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 February 22nd, 2015, 02:02 PM #1 Member   Joined: Jan 2015 From: Norway Posts: 95 Thanks: 1 Lengths and angles a->= 3 b->= square root 2 <_ ( a->. b >) = 45 celsius f) decide t so that (A -> + tb --->) _|_ (2a -> - b ->) The book says the answer is t= - 15/4 The book says p-> _|_ q-> <-> P-> * q-> = 0 But in this case the answer dosent become 0. So i don't understand why _|_ in this case won't become cero Last edited by Zman15; February 22nd, 2015 at 02:05 PM.
February 22nd, 2015, 02:54 PM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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Hello, Zman15!

I'll guess what you mean.

Quote:
 $\vec{a}= 3,\;\;\vec{b}= \sqrt{2}$ <_ ( a->. b >) = 45 Celsius $\;\color{blue}{\text{What does this mean?}}$ f) Decide $t$ so that: $\:(\vec{a}\,+\,t\vec{b})\:\perp\:(2\vec{a}\,-\,\vec{b})$ The book says the answer is $t\,=\, -\frac{15}{4}\;\;$ $\color{blue}{\text{This is wrong!}}$ The book says: $\;\vec{p}\:\perp\:\vec{q} \;\Longleftrightarrow\; \vec{p}\,\cdot\,\vec{q} \:=\:0\;\;$ $\color{blue}{\text{Correct!}}$ But in this case the answer doesn't become 0. $\color{blue}{\text{Of course not! \; Their }t\text{-value is wrong.}}$

$\left(3\vec{i}\,+\,t\sqrt{2}\vec{j}\right)\,\perp\ ,\left(6\vec{i} \,-\,\sqrt{2}\vec{j}\right) \;\;\;\Longleftrightarrow\;\;\;\left(3\vec{i}\,+\, t \sqrt{2}\vec{j}\right)\,\cdot\,\left(6\vec{i} \,-\,\sqrt{2}\vec{j}\right) \:=\:0$

$\;\;\;(3)(6)\,+\,(t\sqrt{2})(-\sqrt{2}) \:=\:0 \;\;\;\Rightarrow\;\;\;18\,-\,2t \:=\:0 \;\;\;\Rightarrow\;\;\;\boxed{t \,=\,9}$

 February 22nd, 2015, 03:21 PM #3 Member   Joined: Jan 2015 From: Norway Posts: 95 Thanks: 1 Thanks for the answer i need a help with an other task : http://i1277.photobucket.com/albums/...psogbql2s2.jpg The figure shows a-> = 2 and b-> = 3 I have a problem with task C) whict says BC> = a + b and decide | BC> | The book says the answer is square root 19, and square root 19 = 4.35. A + B = 2 + 3 = 5. And | BC | = BC ^ 2 = 5 ^ 2 = 25. Which would mean square root of 25 is 5. Has the book given the wrong answer again?

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