Like

Report

If $ a, b $, and $ c $ are not all 0, show that the equation $ ax + by + cz + d = 0 $ represents $ a $ plane and $ \langle a, b, c \rangle $ is a normal vector to the plane.

Hint: Suppose $ a \neq 0 $ and rewrite the equation in the form

$$ a \left ( x + \frac{d}{a} \right) + b (y - 0) + c (z - 0) = 0 $$

$-\frac{d}{a} \mathbf{i}+0 \mathbf{j}+0 \mathbf{k}$

Vectors

You must be signed in to discuss.

Johns Hopkins University

Campbell University

Harvey Mudd College

Boston College

mhm. And the question they're saying if abc are not all zero to show that the equation X plus v bypasses a plus the is equal to zero re presents a plane and abc is a normal vector to the plane. So if is not equal to zero then X plus B by let's see that plus D is equal to zero means that E we take common E. And uh within the common done it is X plus D by a Plus B into Y -0 0 plus Seen to Zed zero equal to zero. So this is the scalar equation of the plane through the point. Okay minus debate. A 00 with normal vector A. B. C. Similarly. Mhm. Mhm. If we put the value of B is not equal to zero so Or if she is not equal to zero then the equation of the plane can be rewritten as equation of the plane can be rewritten as E x minus zero plus B, Y plus being two, Y plus D by B plus see into that minus zero is equal to zero. Or if we put the value of C not equal to zero then the equation can be written as a into x minus zero plus being two, Y minus zero plus Scene two. Yeah they plus D by sea Is equal to zero. So therefore this is the equally scalar equation of a plane. True. The point zero -D by B0 or The zero -D by Sea. People with yeah normal vector. Mhm. A B C. Hence these are the possible equations scalar equations of the plane when abc is not equal to zero, and hence with the normal vector, Abc with the point that the pain passes as minus D by a 000 minus D by b zero or 00 minus D by sea. So this is the answer of the question.