Author Topic: My Specialist Maths Mega-thread :D (Read 17897 times) Tweet Share

BubbleWrapMan · « **Reply #75 on:** September 25, 2012, 10:06:55 pm »

If it proves one is not a linear combination of the other two, then it is legit.

meemz · « **Reply #76 on:** September 26, 2012, 05:26:43 pm »

Quote from: monkeywantsabanana on March 16, 2012, 08:55:29 pm

Given b = 3i+2j-k and c=2i-j+k, find the unit vector perpendicular to b and c

someone answer?
don't we have to use cross product on the calc to solve these questions??
thanks

b^3 · « **Reply #77 on:** September 26, 2012, 05:37:34 pm »

It can still be done with things inside the spesh course.

Lets start with $\underset{\sim}{d}=x\underset{\sim}{i}+y\underset{\sim}{j}+z\underset{\sim}{k}$
Now for $\underset{\sim}{b}$ and $\underset{\sim}{c}$ to be perpendicular we have
$\underset{\sim}{d}.\underset{\sim}{b}=0$ and $\underset{\sim}{d}.\underset{\sim}{c}=0$ .

Now we still need a third bit of information, but we know $\underset{\sim}{d}$ has to be a unit vector, so that it's magnitude has to be 1.
$\sqrt{x^{2}+y^{2}+z^{2}}=1$ .

$\begin{alignedat}{1}3x+2y-z & =0...[1] \\ 2x-y+z & =0...[2] \\ x^{2}+y^{2}+z^{2} & =1...[3] \end{alignedat}$
So we have three equations, and three variables, solve for $x,\:y,\:z$ .

meemz · « **Reply #78 on:** September 27, 2012, 12:44:18 am »

Quote from: b^3 on September 26, 2012, 05:37:34 pm

It can still be done with things inside the spesh course.

Lets start with $\underset{\sim}{d}=x\underset{\sim}{i}+y\underset{\sim}{j}+z\underset{\sim}{k}$
Now for $\underset{\sim}{b}$ and $\underset{\sim}{c}$ to be perpendicular we have
$\underset{\sim}{d}.\underset{\sim}{b}=0$ and $\underset{\sim}{d}.\underset{\sim}{c}=0$ .

Now we still need a third bit of information, but we know $\underset{\sim}{d}$ has to be a unit vector, so that it's magnitude has to be 1.
$\sqrt{x^{2}+y^{2}+z^{2}}=1$ .

$\begin{alignedat}{1}3x+2y-z & =0...[1] \\ 2x-y+z & =0...[2] \\ x^{2}+y^{2}+z^{2} & =1...[3] \end{alignedat}$
So we have three equations, and three variables, solve for $x,\:y,\:z$ .

Legend
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Author Topic: My Specialist Maths Mega-thread :D (Read 17897 times) Tweet Share

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