Question

Suppose u = (1,2,3), v = (2, -3, -4), and w = (1,0,-2). (a) Find u-2v + 3w. (b) Write the vector (3,4,5) as a linear combination of u, v, and w.

Answer 1

Answer with explanation:

We are given vectors u,v and w as follows:

      $u=(1,2,3)\,\ v=(2,-3,-4),\ and\ w=(1,0,-2)$

(a)

Now we are asked to find the value of:

  $u-2v+3w$

The value is calculated as follows:

$u-2v+3w=(1,2,3)-2(2,-3,-4)+3(1,0,-2)\\\\i.e.\\\\u-2v+3w=(1,2,3)+(-2\times 2,-2\times -3,-2\times -4)+(3\times 1,3\times 0,3\times -2)\\\\i.e.\\\\u-2v+3w=(1,2,3)+(-4,6,8)+(3,0,-6)\\\\i.e.\\\\u-2v+3w=(1-4+3,2+6+0,3+8-6)\\\\i.e.\\\\u-2v+3w=(0,8,5)$

                              Hence,

                 $u-2v+3w=(0,8,5)$

(b)

We are asked to represent (3,4,5) as a linear combination of u, v, and w.

Let:

$(3,4,5)=au+bv+cw$

$(3,4,5)=a(1,2,3)+b(2,-3,-4)+c(1,0,-2)$

i.e.

$(3,4,5)=(a,2a,3a)+(2b,-3b,-4b)+(c,0,-2c)\\\\i.e.\\\\(3,4,5)=(a+2b+c,2a-3b,3a-4b-2c)\\\\i.e.\\\\a+2b+c=3----------(1)$

$2a-3b=4-------------(2)$

$3a-4b-2c=5------------(3)$

on multiplying equation(1) by 2 an then adding equation (1) to equation (3)

$5a=11\\\\i.e.\\\\a=\dfrac{11}{5}$

on putting the value of a in equation (2) we get:

$2\times \dfrac{11}{5}-3b=4\\\\i.e.\\\\\dfrac{22}{5}-3b=4\\\\i.e.\\\\3b=\dfrac{22}{5}-4\\\\i.e.\\\\3b=\dfrac{2}{5}\\\\i.e.\\\\b=\dfrac{2}{15}$

Now on putting the value of a and b in equation (1)

$\dfrac{11}{5}+2\times \dfrac{2}{15}+c=3\\\\i.e.\\\\\dfrac{11}{5}+\dfrac{4}{15}+c=3\\\\i.e.\\\\c=3-\dfrac{11}{5}-\dfrac{4}{15}\\\\i.e.\\\\c=\dfrac{3\times 15-11\times 3-4}{15}\\\\i.e.\\\\c=\dfrac{45-33-4}{15}\\\\i.e.\\\\c=\dfrac{8}{15}$

Hence, we have:

$(3,4,5)=\dfrac{11}{5}(1,2,3)+\dfrac{2}{15}(2,-3,-4)+\dfrac{8}{15}(1,0,-2)\\\\i.e.\\\\(3,4,5)=\dfrac{11}{5}u+\dfrac{2}{15}v+\dfrac{8}{15}w$