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The planes $5x+2y+2z=−19$ and $3x+4y+2z=−7$ are not parallel, so they must intersect along a line that is common to both of them. What is the vector parametric equation for this line?

I'm not literate in MathJax, so I'll try to describe what I was able to do. I was able to find the directional vectors and cross them, but the problem was finding the $t$ values. I'm not even fully confident in the directional vectors that I was able to solve for, so I would appreciate a step-by-step solution because I would like to understand this from the ground up.

**I've tried substituting $t$ for $z$, but when trying to solve the system the $z$ components just cancel out, leaving $x$ and $y$, so I'm left with no $t$ values to manipulate.

**Update: I've been told to show my work, so here's my best attempt: $$ \begin{align} 5x+2y+2z=19 &\rightarrow \overrightarrow a=( 5, 2, 2 )\\ 3x+4y+2z=-7 &\rightarrow \overrightarrow b=( 3, 4, 2 ) \end{align}$$ I was told to ignore the constant at the end to solve for the direction vectors, but I'm not completely sure if this is is the correct way of going about it. $$ \overrightarrow a ×\overrightarrow b=( 4,-4,14 ) $$ We were given the formula $$ \overrightarrow r=\overrightarrow r_0+t\overrightarrow v,$$ where

  • $\overrightarrow r_0=( x_0,y_0,z_0 )$ is the initial position vector,
  • $\overrightarrow r = (x,y,z)$ is the final position vector,
  • $\overrightarrow v = (a,b,c)$ are the direction numbers, and
  • $t$ is a parameter.

That is all I've been able to do so far; I tried to solve the system to find the point where they intersected, but the $2z$ on both sides canceled when I tried to substitute $t$ in like some suggested.

Jean Marie
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alipboy
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3 Answers3

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I am not totally sure about the meaning of “vector parametric equation” and therefore my answer may not be what you want.

Consider the system$$\left\{\begin{array}{l}5x+2y+2z=-19\\3x+4y+2z=-7.\end{array}\right.\tag1$$It is equivalent to$$\left\{\begin{array}{l}5x+2y=-19-2z\\3x+4y=-7-2z.\end{array}\right.$$So, you can see it as system of two linear equations in two unknowns. Solving it, you get,$$\left\{\begin{array}{l}x=\frac17(-2z-31)\\y=\frac17\left(-2z+11\right)\end{array}\right.$$So, the solutions of the system $(1)$ are the points of the form$$\left(\frac17(-2t-31),\frac17\left(-2t+11\right),t\right),$$with $t\in\mathbb R$. Take $t=0$; you get the point $\left(-\frac{31}7,\frac{11}7,0\right)$, which belongs to your line. Now, take $t=1$; you get the point $\left(-\frac{33}7,\frac97,1\right)$, which is another point of your line. So, an equation of the line will be$$\left(-\frac{31}7,\frac{11}7,0\right)+t\left(\left(-\frac{33}7,\frac97,1\right)-\left(-\frac{31}7,\frac{11}7,0\right)\right),$$with $t\in\mathbb R$.

José Carlos Santos
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Substitute one of the variables by any constant. Then, after the substitution, compute the remaining two variables. This will give you one point of your straight line.

Then substitute the same initial variable by another constant. Compute the other variables, which will give you the other point.

Once you have two points, then write down your parametric expression/equation.

Good luck.

PS. In general, on rare occasions, this algorithm might not work. Then think just a bit and try it again with a different choice of substitutions.

Wlod AA
  • 2,124
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Say we solve the second equation for $2z$. Get $2z=-7-3x-4y$. Then substitute that into the first equation: $5x+2y-7-3x-4y=-19$.

So, simplifying, we get $x-y=-6$.

To parametrize, set $x=t$. Then $y=t+6$. And $2z=-7-3t-4(t+6)$, so $z=-\dfrac {7t}2-\dfrac {31}2$.

Thus $\{(t,t+6,-\dfrac {7t}2-\dfrac {31}2): t\in\Bbb R\}$ would be a parametrization.

Alternatively, cross the two normals to get a vector parallel to the line, and then use a point: $(3,4,2)×(5,2,2)=(4,4,-14)$.

$(0,6,-\dfrac {31}2)+t(4,4,-14)$ would then be another parametrization.