Question

Vector problem:

Answer 1

Am I correct until there?

Answer 2

that's what I got for a, -7

Answer 3

8

Answer 4

Sorry, i got confused with the page

Answer 5

Yeah, you are correct a = -7

Answer 6

So c. i is direct...but what should I do for ii?

Answer 7

yeah, since AB dot AC=0

Answer 8

can we get that statement for ii in englihs please?

Answer 9

after the question above, what is the value to "a" if q = 1.2

Answer 10

Thanks

Answer 11

I am assuming there is no calculator allowed?

Answer 12

For this test we can

Answer 13

I could try using dot product again...or using the cos of last excercise

Answer 14

if you are using a graphing calculator, you can just solve the equation for a.

Answer 15

I thought that the rule for cos was:
\[\cos \alpha=\frac{ v*w }{ \left| v \right|\left| w \right|}\]

Why is it different here?

Answer 16

u*w is the same, but the denominator is differnet

Answer 17

It isn't different, they already made this product

Answer 18

given i,
\(0=2a+14-\sqrt{14a^2+280}*\cos(1.2)\)

Answer 19

sqrt14 * sqrt20+a...why is it squared on the exam?

Answer 20

because they calculated the magnitude

Answer 21

AB= (1 3 2)
AC= (2 4 a)
\[|AB|=\sqrt{14}\]
\[|AC|=\sqrt{20+a^2}\]
\[|AB|*|AC|=\sqrt{280+14a^2}\]
\[AB\cdot AC=14+2a\]
therefore\[cos(p)=\frac{14+2a}{\sqrt{280+14a^2}}\]

Answer 22

AB = (1 3 2)
AC = (2 4 a)
|AB|=\(\sqrt{1^2+3^2+2^2}=\sqrt{14}\)
|AC| = \(\sqrt{2^2+4^2+a^2}=\sqrt{20+a^2}\)
|AB||AC|=\(\sqrt{14(20+a^2)}=\sqrt{a^2+280}\)

Answer 23

Oh...I missed one square...sorry :/ Ok, I will solve this then

Answer 24

so just subsitute 1,2 for q and write the equation such that you can input into the calculator and find zeros
\(\huge 0=2a+14-\sqrt{14a^2+280}*\cos(1.2)\)

Answer 25

I am in that :) thanks

Answer 26

Good luck on your IB exam :)

Answer 27

Good luck ;)

Answer 28

Thanks both of you :) really

Answer 29

You're welcome ;)

Answer 30

a=-3.25 right ?

Answer 31

http://www.wolframalpha.com/input/?i=0%3D2a%2B14-sqrt%2814a%5E2%2B280%29*cos%281.2%29
it appears so...

Answer 32

Thanks a lot, really

Answer 33

anytime :)