EunoR

2021-02-09

Determine whether the given

rogreenhoxa8

Skilled2021-02-10Added 109 answers

The given linear equation system

$2{x}_{1}+{x}_{2}+{x}_{3}=3$

$-2{x}_{1}+{x}_{2}+{x}_{3}=1$

The given system of linear equations either represents coincident planes, parallel planes, or coincident planes with a line at their intersection.

Let ${n}_{1}=(2,1,1)$ and ${n}_{2}=(-2,1,-1)$ be normal vectors of both the equations. Then n, and na are not parallel.

Let ${x}_{3}=t$, then replace ${x}_{3}=t$ in the system of linear equations $2{x}_{1}+{x}_{2}+t=3\text{}\text{}\text{}(1)$

$-2{x}_{1}+{x}_{2}+t=1\text{}\text{}\text{}(2)$

$(1)+(2)\Rightarrow 2{x}_{2}+2t=4\Rightarrow 2{x}_{2}=4-2t\Rightarrow {x}_{2}=2-t$ Replace ${x}_{2}=2-t$ and ${x}_{3}=t$ in the firs equation

$2{x}_{1}+2-t+t=3$

$2{x}_{1}=3-2$

${x}_{1}=\frac{1}{2}$

Hence, the plane intersect in a line and the parametric equations are

${x}_{1}=\frac{1}{2},{x}_{2}=2-t\text{and}{x}_{3}=t$

Jeffrey Jordon

Expert2021-10-27Added 2607 answers

Answer is given below (on video)

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