1. ## A Vector Problem

Consider the planes $\displaystyle 5x+4y+1z = 1$ and $\displaystyle 5x+1z = 0$

(A) Find the unique point P on the y-axis which is on both planes. (__, __, __)

(B) Find a unit vector U with positive first coordinate that is parallel to both planes. ___I + ___J + ___K

(C) Use the vectors found in parts (A) and (B) to find a vector equation for the line of intersection of the two planes, r(t) = ___I + ___J + ___K
I was able to find part A by setting x and z to 0 and solve for y:
$\displaystyle 5(0) + 4y + 1(0) = 1$
$\displaystyle y = \frac{1}{4} => (0, \frac{1}{4}, 0)$

Need help solving part B and C

2. Originally Posted by viet
I was able to find part A by setting x and z to 0 and solve for y:
$\displaystyle 5(0) + 4y + 1(0) = 1$
$\displaystyle y = \frac{1}{4} => (0, \frac{1}{4}, 0)$

Need help solving part B and C
The equations of the two planes are given in normal form with

$\displaystyle \overrightarrow{n_1} = (5,4,1)$ and

$\displaystyle \overrightarrow{n_2} = (5,0,1)$

The vector which is perpendicular to both normal vectors must be parallel to both planes indicating the direction of the line of intersection:

$\displaystyle \vec v = \overrightarrow{n_1} \times \overrightarrow{n_2} = (4,0,-20)$

$\displaystyle |\vec v| = \sqrt{16+400} = \sqrt{416}=4 \cdot \sqrt{26}$

The unit vector $\displaystyle \vec U = \dfrac{(4,0,-20)}{4 \cdot \sqrt{26}}=\left(\dfrac1{\sqrt{26}}\ ,\ 0\ , -\dfrac5{\sqrt{26}}\right)$

You already know that the line of intersection passes through the point $\displaystyle \left( 0, \dfrac14 ,0\right)$

Therefore

$\displaystyle \vec r = \left( 0, \dfrac14 ,0\right) + t \cdot \left(\dfrac1{\sqrt{26}}\ ,\ 0\ , -\dfrac5{\sqrt{26}}\right)$