Find an equation of plane that contains line $x=-2+3t$, $y=4+2t$, $z=3-t$
and is perpendicular to the plane $x-2y+z=5$.
Give final answer in the form of $ax+by+cz=d$.
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Widawensen
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Desmond Tan
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OK, the normal vector of the plane should be perpendicular to the line and also to the normal vector of the other plane. Do you know how to find a vector perpendicular to the two given vectors? – Ivan Neretin Feb 27 '17 at 20:13
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<3,2,-1> x <1,-2,1> = <0,-4,-8> – Desmond Tan Feb 27 '17 at 20:26
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That's right. Here is your normal vector. (You might want to multiply it by -1 to look more "natural".) Now take any point on the line (and hence on the plane) and use it to find $d$. – Ivan Neretin Feb 27 '17 at 20:30
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Oh! Thanks for the guidance, it helped me a lot. – Desmond Tan Feb 27 '17 at 20:33