$\begingroup$

How do you convert equations of planes from cartesian to vector form?

For example, $7x + y + 4z = 31$ that passes through the point $(1,4,5)$

is $(1,4,5) + s(4,0,-7) + t(0,4,-1)$ , $s$,$t$ in $\mathbb{R}$

$\endgroup$

2 Answers

$\begingroup$

Equation of a plane in vector form is like $$(\vec r -\vec a)\cdot \hat n=0$$ or, $$\vec r \cdot \hat n=\vec a\cdot \hat n$$ or, $$\vec r \cdot \hat n=d$$

So you should proceed as $$7x+y+4z=31$$ or, $$(x\hat i+y \hat j+z \hat k)\cdot(7\hat i+ \hat j+ 4\hat k)=(\hat i+4 \hat j+5 \hat k)\cdot(7\hat i+ \hat j+4 \hat k)$$ or, $$\vec r\cdot(7\hat i+ \hat j+ 4\hat k)=(\hat i+4 \hat j+5 \hat k)\cdot(7\hat i+ \hat j+4 \hat k)$$ This is 1 way of converting cartesian to polar. The one in your question is another.

Observe the position vector in your question is same as the point given and the other 2 vectors are those which are perpendicular to normal of the plane.Now the normal has been found out. Use simple tricks like trial and error to find the d.c.s of the vectors. For example, put the x-component 0 and then go on or make the y-component 0 and proceed. The equation may not be same but it will equally represent the same straight line.

$\endgroup$ 1 $\begingroup$

ax + by + cz = k => z = (k -ax - by)/c => ( x , y , z ) = (x , y , (k - ax -by)/c)

= x(1, O , -a/c) + y(O , 1 , -b/c) + ( O , O , k/c)

= s(c , O , -a) + t( O , c , -b) + ( O , O , k/c)

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy