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.

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.

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

.

Thus, the wanted vector is . Therefore,

.

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

where is a real number.

109 trust me

step-by-step explanation:

idk what the heck he answer is, i’m just putting this here so i finish my account