Help with checking Vector Question

**mathematicallyretarded** · May 16th 2010, 03:28 PM

Hey guys I am not sure if i am on the right track here so i need some advice

Question is:

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

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

My Solution:

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

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

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

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

Answer? $\frac{11}{2}i-\frac{19}{12}j$

**sa-ri-ga-ma** · May 16th 2010, 05:36 PM

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.

**mathematicallyretarded** · May 16th 2010, 06:18 PM

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 $<br /> 3{\bf i}+(-4+\frac{9}{2}){\bf j}=3{\bf i}+\frac{1}{2}{\bf j}<br />$

now i will do the (iii) one again

**mathematicallyretarded** · May 16th 2010, 06:28 PM

Ok would the (iii) one be

$\frac{5}{2}i+\frac{7}{6}j$

**Soroban** · May 16th 2010, 07:40 PM

Hello, mathematicallyretarded!

You are making hard work out of it.

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

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

Express the following vectors in their simplest form:

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

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

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

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

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

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

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

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

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

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

**mathematicallyretarded** · May 17th 2010, 01:22 AM

Thanks! I have to learn to keep it simple

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