# Thread: finding the equation of the line which pasees..

1. ## finding the equation of the line which pasees..

through the point (4,-3.2) and is parallel to the line of intersection of the planes: i) 2x +3y + 2z = 6 and ii) 4x - 3y + z =11

pre.s. i pologise if this does n't belong to this forum and would appreciate your help directing my question to the right forum in this case

Points P(0,1,3/2), Q(2,0,1), Z(1,4/3,0) are on the first plane.
vector PQ is <2,-1,-1/2> and QZ is <-1,4/3,-1> by forming their cross product I find the normal to the first plane to be <1,3,2> .

Points P'(0,1,14), Q'(5/2,0,1), Z'(1,-7/3,0) are on the second plane.
vector P'Q' is <5/2,-1,-13> and Q'Z' is <-3/2,-7/3,-1> by forming their cross product I find the normal to the second plane to be <-4,3,-1>

By forming the cross product of the normals I find the direction vector of the line i am interested in to be <3,1,-9>

and hence the cartesian equation of the wanted line to be
(x-4)/3 = y + 3 = (z-2)/-9

However the book gives (x-4)/3 = (y + 3) = (z-2)/-6

I have gone through it a miriad times and can't find my mistake I would absolutely appreciate it if I can get some help

2. 2x + 3y + 2z = 6
X intercept occurs when y and z = 0, so 2x = 6, x = 3
Y intercept occurs when x and z = 0, so 3y = 6, y = 2
Z intercept occurs when x and y = 0, so 2z = 6, z = 3

For the first question, in vector form, we have:
r = ro + tv
= (4i - 3j + 2k) + t (3i + 2j + 3k)
And for the second question, the line in vector form would be:
= (4i - 3j + 2k) + t (11/4i + -11/3j + 11z)

3. Emian, I don't know what you have done, but the question is asking for one equation... you have two... are you sure you have solved the same exercise
?

4. Originally Posted by pepsi
through the point (4,-3.2) and is parallel to the line of intersection of the planes: i) 2x +3y + 2z = 6 and ii) 4x - 3y + z =11
The book gives (x-4)/3 = (y + 3) = (z-2)/-6
If you have given the correct equations for the planes, then the given answer is off.
The direction vector of the line of intersection is $
\frac{{\left\langle {2,3,2} \right\rangle \times \left\langle {4, - 3,1} \right\rangle }}
{3} = \left\langle {3,{\color{blue}2}, - 6} \right\rangle$
.

The direction number in blue is different.

5. Originally Posted by pepsi
through the point (4,-3.2) and is parallel to the line of intersection of the planes: i) 2x +3y + 2z = 6 and ii) 4x - 3y + z =11

pre.s. i pologise if this does n't belong to this forum and would appreciate your help directing my question to the right forum in this case

Points P(0,1,3/2), Q(2,0,1), Z(1,4/3,0) are on the first plane.
vector PQ is <2,-1,-1/2> and QZ is <-1,4/3,-1> by forming their cross product I find the normal to the first plane to be <1,3,2> .
You don't need to find two vectors in the plane and take their cross product to find a normal vector to a plane. A normal vector to the plane Ax+ By+ Cz= D is <A, B, C> so a normal to the plane 2x+3y+ 2z= 6 is <2, 3, 2>, not <1, 3, 2>. Your vectors PQ and QZ are correct but their cross product is <5/3, 5/2, 5/3>= (5/6)<2, 3, 2>.

Points P'(0,1,14), Q'(5/2,0,1), Z'(1,-7/3,0) are on the second plane.
vector P'Q' is <5/2,-1,-13> and Q'Z' is <-3/2,-7/3,-1> by forming their cross product I find the normal to the second plane to be <-4,3,-1>
Similarly, a normal to 4x-3y+ z= 11 is <4, -3, 1>, equivalent to <-4, 3, -1> so that is correct.

By forming the cross product of the normals I find the direction vector of the line i am interested in to be <3,1,-9> and hence the cartesian equation of the wanted line to be
(x-4)/3 = y + 3 = (z-2)/-9

However the book gives (x-4)/3 = (y + 3) = (z-2)/-6

I have gone through it a miriad times and can't find my mistake I would absolutely appreciate it if I can get some help

6. I much appreciate you help, its been real helpful..