1. ## Vector questions

1. Show a^2 b^2 = a.b a.b + a ^ b . a ^ b

2. A plane with position vectors a, b and c, find the distance of a fourth, with position vector r, from the plane.

3. A plane with the point (1, -2, 1) and the line r = (1+3?)i + (?-1)j - 2?k. Find the distance of the plane from the origin.

If someone could start me off in the right direction that would be great.

2. Hello,

2. Let V be the volume of the tetrahedron with vertices a, b, c, r.
Let S be the area of the triangle with vertices a, b, c.
Then, the distance d is V=(1/3)Sd.
V and S can be computed using external products.

OR,

Draw a perpendicular line from r to the plane, meeting the plane at the point h. You can express h as h=sa+tb+uc where s+t+u=1.
Determine s, t, u by (r-h).(a-b)=0, (r-h).(b-c)=0.
The distance d is the length of the vector r-h.

3. Find 2 points on the line and Use 2.

Bye.

3. Originally Posted by Happy Dancer
...

3. A plane with the point (1, -2, 1) and the line r = (1+3?)i + (?-1)j - 2?k. Find the distance of the plane from the origin.

If someone could start me off in the right direction that would be great.
Is the questionmark meant to be the variable, for instance, t?

If so, re-write the equation of the line to:

$\begin{array}{l}x=1+3t \\ y=-1-r \\ z = 0-2t\end{array}~\implies~(x,y,z)=(1,-1,0)+t \cdot (3,-1,-2)$

That means you have one point of the plane A(1,-1,0) and one vector which spans the plane $\vec v=(3,-1,-2)$. The second span-vector is $\vec u = \overrightarrow{AP} = (1,-2,1)-(1,-1,0)=(0,-1,1)$

Therefore the plane has the equation:

$px,y,z)=(1,-1,0)+t \cdot (3,-1,-2)+s\cdot (0,-1,1)" alt="px,y,z)=(1,-1,0)+t \cdot (3,-1,-2)+s\cdot (0,-1,1)" />

You easily can prove that the origin lies in this plane $\left(r=-\frac13~\wedge~ s=-\frac23 \right)$

that means the distance of the plane frome the origin is zero.