Question

Z B Zo e yo Zc у A Xc yo For the diagram above, given that the coordinates of A are (0,0,0)m, B are (0, -3, 4)m and C are (2,3,6) m, determine the following: The x component of the position vector, l'AB- Please give this answer in metres (enter the unit "m" with the answer) to three significant figures. Answer: The y component of the position vector, l'AB- Please give this answer in metres (enter the unit "m" with the answer) to three significant figures Answer:

Answer 1

LAB B (X®, Ys, Ze) A (XA, Ya, Za)

The position vector is expressed between two points in 3D space, the general equation for the position vector is,

  

TAB тв ТА

For the points A (0,0,0) and B (0,-3,4)

or

TAB 0i – 3] + 4k

From the position vector of r_AB

(1) The x component of the vector r_AB is 0 m.

$r_{AB} ={\color{Blue} 0 } i -3 j +4k$

(2) The y component of the vector r_AB is -3 m.

$r_{AB} =0 i +{\color{Blue} (-3)} j +4k$

(3) The z component of the vector r_AB is 4 m.

$r_{AB} =0 i -3 j +{\color{Blue} 4}k$

(4) The magnitude of a vector is the square root of sum of the squares of the x,y and z components.

|ТАВ (0:2)2 + ( y + (z)2

The magnitude of the vector r_AB is calculated as follows,

TAB 0i – 3] + 4k

$\left |r_{AB} \right | =\sqrt{(0)^{2} +(-3)^{2} +(4)^{2}}$

$\left |r_{AB} \right | =\sqrt{ 9+ 16}$

$\left |r_{AB} \right | =\sqrt{ 25}$

$\left |r_{AB} \right | = 5$

The magnitude of the vector r_AB is 5 m.

For the points A (0,0,0) and C (2,3,6)

The positon vectort is

From the position vector of r_AC

(5) The x component of the vector r_AC is 2 m.

$r_{AC} ={\color{Blue} 2} i +3 j +6 k$

(6) The y component of the vector r_AC is 3 m.

$r_{AC} =2 i +{\color{Blue} 3} j +6 k$

(7) The z component of the vector r_AC is 6 m.

$r_{AC} =2 i +3 j +{\color{Blue} 6} k$

(8) The magnitude of a vector is the square root of sum of the squares of the x,y and z components.

$\left |r_{AC} \right | =\sqrt{(x)^{2} +\left (y \right )^{2} +(z)^{2}}$

The magnitude of the vector r_AC is calculated as follows,

$\left |r_{AC} \right | =\sqrt{(2)^{2} +(3)^{2} +(6)^{2}}$

$\left |r_{AC} \right | =\sqrt{ 4+9+ 36}$

$\left |r_{AC} \right | =\sqrt{ 49}$

$\left |r_{AC} \right | = 7$

The magnitude of the vector r_AB is 7 m.

(9) The angle between two vectors is found by taking dot product of the vector

$\vec{a} .\vec{b} =\left | \vec{a} \right | . | \vec{b}\, |cos (\theta)$

On rearranging,

$cos (\theta) =\frac{\vec{a} .\vec{b} }{ \left | \vec{a} \right | . | \vec{b}\, |}$

$\theta = cos^{-1}\left \{ \frac{ \vec{a} .\vec{b}}{|\vec{a}|.|\vec{b}| } \right \}$

using the above equation on r_AB and r_AC

$\theta = cos^{-1}\left \{ \frac{ \vec{r_{AB} }.\, \, \vec{r_{AC}}}{|\vec{r_{AB}| }.\, \, |\vec{r_{AC}}| } \right \}$

$\vec{r_{AB} }.\, \, \vec{r_{AC}}=(0 i -3 j +4k).(2 i +3 j +6 k)$

$\vec{r_{AB} }.\, \, \vec{r_{AC}}=(0 i -3 j +4k).(2 i +3 j +6 k)=(0 \times 2)+ (-3 \times 3)+(4 \times 6)$

$\vec{r_{AB} }.\, \, \vec{r_{AC}}=(0)+ (-9)+(24)$

$\vec{r_{AB} }.\, \, \vec{r_{AC}}=15$

from (4) and (8)

$\left |r_{AB} \right | = 5$

$\left |r_{AC} \right | = 7$

$|\vec{r_{AB}| }.\, \, |\vec{r_{AC}}| = 5 \times 7 =35$

$\theta = cos^{-1}\left \{ \frac{ \vec{r_{AB} }.\, \, \vec{r_{AC}}}{|\vec{r_{AB}| }.\, \, |\vec{r_{AC}}| } \right \}$

$\theta = cos^{-1}\left \{ \frac{15}{35} \right \}$

$\theta = 64.62^{\circ}$

The angle between r_AB and r_AC is 64.62 ^o