Question

Consider the planes 4x+3y+4z=1 and 4x+4z=0. (A) Find the unique point P on the y−axis which is on both planes. ( 0 equation editorEquation Editor , 1/3 equation editorEquation Editor , 0 equation editorEquation Editor ) (B) Find a unit vector u with positive first coordinate that is parallel to both planes. u= 12/288^0.5 equation editorEquation Editor i + 0 equation editorEquation Editor j + -12/288^0.5 equation editorEquation Editor k (C) Use parts (A) and (B) to find a vector equation for the line of intersection of the two planes.

Answer 1

Answer:

Step-by-step explanation:

A) From the order of the exercise we already know that the intersection points lies on the Y-axis, so its coordinates are P(0;y;0). In order to find it, we only need to substitute the equation 4x+4z=0 into the equation 4x+3y+4z=1. Then,

1=4x+3y+4z = 3y + (4x+4z)= 3y+0.

From the expression above it is easy to obtain that y=1/3, and the intersection point is P(0;1/3;0).

B) To obtain the parallel vector to both planes we use the cross product of the normal vector of the planes.

$\left[\begin{array}{ccc}i&j&k\\4&3&4\\4&0&4\end{array}\right] = 12i-0j+12k$

As we want a unit vector, we must calculate the modulus of u:

$|u|=\sqrt{12^2+0^2+12^2} = \sqrt{2\cdot 12^2}=12\sqrt{2}$.

Thus, the wanted vector is $\frac{u}{12\sqrt{2}}$. Therefore,

$u = \frac{1}{\sqrt{2}}i-\frac{1}{\sqrt{2}}k = \frac{\sqrt{2}}{2}i-\frac{\sqrt{2}}{2}k$.

C) In order to obtain the vector equation of the intersection line of both planes, we just need to put together the above results.

$r = \frac{1}{3}j +\lambda \left( \frac{\sqrt{2}}{2}i- \frac{\sqrt{2}}{2}k \right)$

where $\lambda$ is a real number.