# vector problems

• May 14th 2008, 03:07 PM
JMV
vector problems
ive been working on about 20 questions of vectors.. but these 5 i cant seem to get.. any help would be much appreciated!

1. Deterine the components of a vector with tail at P (1,1,1), magnitude 3, and in the same direction as a= (-3,-4, 12)

2. The vector v has a magnitude 10 units, positive and equal x and y components, and a z component of 4. Determine v.

3. the vector u= (1,-2,1) Determine 3 different non-collinear vectors that are perpendicular to u. Then show that the 3 vectors determined in the first part of the question are coplanar.

4. Prove that (a-b) x (a+b)= 2(axb)

5. Find the total work done by a 15 N force, F, in the direction of the vector (1,2,2) when it moves particle from O (0,0,0) to P(1,-3,4) and then from P to Q(7,2,5). The distance is measured in metres.
• May 14th 2008, 11:12 PM
earboth
Quote:

Originally Posted by JMV
...
1. Deterine the components of a vector with tail at P (1,1,1), magnitude 3, and in the same direction as a= (-3,-4, 12)

2. The vector v has a magnitude 10 units, positive and equal x and y components, and a z component of 4. Determine v.

3. the vector u= (1,-2,1) Determine 3 different non-collinear vectors that are perpendicular to u. Then show that the 3 vectors determined in the first part of the question are coplanar.
...

to #1:

$\displaystyle |\vec a| = \sqrt{9+16+144}=13$

Therefore a vector which has the same direction as $\displaystyle \vec a$ and has the magnitude 3 has the components: $\displaystyle \left(-\frac9{13}~,~-\frac{12}{13}~,~\frac{36}{13}\right)$

to #2:
$\displaystyle \vec v=(x, x, 4)~\implies~|\vec v|=\sqrt{x^2+x^2+16}$

And therefore you have to solve for x:

$\displaystyle 2x^2+16=100~\implies~x=\sqrt{84}$ (since only the positive value should be used). That means:

$\displaystyle \vec v=\left(\sqrt{84}~,~\sqrt{84}~,~4\right)$

to #3:
Let $\displaystyle \vec v \perp \vec u~\implies~\vec v \cdot \vec u = 0$. Let $\displaystyle \vec v = (r, s, t)$

Then you know:

$\displaystyle (1, -2, 1) \cdot (r,s,t)=r-2s+t$

$\displaystyle r-2s+t=0~\implies~t=2s-r$ and therefore the vector

$\displaystyle \vec v = (r, s, 2s-r)~,~r,s \in \mathbb{R}$ is perpendicular to $\displaystyle \vec u$

Actually this is the equation of a plane passing through the origin with $\displaystyle \vec u$ as it's normal vector.

Choose numbers for r and s to build vectors and do the required examinations.