Chapter 11.4, Problem 26E

To determine

To find: The general form of the equation of a place passing through the given point and perpendicular to the specified vector.

Answer to Problem 26E

$-x+y-2z+12=0$

Explanation of Solution

Given information:

A plane is passing through point $(0,0,6)$ and is perpendicular to the vector $x=1-t,y=2+t,z=4-2t$ .

Concepts used:

A plane containing the point $P(x_1,y_1,z_1)$ perpendicular to the normal vector $\text{v}=\langle a,b,c\rangle$

can be represented by standard form of the equation of a plane −

$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$ .

Calculation:

Let plane is passing through the points $P$

$(0,0,6)$ and is perpendicular to the vector $x=1-t,y=2+t,z=4-2t$ .

So, $x_1=0,y_1=0,z_1=6$ .

Line is perpendicular to the normalvector $x=1-t,y=2+t,z=4-2t$ .

So, direction numbers are $a=-1,b=1,c=-2$ .

Therefore, general form of the equation of plane will be given by −

$\begin{array}{l}a(x-x_1)+b(y-y_1)+c(z-z_1)=0\\ \Rightarrow -1(x-0)+1(y-0)-2(z-6)=0\\ \Rightarrow -x+y-2z+12=0\end{array}$

Hence, desired equation of plane is $-x+y-2z+12=0$ .