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Thread: Vectors (2)

  1. #1
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    Vectors (2)

    Q1: The points A, B and C have position vectors a, b and c respectively, relative to the origin O.

    Show that the area of triangle OAB is $\displaystyle \frac{1}{2} \sqrt{|a|^2 |b|^2 - (a\cdot b)}units^2$

    (The "a" and "b" above are vectors)

    Q2: The points A, B and C have position vectors a, b and c respectively, relative to the origin O.

    Find the length of projection of AB on AC if a = i - 5j + 3k, b = 2j -4k and c = i + j

    My answer for Q2 is $\displaystyle \frac{5}{\sqrt{29}}$...is it correct?

    Thank you for helping!
    Last edited by Tangera; Jan 10th 2009 at 06:23 AM.
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  2. #2
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    Scalar product

    Hello Tangera
    Quote Originally Posted by Tangera View Post
    Q1: The points A, B and C have position vectors a, b and c respectively, relative to the origin O.

    Show that the area of triangle OAB is $\displaystyle \frac{1}{2} \sqrt{|a|^2 |b|^2 - (a\cdot b)}units^2$

    (The "a" and "b" above are vectors)

    Q2: The points A, B and C have position vectors a, b and c respectively, relative to the origin O.

    Find the length of projection of AB on AC if a = i - 5j + 3k, b = 2j -4k and c = i + j

    My answer for Q2 is $\displaystyle \frac{5}{\sqrt{29}}$...is it correct?

    Thank you for helping!
    Q1 The scalar, or dot, product of two vectors $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$ is $\displaystyle |\vec{a}||\vec{b}|\cos\theta$, where $\displaystyle \theta$ is the angle between the vectors.

    Now the area of the triangle OAB $\displaystyle = \frac{1}{2}OA.OB.\sin\theta = \frac{1}{2}|\vec{a}||\vec{b}|\sin\theta$

    $\displaystyle = \frac{1}{2}|\vec{a}||\vec{b}|\sqrt{1-\cos^2\theta}$

    $\displaystyle = \frac{1}{2}\sqrt{|\vec{a}|^2|\vec{b}|^2-|\vec{a}|^2|\vec{b}|^2\cos^2\theta}$

    $\displaystyle = \frac{1}{2}\sqrt{|\vec{a}|^2|\vec{b}|^2-(\vec{a}.\vec{b})^2}$

    Q2 Geometrically, the dot product of $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$ is the length of one vector multiplied by the length of the projection of the other vector onto it. So the length of the projection of $\displaystyle \vec{x}$ onto $\displaystyle \vec{y}$ is given by

    $\displaystyle \frac{\vec{x}.\vec{y}}{|\vec{y}|}$

    So here we want $\displaystyle \frac{\vec{AB}.\vec{AC}}{AC}$

    = $\displaystyle \frac{(\vec{b} - \vec{a}).(\vec{c}-\vec{a})}{|\vec{c}-\vec{a}|}$

    I make the answer $\displaystyle \frac{21\sqrt{5}}{5}$. Would you like to have another look at it, and get back to me?

    Grandad
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  3. #3
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    Hello, Tangera!

    I got a different answer for #2 . . .


    2) The points $\displaystyle A, B, C$ have position vectors $\displaystyle \vec a, \vec b, \vec c,$, respectively,
    relative to the origin $\displaystyle O.$

    $\displaystyle \text{Find the length of projection of }\overrightarrow{AB}\text{ on }\overrightarrow{AC}\text{ if:}$ .$\displaystyle \begin{array}{ccc}\vec a &= &\vec i - 5\vec j + 3\vec k \\ \vec b &=& 2\vec j -4\vec k \\ \vec c &=& \vec i + \vec j \end{array}$

    We have: .$\displaystyle \begin{array}{ccc}\vec a &=& \langle 1,\text{-}5,3\rangle \\ \vec b &=& \langle0,2,\text{-}4\rangle \\ \vec c &=& \langle 1,1,0 \rangle \end{array}$ . $\displaystyle \Rightarrow \quad \begin{array}{ccc}\overrightarrow{AB} &=& \langle\text{-}1,7,\text{-}7\rangle \\ \overrightarrow{AC} &=& \langle0,6,\text{-}3\rangle \end{array}$


    Formula: .$\displaystyle \text{proj}_{\vec v}\vec u \:=\:\frac{\vec u \cdot\vec v}{|\vec v|^2}\,\vec v $

    We have: .$\displaystyle \text{proj}_{\overrightarrow{AC}}\overrightarrow{A B} \:=\:\frac{\langle\text{-}1,7,\text{-}7\rangle\cdot\langle0,6,\text{-}3\rangle}{(\sqrt{0^2+6^3+3^2})^2}\,\langle0,6,\te xt{-}3\rangle$ .$\displaystyle = \;\frac{63}{45}\,\langle 0,6,\text{-}3\rangle $


    Its length is: .$\displaystyle \left|\frac{7}{5}\langle 0,6,\text{-}3\rangle\right| \;=\;\frac{7}{5}\sqrt{0+36+9} \;=\;\frac{7}{5}\sqrt{45} \;=\;\boxed{\frac{21}{5}\sqrt{5}}$


    But check my work . . . please!
    .
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