How do you find the vector parametrization of the line of intersection of two planes $2 x - y - z = 5$ $2x - y - z = 5$ and $x - y + 3 z = 2$ $x - y + 3z = 2$ ?

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Answer 1 · 2015-03-22T12:18:52

One answer could be:

$x = t$ $x=t$

$z = \frac{1}{4} t - \frac{3}{4}$ $z=1/4t-3/4$

$y = \frac{7}{4} t - \frac{17}{4}$ $y=7/4t-17/4$ .

Let's solve the system of the two equations, explaining two letters in function of the third:

$2 x - y - z = 5$ $2x-y-z=5$

$x - y + 3 z = 2$ $x-y+3z=2$

So:

$y = 2 x - z - 5$ $y=2x-z-5$

$x - (2 x - z - 5) + 3 z = 2 \Rightarrow x - 2 x + z + 5 + 3 z = 2 \Rightarrow$ $x-(2x-z-5)+3z=2rArrx-2x+z+5+3z=2rArr$

$4 z = x - 3 \Rightarrow z = \frac{1}{4} x - \frac{3}{4}$ $4z=x-3rArrz=1/4x-3/4$

so:

$y = 2 x - (\frac{1}{4} x - \frac{3}{4}) - 5 \Rightarrow y = 2 x - \frac{1}{4} x + \frac{3}{4} - 5$ $y=2x-(1/4x-3/4)-5rArry=2x-1/4x+3/4-5$

$y = \frac{7}{4} x - \frac{17}{4}$ $y=7/4x-17/4$ .

SO:

$z = \frac{1}{4} x - \frac{3}{4}$ $z=1/4x-3/4$

$y = \frac{7}{4} x - \frac{17}{4}$ $y=7/4x-17/4$

and if we put $x = t$ $x=t$ :

$x = t$ $x=t$

$z = \frac{1}{4} t - \frac{3}{4}$ $z=1/4t-3/4$

$y = \frac{7}{4} t - \frac{17}{4}$ $y=7/4t-17/4$ .

How do you find the vector parametrization of the line of intersection of two planes 2x−y−z=52x - y - z = 5 and x−y+3z=2x - y + 3z = 2?