# Vector

• Oct 6th 2012, 11:53 AM
alexander9408
Vector
A straight line is given as 2(x+4) = -2(y-5) = -3(z-6),

(a)determine the vector equation of the straight line. The answer I found is r = -4i +5j +6k + t(3i -3j +2k), is this correct?
(b)find the coordinate of a point Q that lies at the straight line such the |(OQ) ⃗ |= 3. I can't get this one, someone please guide me, thanks.
• Oct 6th 2012, 12:20 PM
Plato
Re: Vector
Quote:

Originally Posted by alexander9408
A straight line is given as 2(x+4) = -2(y-5) = -3(z-6),
(a)determine the vector equation of the straight line. The answer I found is r = -4i +5j +6k + t(3i -3j +2k), is this correct?
(b)find the coordinate of a point Q that lies at the straight line such the |(OQ) ⃗ |= 3. I can't get this one, someone please guide me, thanks.

I get $<-4,5,6>+t<3,-3,-2>$. So check that 2k.
• Oct 6th 2012, 01:16 PM
HallsofIvy
Re: Vector
Quote:

Originally Posted by alexander9408
A straight line is given as 2(x+4) = -2(y-5) = -3(z-6),

(a)determine the vector equation of the straight line. The answer I found is r = -4i +5j +6k + t(3i -3j +2k), is this correct?

With t= 1, r= -4i+ 5j+ 6k+ 3i- 3j+ 2k= -i+ 2j+ 8k or the point (-1, 2, 8). 2(x+ 4)= 2(-1+ 4)= 6, -2(y- 5)= -2(2- 5)= 6, and -3(z- 6)= -3(8- 6)= -6. Those are NOT equal so the point (-1, 2, 8) on your line is not on the given line. They are not the same line.

Quote:

(b)find the coordinate of a point Q that lies at the straight line such the |(OQ) ⃗ |= 3. I can't get this one, someone please guide me, thanks.
I presume you know that the distance from Q= (x, y, z) to O= (0, 0, 0) is $\sqrt{x^2+ y^2+ z}$ so, setting that equal to 3 and squaring, $x^2+ y^2+ z^2= 9$.

Since all of 2(x+ 4), -2(y+ 5), and -3(z- 6) are equal, set them all equal to t: 2(x+ 4)= t so x= -4+ (1/2)t. -2(y+ 5)= t so y= -5- (1/2)t. -3(z- 6)= t so z= 6- (1/3)t. Replace x, y, and z in $x^2+ y^2+z^2= 9$ to get a single equation in t.
• Oct 6th 2012, 08:46 PM
Soroban
Re: Vector
Hello, alexander9408!

Quote:

$\text{A straight line is given as: }\:2(x+4) \:=\: \text{-}2(y-5) \:=\: \text{-}3(z-6)$

$\text{(a) Determine the vector equation of the straight line.}$

To get the equations in Symmetric Form, divide by -6:

. . $\frac{x+4}{3} \;=\;\frac{y-5}{-3} \;=\;\frac{z-6}{-2}$

We have a point $(\text{-}4,5,6)$ and the direction vector $\vec v \:=\:\langle 3,\text{-}3,\text{-}2\rangle$

Write the vector equation.

Quote:

$\text{(b) Find the coordinates of a point }Q\text{ on the line such that: }\,|\overrightarrow{OQ}| = 3.$

There are two solutions.

I used the Parametric Form of the line: . $\begin{Bmatrix}x &=& \text{-}4 + 3t \\ y &=& 5 - 3t \\ z &=& 6 - 2t \end{Bmatrix}$

Then: . $|\overrightarrow{OQ}| \,=\,3 \quad\Rightarrow\quad \sqrt{(\text{-}4+3t)^2 + (5-3t)^2 + (6-2t)^2} \;=\;3$

. . $16-24t + 9t^2 + 25 - 30t + 9t^2 + 36 - 24t + 4r^2 \;=\;9$

. . $22t^2 - 78t + 68 \;=\;0 \quad\Rightarrow\quad 11t^2 - 39t + 34 \;=\;0$

. . $(11t-17)(t-2) \:=\:0 \quad\Rightarrow\quad t \;=\;\tfrac{17}{11},\,2$

If $t \,=\,\tfrac{17}{11}:\;Q(x,y,z) \;=\;\left(\tfrac{7}{11},\,\tfrac{4}{11},\,\tfrac{ 32}{11}\right)$

If $t\,=\,2:\;Q(x,y,z) \;=\;(2,\,\text{-}1,\,2)$