Is My Vector Equation for a Plane Correct Without a Diagram?

[kelvin macks]

Homework Statement

for this question (photo 1), i am not sure whether this is type 1 (as the type in photo 2) or type 2 ( as in photo 3 ). the question didnt provide a diagram, this is making me confused. so i did it another way on the right , (using pencil ). is my working acceptable ?

Homework Equations

The Attempt at a Solution

 

[haruspex]

The vector equation can be in parametric or non-parametric form. In the third picture, the first vector form is the parametric one, while the second form, with a normal vector and a distance, is the non-parametric form. So I would guess (a) is asking for the n, d form.

 

[ehild]

You are supposed to draw a picture to your solution!
Sorry I can not decipher your pencilled note.

The parametric form of a plane uses two vectors in the plane ($\vec b$ and $\vec c$) , and a point of the plane (A, its position vector is $\vec a$). The position vector $\vec r$ of an arbitrary point R of the plane is obtained as the sum $$\vec r = \vec a +μ \vec b + λ \vec c$$
In the problem, A, B, C mean points of the plane, with position vectors (1,1,4), (3,0,-1) and (2,-2,0). Vectors connecting A to B and A to C $, \vec b=\vec{AB}$ and $\vec c=\vec{AC}$, lie in the plane. You have to add their linear combination to the vector $\vec a = (1,1,4)$

The solution in photo 1 is like in photo 2, only Q stands for B and P stands for C.

ehild

 

[kelvin macks]

You are supposed to draw a picture to your solution!
Sorry I can not decipher your pencilled note.

The parametric form of a plane uses two vectors in the plane ($\vec b$ and $\vec c$) , and a point of the plane (A, its position vector is $\vec a$). The position vector $\vec r$ of an arbitrary point R of the plane is obtained as the sum $$\vec r = \vec a +μ \vec b + λ \vec c$$
In the problem, A, B, C mean points of the plane, with position vectors (1,1,4), (3,0,-1) and (2,-2,0). Vectors connecting A to B and A to C $, \vec b=\vec{AB}$ and $\vec c=\vec{AC}$, lie in the plane. You have to add their linear combination to the vector $\vec a = (1,1,4)$

The solution in photo 1 is like in photo 2, only Q stands for B and P stands for C.

ehild

how do we know that a is connected to b and b is connected to c ? or a is connected to b and a is connected to c? in photo 3 , a is connected to b and b is connected to c.. that's why I'm wondering whether i can apply the same concept for the question in photo 1.

 

[kelvin macks]

You are supposed to draw a picture to your solution!
Sorry I can not decipher your pencilled note.

The parametric form of a plane uses two vectors in the plane ($\vec b$ and $\vec c$) , and a point of the plane (A, its position vector is $\vec a$). The position vector $\vec r$ of an arbitrary point R of the plane is obtained as the sum $$\vec r = \vec a +μ \vec b + λ \vec c$$
In the problem, A, B, C mean points of the plane, with position vectors (1,1,4), (3,0,-1) and (2,-2,0). Vectors connecting A to B and A to C $, \vec b=\vec{AB}$ and $\vec c=\vec{AC}$, lie in the plane. You have to add their linear combination to the vector $\vec a = (1,1,4)$

The solution in photo 1 is like in photo 2, only Q stands for B and P stands for C.

ehild

sorry , for photo 2, i can't understand how can the plane is parallel to vector b and vector c .. can you draw me a better diagram. i can't imagine

 

[ehild]

b and c are vectors lying in the plane. A vector is parallel with a plane if all its points are at the same distance from the plane. It the line lies in the plane, all points are at zero distance of the plane. So the line is parallel with the plane.

ehild