Equation of a line that goes through $A(-3,-7,-5)$ and $B(2,3,0)$ and find $C(x, -1, z)$ on the same line

Question

Problem: Find the equation of a line that passes through $A(-3,-7,-5)$ and $B(2,3,0)$ and find $C(x, -1, z)$ on the same line.

I have completely forgotten how to solve this and I've been reading about how to solve this and it's super confusing. How do I find the equation of this line? I've found the parametric, vector and symmetric forms for the equation, but I don't have a clue how to actually solve it.

Answer 1

vector between A and B is $$\overrightarrow{AB}=5i+10j+5k$$ so the parametric equation of line is $$\frac{x-x_0}{v_x}=t$$ $$\frac{x+3}{5}=t$$ $$\frac{y-y_0}{v_y}=t$$ $$\frac{y+7}{10}=t$$ $$\frac{z-z_0}{v_z}=t$$ $$\frac{z+5}{5}=t$$ now use the point C(x,-1,z)

at $y=-1$ $$\frac{-1+7}{10}=t$$ $$t=0.6$$ then you can use the value of $t$ to find the value of $x$ and $z$ $$\frac{x+3}{5}=0.6$$ $$x=0$$ $$\frac{z+5}{5}=0.6$$ $$z=-2$$

Answer 2

By definition, if $P(x,y,z)$ belongs to the line $AB$, then $\exists \lambda \in \mathbb{R}$ such that $\vec{AP}=\lambda \vec{AB}$.

In your case that means the following set of equations for $C(x,-1,z)$:

$x+3=5\lambda$,

$y+7=6=10 \lambda$ and

$z+5=5\lambda$

We have $\lambda=\dfrac{3}{5}$ hence $C=(0,-1,-2)$...

Answer 3

The equation of a line that passes through $(x_1,y_1,z_1),(x_2,y_2,z_2)$ can be expressed as $$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}.$$ Then, set $y=-1$ to find $x,z$.

Answer 4

In general, the equation of the line passing trough the points $\color{blue}{(x_{1}, y_{1}, z_{1})}$ & $\color{blue}{(x_{2}, y_{2}, z_{2})}$ is given as $$\color{blue}{\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}=k \space \text{(any arbitrary constant)}}$$ As per your question, the equation of the line passing through the points $\color{blue}{A(-3, -7, -5)}$ & $\color{blue}{B(2, 3, 0)}$ s given as $$\frac{x-(-3)}{2-(-3)}=\frac{y-(-7)}{3-(-7)}=\frac{z-(-5)}{0-(-5)} $$ $$\frac{x+3}{5}=\frac{y+7}{10}=\frac{z+5}{5} $$ $$\implies \color{red}{\frac{x+3}{1}=\frac{y+7}{2}=\frac{z+5}{1}=k }$$ Now, the consider a general point on the above line as $\color{blue}{((k-3), (2k-7), (k-5) )}$ & compare its y-coordinate with that of the given point $\color{blue}{C(x, -1, z)}$ we get $$2k-7=-1 $$$$\implies k=\frac{6}{2}=3$$ Hence, the point $\color{blue}{C(x, -1, z)}$ on the above line is obtained by substituting $k=3$ in the general point as $\color{blue}{((3-3), (2\times 3-7), (3-5) )}\equiv \color{red}{(0, -1, -2)}$