_3k\)
next compare components both sides
\[-3i-j+6k=\lambda_1(-i+2j+3k)+\lambda_2(3i+j-2k)+\lambda_3(i+5j+2k)\]
whats the coefficient of \(\large i\) on left hand side ?
-3
whats the coefficient of \(\large i\) on right hand side ?
There are three values for \(i\) : -1,3,1
what about \(\lambda \)'s ? they are also attached to them right ?
i is the unit vector (1,0,0) hmm
\[-3i-j+6k=\lambda_1(-i+2j+3k)+\lambda_2(3i+j-2k)+\lambda_3(i+5j+2k) \] \[-3\color{Red}{i}-\color{Red}{j}+6\color{Red}{k} = \color{red}{i} (-\lambda_1 + 3\lambda_2 + \lambda_3) + \color{Red}{j}(2\lambda_1 + \lambda_2 + 5\lambda_3) + \color{Red}{k}(3\lambda_1 -2\lambda_2 + 2\lambda_3)\]
Now compare, you will get 3 equations and you can solve the 3 unknowns
\(\large -3 = -\lambda_1 + 3\lambda_2 + \lambda_3\) \(\large -1 = 2\lambda_1 + \lambda_2 + 5\lambda_3\) \(\large 6 = 3\lambda_1 -2\lambda_2 + 2\lambda_3\)
solve those 3 equations using any of your favorite methods
How do I solve them .. using elimination method ? @ganeshie8
that works
ok
Is there any simpler method..
you should get : http://www.wolframalpha.com/input/?i=solve++-3+%3D+-x+%2B+3y+%2B+z%2C+-1+%3D+2x+%2B+y+%2B+5z%2C6+%3D+3x+-2y+%2B+2z
i have used x,y,z instead of lambdas
I don't think there will be any simpler method this time :) you will have to solve those 3 equations and get above answer
OK
Thank you @ikram002p and @ganeshie8
np :)
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