Vectors (plane and line)

**hiy312** · June 16th 2012, 12:59 AM

Hi all, I have difficulty with these five questions about vector.

1. Find the coordinates of mirror image of A (3,1,2) when reflected in the plane x+2y+z=1.
(Do I have to find the foot of normal? But after that what should I do?)

2. Do the line through (3,4,-1) and normal to x+4y-z= -2 intersect any of the coordinate axes?
(After I find the equation of line how do I know if it intersects any axes?)

3. Show that the lines x -1 = (y-2)/2 = z+3 and x+1 = y-3 = 2z+5 are coplanar and find the equation of the plane which contains them.

4. A(1,2,-3/2) lies on the plane x+2y-2z=8. Find the coordinates of B such that vector AB is normal to the plane and 6 units from it.

The test is about to come and I am so worried about it...I hope someone can help me with solutions
Also is there any tips for the vector test? I am very bad in doing vectors...

**biffboy** · June 16th 2012, 01:31 AM

for no2 intersects x axis if there is a point on it with y=
0 and z=0
Similarly for other 2 axes

**Soroban** · June 16th 2012, 07:59 AM

Hello, hiy312!

$\text{3. Show that the lines }\:\begin{Bmatrix}x -1 &=& \dfrac{y-2}{2} &=& z+3 \\ x+1 &=& y-3 &=& 2z+5\end{Bmatrix}\,\text{ are coplanar}$

$\text{and find the equation of the plane which contains them.}$

Two lines are coplanar if they intersect.

We have: . $\begin{Bmatrix}x &=& t+1 \\ y &=& 2t+2 \\ z &=& t-3 \end{Bmatrix}\quad \begin{Bmatrix}x &=& u-1 \\ y &=& u+3 \\ z &=& \frac{u-5}{2}\end{Bmatrix}$

Equate coordinates: . $\begin{Bmatrix} t+1 &=& u-1 && \Rightarrow && t-u &=& -2 \\ 2t + 2 &=& u+3 && \Rightarrow && 2t - u &=& 1 \\ t-3 &=& \frac{u-5}{2} && \Rightarrow && t - \frac{u}{2} &=& \frac{1}{2} \end{Bmatrix}$

Solve the system of equations: . $t = 3,\;u = 5$

The lines intersect at $P(4,8,0)$ . . . They are coplanar.

The normal of the plane is perpendicular to both lines.
The lines have vectors: . $\langle 1,2,1\rangle\,\text{ and }\,\langle 1,1,\tfrac{1}{2}\rangle$

Hence: . $\vec n \;=\;\begin{vmatrix} i&j&k \\ 1&2&1 \\ 1&1&\frac{1}{2}\end{vmatrix} \;=\;i(1-1) - j(\tfrac{1}{2}-1) + k(1-2) \;=\;0i + \tfrac{1}{2}j - k$

. . . . . . $\vec n \;=\;\langle 0,\,\tfrac{1}{2},\,\text{-}1\rangle \;=\;\langle 0,\,1,\,\text{-}2\rangle$

The equation of the plane through $(4,8,0)$ with $\vec n \,=\,\langle 0,1,\text{-}2\rangle$ is:

. . $0(x-4) + 1(y-8) - 2(z-0) \:=\:0 \quad\Rightarrow\quad y - 2z -8 \;=\;0$

**hiy312** · June 16th 2012, 06:06 PM

Originally Posted by biffboy

for no2 intersects x axis if there is a point on it with y=
0 and z=0
Similarly for other 2 axes

Sorry but I really dont understand.. I think I need a solution

**hiy312** · June 16th 2012, 06:06 PM

Originally Posted by Soroban

Hello, hiy312!

Two lines are coplanar if they intersect.

We have: . $\begin{Bmatrix}x &=& t+1 \\ y &=& 2t+2 \\ z &=& t-3 \end{Bmatrix}\quad \begin{Bmatrix}x &=& u-1 \\ y &=& u+3 \\ z &=& \frac{u-5}{2}\end{Bmatrix}$

Equate coordinates: . $\begin{Bmatrix} t+1 &=& u-1 && \Rightarrow && t-u &=& -2 \\ 2t + 2 &=& u+3 && \Rightarrow && 2t - u &=& 1 \\ t-3 &=& \frac{u-5}{2} && \Rightarrow && t - \frac{u}{2} &=& \frac{1}{2} \end{Bmatrix}$

Solve the system of equations: . $t = 3,\;u = 5$

The lines intersect at $P(4,8,0)$ . . . They are coplanar.

The normal of the plane is perpendicular to both lines.
The lines have vectors: . $\langle 1,2,1\rangle\,\text{ and }\,\langle 1,1,\tfrac{1}{2}\rangle$

Hence: . $\vec n \;=\;\begin{vmatrix} i&j&k \\ 1&2&1 \\ 1&1&\frac{1}{2}\end{vmatrix} \;=\;i(1-1) - j(\tfrac{1}{2}-1) + k(1-2) \;=\;0i + \tfrac{1}{2}j - k$

. . . . . . $\vec n \;=\;\langle 0,\,\tfrac{1}{2},\,\text{-}1\rangle \;=\;\langle 0,\,1,\,\text{-}2\rangle$

The equation of the plane through $(4,8,0)$ with $\vec n \,=\,\langle 0,1,\text{-}2\rangle$ is:

. . $0(x-4) + 1(y-8) - 2(z-0) \:=\:0 \quad\Rightarrow\quad y - 2z -8 \;=\;0$

Thank you so much! It helps me a lot!

**hiy312** · June 16th 2012, 06:07 PM

Can anyone teach me the other three questions please?

**AlexReed** · June 16th 2012, 07:07 PM

For question 1, after you have found the foot of normal line, you can use such an equation: the vector formed by pointing from foot to point A is equal to the vector formed by pointing from image to the foot.

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