Two lines are given:
1) $$\frac{x-2}{3} = \frac{y+1}{1} = -\frac{z}{4}$$ 2) $x = t$, $y = 2$ $z = t +1$ So I guess I have to change the first one, based on this?
1) Do the lines cross, or not?
2) Find the angle, between first line, and plane $2x + y − z − 4 = 0$
3) Find the equation of plane, in which there is the first line and the point $A(1;-1;2)$
I tried, but I even forgot how to do the first one.
Thank you in advance.
On
First question:
It is simpler to mix Cartesian equation and parametric representation (as they are): from the parametric representation of the second line, you know the lines intersect if an only if: $$\frac{t-2}3=3=-\frac{t+1}4,$$ whence $t=11$ from the first equation, $t=-13$ from the second. So there is no intersection point.
Third question:
A plane passing through the first line belongs to the pencil of planes defined by the Cartesian equations of the line: this line is the intersection of the planes: $$x-3y-5=0\quad\text{and}\quad4y+z+4 =0,$$ so a plane of the pencil they define has equation: $$\lambda(x-3y-5)+\mu(4y+z+4)=0\qquad\text{for some}\enspace\lambda,\mu\in\mathbf {R}.$$ It passes through the point $(1,-1,2)$ if $$-\lambda+2\mu=0$$ We'll choose $\mu=1$, whence $\lambda=2$, so an equation of the plane through the first line and the point $(1,-1,2)$ is $$2x-2y+z-6=0.$$
Start by converting the $2$ lines into vector equations: $$\begin{bmatrix}x \\ y \\ z \end{bmatrix}=\begin{bmatrix}2 \\ -1 \\ 0 \end{bmatrix}+\mu\begin{bmatrix}3 \\ 1 \\ -4 \end{bmatrix}\tag{1}$$
$$\begin{bmatrix}x \\ y \\ z \end{bmatrix}=\begin{bmatrix}0 \\ 2 \\ 1 \end{bmatrix}+t\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}\tag{2}$$
$\color{red}{1.}$ If the two lines cross then there is a value of $\mu$ and $t$ that will satisfy both line equations. You can find this by equating the two lines such that $$\begin{bmatrix}2 \\ -1 \\ 0 \end{bmatrix}+\mu\begin{bmatrix}3 \\ 1 \\ -4 \end{bmatrix}=\begin{bmatrix}0 \\ 2 \\ 1 \end{bmatrix}+t\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}$$
This implies $$ 2+3\mu=t\tag{a}$$
$$ \mu-1=2\implies \mu =3\tag{b}$$ substitute $(b)$ into $(a)$ $\implies t=11$
For the $z$-component $$-4\mu=t+1$$
With $\mu =3 \implies -4 \times 3=-12=$ LHS of $z$-component equation.
With $t =11 \implies 11+1 =12=$ RHS of $z$-component equation
so $-12\ne 12$
Since LHS $\ne$ RHS we conclude that the lines do not intersect (cross) at any point.
$\color{lime}{2.}$ To find the angle between the line and the plane you must first find the angle between the normal and the line, start by converting the plane into vector form $$\begin{bmatrix}x \\ y \\ z \end{bmatrix}\cdot\begin{bmatrix}2 \\ 1 \\ -1 \end{bmatrix}=4$$
the normal is given by $\vec A=\begin{bmatrix}2 \\ 1 \\ -1 \end{bmatrix}$ and the direction vector of the first line is $\vec B=\begin{bmatrix}3 \\ 1 \\ -4 \end{bmatrix}$
Now the angle between the normal vector of the plane and line direction vector is given by the dot product
$\theta=\cos^{-1}\left(\cfrac{\vec A \cdot \vec B}{\mid \vec A\mid\mid \vec B\mid}\right)=\cos^{-1}\left(\cfrac{\begin{bmatrix}2 \\1 \\-1 \end{bmatrix} \cdot \begin{bmatrix}3 \\1 \\-4 \end{bmatrix}}{\sqrt{(2)^2+(1)^2+(-1)^2} \sqrt{(3)^2+(1)^2+(-4)^2}}\right)=\cos^{-1}\left(\cfrac{6+1+4}{\sqrt{6} \sqrt{26}}\right)=\cos^{-1}\left(\cfrac{11}{\sqrt{6} \sqrt{26}}\right)\approx 28.3^{\circ}$
Subtract this angle from $90^{\circ}$ to get the angle between the line and the plane.
So angle between line and plane is $90^{\circ}-28.3^{\circ}=\color{blue}{61.7^{\circ}}$.
$\color{#F80}{3.}$ To find the equation of the plane containing the first line, you need a position vector of a point that lies on that line which is $\begin{bmatrix}2 \\ -1 \\ 0 \end{bmatrix}$ and the direction vector of the line itself which is $\begin{bmatrix}3 \\ 1 \\ -4 \end{bmatrix}$
Now compute the cross-product of these two vectors to get the normal $\vec{n}$ to the plane:
$\begin{bmatrix}2 \\ -1 \\ 0 \end{bmatrix}\times\begin{bmatrix}3 \\ 1 \\ -4 \end{bmatrix}=\begin{bmatrix}i & j & k \\2 & -1 & 0 \\3 & 1 & -4 \end{bmatrix}=\begin{bmatrix}4 \\ 8 \\ 5 \end{bmatrix}=\vec{n}$
Now use the general vector equation of the plane:$$\vec{r}\cdot\vec{n}=\vec{a}\cdot\vec{n}$$ where $\vec{r}$ is the position vector relative to the origin of any point $(x,y,z)$ in the plane and $\vec{a}$ is
a known point that lies on the plane. Taking $\vec{a}$ to be $\begin{bmatrix}2 \\ -1 \\ 0 \end{bmatrix}$, the equation of the plane becomes $$\begin{bmatrix}x \\ y \\ z \end{bmatrix}\cdot\begin{bmatrix}4 \\ 8 \\ 5 \end{bmatrix}=\begin{bmatrix}2 \\ -1 \\ 0 \end{bmatrix}\cdot\begin{bmatrix}4 \\ 8 \\ 5 \end{bmatrix}=2\times 4-1\times 8+0\times 5=0$$
In Cartesian coordinates the equation of the plane is $$\color{blue}{4x+8y+5z=0}$$
To find the equation of the plane containing the second line, you need a position vector of a point
that lies on that line which is $\begin{bmatrix}0 \\ 2 \\ 1 \end{bmatrix}$ and the direction vector of the line itself which is $\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}$.
Now computing the cross-product of these: $\vec{n}=\begin{bmatrix}0 \\ 2 \\ 1 \end{bmatrix}\times\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}=\begin{bmatrix}i & j & k \\0 & 2 & 1 \\1 & 0 & 1 \end{bmatrix}=\begin{bmatrix}2 \\ 1 \\ -2 \end{bmatrix}$
Now substituting $\vec a =\begin{bmatrix}0 \\ 2 \\ 1 \end{bmatrix}$ and $\vec n =\begin{bmatrix}2 \\ 1 \\ -2 \end{bmatrix}$ into the equation of the plane gives
$$\begin{bmatrix}x \\ y \\ z \end{bmatrix}\cdot\begin{bmatrix}2 \\ 1 \\ -2 \end{bmatrix}=\begin{bmatrix}0 \\ 2 \\ 1 \end{bmatrix}\cdot\begin{bmatrix}2 \\ 1 \\ -2 \end{bmatrix}=0\times 2+2\times 1+1\times -2=0$$
In Cartesian coordinates the equation of the plane is $$\color{blue}{2x+y-2z=0}$$