Vectors (2)

**Tangera** · January 10th 2009, 06:09 AM

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 $\frac{1}{2} \sqrt{|a|^2 |b|^2 - (a\cdot b)}units^2$

(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 $\frac{5}{\sqrt{29}}$ ...is it correct?

Thank you for helping!

**Grandad** · January 10th 2009, 07:15 AM

Hello Tangera

Originally Posted by Tangera

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 $\frac{1}{2} \sqrt{|a|^2 |b|^2 - (a\cdot b)}units^2$

(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 $\frac{5}{\sqrt{29}}$ ...is it correct?

Thank you for helping!

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

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

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

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

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

Q2 Geometrically, the dot product of $\vec{a}$ and $\vec{b}$ 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 $\vec{x}$ onto $\vec{y}$ is given by

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

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

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

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

Grandad

**Soroban** · January 10th 2009, 08:59 AM

Hello, Tangera!

I got a different answer for #2 . . .

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

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

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

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

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

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

But check my work . . . please!
.

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