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Given that the non-zero vectors a and b are non-parallel and that
3a+t(2a-3b) = a+b+s(a+2b),
find the value of t and of s

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Answer 1 · 2018-03-19T11:50:21

$s = \frac{4}{7}, t = - \frac{5}{7}$ $s = 4/7, t = -5/7$

If $\to a, \to b$ $vec a, vec b$ are non parallel then from

$3 \to a + t (2 \to a - 3 \to b) = \to a + \to b + s (\to a + 2 \to b)$ $3 vec a+t(2 vec a-3vec b) = vec a+vec b+s(vec a+2 vec b)$

we conclude

$(2 + 2 t - s) \to a - (3 t + 1 + 2 s) \to b = \to 0$ $(2 +2t-s)vec a -(3t+1+2s) vec b = vec 0$

and then

${(2 +2t-s=0),(3t+1+2s=0):}$

and solving gives

$s = 4/7, t = -5/7$