Question

Obtain the equation of the plane passing through the line x-2/2=-(y+1)/1 z-3/4 and
which is perpendicular to the plane x+2y+z=4

Answer 1

Answer on Question #46170 – Math – Analytic Geometry

Question. Obtain the equation of the plane $Q$ passing through the line

$L:\ \frac{x-2}{2}=-\frac{y+1}{1}=\frac{z-3}{4}$

and which is perpendicular to the plane $P:\ x+2y+z=4$.

Solution. We have that

- the line $L$ passes through a point $A(2,-1,3)$ in the direction of the vector $l(2,-1,4)$,

- the normal vector of the plane $P$ has coordinates $p(1,2,1)$.

Let $n(a,b,c)$ be normal vector of the plane $Q$ passing through the line $L$ and perpendicular to $Q$. Then $Q$ passes through point $A$, whence its equation has the following form:

$a(x-2)+b(y+1)+c(z-3)=0.$

Notice that $n$ must be perpendicular to both vectors $l(2,-1,4)$ and $p(1,2,1)$, and therefore we can choose $n$ to be the cross product of these vectors:

$n=l\times p.$

Thus

\[ n=l\times p=(2,-1,4)\times(1,2,1)=\left(\begin{vmatrix}-1&4\\

2&1\end{vmatrix},\begin{vmatrix}4&2\\

1&1\end{vmatrix},\begin{vmatrix}2&-1\\

1&2\end{vmatrix}\right) \]

$=(-1\cdot 1-2\cdot 4,\ 4\cdot 1-1\cdot 2,\ 2\cdot 2-1\cdot(-1))=(-9,2,5)\,.$

Hence $Q$ has the following equation:

$-9(x-2)+2(y+1)+5(z-3)=0$

$-9x+18+2y+2+5z-15=0$

$-9x+2y+5z+5=0.$

Answer. $-9x+2y+5z+5=0$.

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