Help with checking Vector Question

• May 16th 2010, 03:28 PM
mathematicallyretarded
Help with checking Vector Question
Hey guys I am not sure if i am on the right track here so i need some advice

Question is:

if $\displaystyle \vec{a}=3i-2j$ , $\displaystyle \vec{b}=-4i+4j$ , and $\displaystyle \vec{c}=6i+-9j$ express the following vectors in their simplest form:

(i)$\displaystyle -5\vec{b}$
(ii)$\displaystyle 2\vec{a}-\frac{1}{2}\vec{c}$
(iii)$\displaystyle \frac{2}{3}\vec{a}-\frac{1}{2}\vec{b}-\frac{1}{4}\vec{c}$

My Solution:

(i) $\displaystyle -5\left(-4i+4j \right)=20i+\left(-20j \right)$
(ii) $\displaystyle 2(3i-2j)-\frac{1}{2}(6i+(-9)j)$
$\displaystyle = 6i-4j-3i+(-\frac{9}{2})j$
$\displaystyle = (6-3=3i) -4j-\frac{9}{2}$
$\displaystyle = -\frac{4}{1}-\frac{9}{2}$
$\displaystyle =-\frac{8}{2}-\frac{9}{2}$
$\displaystyle =-\frac{17}{2}$

Answer? $\displaystyle 3i-\frac{17}{2}j$

(iii) $\displaystyle \frac{2}{3}(3i-2j)-\frac{1}{2}(-4i+4j)-\frac{1}{4}(6i+(-9)j)$
$\displaystyle (2i-\frac{4}{3}j)-(-2i+2j)-(\frac{3}{2}i+(-\frac{9}{4}j)$
$\displaystyle 2i+2i-\frac{3}{2}i$
$\displaystyle = \frac{4}{1}i+\frac{3}{2}i$
$\displaystyle = \frac{8}{2}+\frac{3}{2}$
$\displaystyle =\frac{11}{2}i$

$\displaystyle -\frac{4}{3}j+\frac{2}{1}j-\frac{9}{4}j$
$\displaystyle = -\frac{16}{12}j+\frac{24}{12}j-\frac{27}{12}j$
$\displaystyle = -\frac{19}{12}j$

Answer? $\displaystyle \frac{11}{2}i-\frac{19}{12}j$
• May 16th 2010, 05:36 PM
sa-ri-ga-ma
while removing the brackets, you have to follow the following rule.

-2(a-b) = -2a + 2b

-2(a+b) = -2a - 2b

Therefore your second the third problems are wrong. Correct it.
• May 16th 2010, 06:18 PM
mathematicallyretarded
Quote:

Originally Posted by sa-ri-ga-ma
while removing the brackets, you have to follow the following rule.

-2(a-b) = -2a + 2b

-2(a+b) = -2a - 2b

Therefore your second the third problems are wrong. Correct it.

ok so (ii) answer is $\displaystyle 3{\bf i}+(-4+\frac{9}{2}){\bf j}=3{\bf i}+\frac{1}{2}{\bf j}$

now i will do the (iii) one again
• May 16th 2010, 06:28 PM
mathematicallyretarded
Ok would the (iii) one be

$\displaystyle \frac{5}{2}i+\frac{7}{6}j$
• May 16th 2010, 07:40 PM
Soroban
Hello, mathematicallyretarded!

You are making hard work out of it.

This is basically Algebra I . . . combining like terms.

Quote:

Given: .$\displaystyle \begin{array}{ccc}\vec a &=&3i-2j \\ \vec b &=&-4i+4j \\ \vec c &=&6i-9j\end{array}$

Express the following vectors in their simplest form:

$\displaystyle (i) \;-5\vec{b}$

$\displaystyle -5\vec b \;\;=\;\;-5(-4i + 4j) \;\;=\;\;20i - 20j$

Quote:

$\displaystyle (ii) \;\;2\vec a -\tfrac{1}{2}\vec c$

$\displaystyle 2\vec a - \tfrac{1}{2}\vec c \;\;=\;\;2(3i - 2j) - \tfrac{1}{2}(6i-9j)$

. . . . . .$\displaystyle =\;\;6i - 4j - 3i + \tfrac{9}{2}j$

. . . . . .$\displaystyle =\;\;3i + \frac{1}{2}j$

Quote:

$\displaystyle (iii)\;\;\tfrac{2}{3}\vec a -\tfrac{1}{2}\vec b - \tfrac{1}{4}\vec c$

$\displaystyle \tfrac{2}{3}\vec a - \tfrac{1}{2}\vec b - \tfrac{1}{4}\vec c \;\;=\;\;\tfrac{2}{3}(3i - 2j) - \tfrac{1}{2}(-4i + 4j) - \tfrac{1}{4}(6i - 9j)$

. . . . . . . . .$\displaystyle =\;\;2i - \tfrac{4}{3}j + 2i - 2j - \tfrac{3}{2}i + \tfrac{9}{4}j$

. . . . . . . . .$\displaystyle =\;\;\frac{5}{2}i - \frac{13}{12}j$

• May 17th 2010, 01:22 AM
mathematicallyretarded
Thanks! I have to learn to keep it simple