# How would I go about finding a vector equation of a hyperplane 3x_1+x_2−2x_3+4x_4=5

How would I go about finding a vector equation of a hyperplane
$3{x}_{1}+{x}_{2}-2{x}_{3}+4{x}_{4}=5$
I know I need to find 3 vectors and another vector that shifts from the origin in the form
$r\stackrel{\to }{a}+s\stackrel{\to }{b}+t\stackrel{\to }{c}+\stackrel{\to }{d}$
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SoroAlcommai9
Hint: Take ${x}_{1},{x}_{3},{x}_{4}$ to be free variables. Then,
${x}_{1}={x}_{1}$
${x}_{2}=5-3{x}_{1}+2{x}_{3}-4{x}_{4}$
${x}_{3}={x}_{3}$
${x}_{4}={x}_{4}$
Now write this in vector notation.