1.) Find an equation for the plane that passes through the origin (0, 0, 0) and is parallel to the plane - x + 3y - 2z = 6
2.) Find the distance between the point (-1, - 2, 0) and the plane 3x - y + 4z =-2
Q1
For the line to be parallel to the line "-x+3y-2z=6" .Then,
the would be "-x+3y-2z=k" . To get k,
Since the line passes through the origin.
"k=0"
The line is therefore, "-x+3y-2z=0"
Q2
"(-1,-2,0)=(x_o,y_o,z_o)\\\\\nThe\\ distance\\ from\\ the\\\\ equation\\ Ax+By+Cz+D=0\\ is\\\\\nd=\\dfrac{|Ax_o+By_o+Cz_o+D|}{\\sqrt{A^2+B^2+C^2}}\\\\\nA=3,B=-1,C=4,D=2\\\\\n\nd=\\dfrac{|3(-1)-1(-2)+4(0)+2|}{\\sqrt{3^2+(-1)^2+4^2}}\\\\\nd=\\dfrac{1}{\\sqrt{26}}unit"
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