# Math Help - so confused, vector length exam tommorow!

1. ## so confused, vector length exam tommorow!

given length BD is 15m and vector equation of line BD is

(-1) (15) i.e +lamda(15,-20,0)
(-7) + lamda (-20)
(11) (0)
find corordinates of D which is in mark scheme (8,-19,11)
This question is so easy but im so annoyed i cant understand it. I wouldnt have thought 15 comes from pythagarus of each point and someow thats incorporated, Obviously not as the mark scheme tells me to get lamda as 3 or 3/5 as appropriate(HOW) then sub in! where does lamda come from....????

2. Originally Posted by i_zz_y_ill
given length BD is 15m and vector equation of line BD is

(-1) (15) i.e +lamda(15,-20,0)
(-7) + lamda (-20)
(11) (0)
find corordinates of D which is in mark scheme (8,-19,11)
This question is so easy but im so annoyed i cant understand it. I wouldnt have thought 15 comes from pythagarus of each point and someow thats incorporated, Obviously not as the mark scheme tells me to get lamda as 3 or 3/5 as appropriate(HOW) then sub in! where does lamda come from....????
The equation of a line is written as:

$\underbrace{(x,y,z)}_{coordinates\ of \ points \ on \ the \ line}=\underbrace{(a,b,c)}_{startingpoint} + t \cdot \underbrace{(u,v,w)}_{direction\ vector}$. So you get:

$\begin{array}{l}x=a+t \cdot u \\
y=b+t \cdot v \\
z=c+t \cdot w
\end{array}$

Instead of t some authors prefer $\lambda, ~ \mu$

Obviously B(-1, -7, 11) and the direction vector is $\vec u=(15, -20,0)$

So you can calculate the coordinates of D by:

$D(-1+\lambda \cdot 15~,~ -7 +\lambda \cdot (-20)~~,~11+\lambda \cdot 0)$

Now calculate $\overrightarrow{BD}=(15\lambda~,~-20\lambda~,~0)$

The length of $|\overrightarrow{BD}|=\sqrt{625\lambda^2}=15$

Solve for $\lambda$ and you'll get: $\lambda=\frac35~\vee~\lambda=-\frac35$

Plug in this value into the equation of the line to calculate the coordinates of D:

$D_1(8, -19,11)$ or $D_2(-10, 5, 11)$