Obtain an equation of the plane in the form ##px+qy+rz=d##

[chwala]

Homework Statement

see attached

Relevant Equations

vectors

The solution is here;

Now to my comments,
From literature, the cross product of two vectors results into a vector in the same dimension. A pointer to me as i did not know the first step. With that in mind and using cross product, i have

$(1-1)i - (-1-1)j+(1+1)k =0i+2j +2k$ as shown in ms attachment.

Now the second part is the reason of this post. My take on that is,

$r=(λ +μ)i + (μ-4-λ)j + (λ-3-μ)k =x+y+z$

Now

$λ +μ = 0, ⇒ λ = -μ$

Therefore,
$μ-4+μ=y$
$-μ-3-μ=z$

on adding the above two equations,

$-4+-3=y+z$
$-7=y+z$

unless there is a better approach or simpler...have a great day.

 

[Orodruin]

The general form of the equation for a plane is
$$
\vec n \cdot \vec r = C,
$$
where $C$ is a constant and $\vec n$ is normal to the plane. You have first found $\vec n$ so that is fine, but after that you are inserting the general form of any parametrised point on the plane. This is unnecessarily complicated. You know that the expression above should equal the same constant $C$ regardless of the point so using a single point in the plane will give you $C$. The easiest is to just take the point corresponding to $\mu =\lambda = 0$, i.e., $\vec r = -4\hat j - 3\hat k$. Doing so with $\vec n = \hat j + \hat k$ results in
$$
C = -4 \cdot 1 - 3 \cdot 1 = -7
$$
and therefore
$$
\vec n \cdot \vec r = y + z = -7
$$

 

[Orodruin]

Also note that there is no requirement that $\mu + \lambda = 0$, but this just describes a line in the plane. The values of $\mu$ and $\lambda$ are arbitrary, but the point is that you will get the same constant $C$ regardless of the values of $\mu$ and $\lambda$.

 

[chwala]

This is also related part of question,

The solution is here - quite clear to me

but nothing wrong with me having,
$y+z-7=0$
$-5x+3y+5z-4=0$

setting $x=0$ gives

$y=-\dfrac{39}{2}$ and $z=\dfrac{25}{2}$

Thus,

$r+λ(n_1 ×n_2) = (0, -19.5, 12.5) +λ(2,-5,5)$

 

[Orodruin]

Sure, it is just a different parametrisation. Setting $\lambda = \lambda' - 7/2$ should give you the parametrisation given in the answer key.

 

[chwala]

This last part is also related to the question- hmmmm i have no idea how they did this but i will still share my thoughts.

Mythoughts,

Distance between a point and a plane is given by,

$D=\dfrac{|ax_0 +by_o+cz_0 +d|}{\sqrt{a^2+b^2+c^2}}=\sqrt{2}$

Now could they have used ,

$x=a +λ (1-b-a)$

$y=a+λ(b-a)$

$z=(a-7)+λ(b-a-7)$

when $λ=0$ I will then have,

$D=\dfrac{|0 +a+a-7+7|}{|\sqrt{0^2+1^2+1^2}|}=\sqrt{2}$

If that is the case then its time for me to have hot coffee!