Answer to Question #211907 in Analytic Geometry for Faith

Question

1.find an equation for the plane that passes through the origin (0,0,0) and is parall to the plane -x+3y-2z=6.

2.find the distance between the point (-1,-2,0) and the plane 3x-y+4z=-2.

3.find the components of a unit vector satisfying V.<3,-1>=0.

Answer 1

Q1. Since we need to find the plane which is parallel to the given plane $-x+3y-2z=6$ . I can write it as, $\vec{r}.(-\hat{i}+3\hat{j}-2\hat{k}) = 6$

Since planes are parallel so their normal vector will be parallel.

So the equation will be, $-x+3y-2z=c$ where c is constant.

since, plane passes through origin, then $c = 0$

So, the equation will be, $-x+3y-2z=0$

Q2. Distance of the point $(x_1,x_2,x_3)$ from the plane $Ax+By+Cz+D = 0$ is given by, $d = \frac{|Ax_1+Bx_2+Cx_3+D|}{\sqrt{A^2+B^2+C^2}}$

Simply putting values, we get, $d = \frac{|(-1\times 3)+2+(4\times 0)+2|}{\sqrt{9^2+(-1)^2+4^2}} = \frac{|-3+2+2|}{\sqrt{14}}= \frac{1}{\sqrt{14}}$ units.

Q3. Given condition is, $V.<3,-1>=0$

Let $V = <a,b>$

Since, $<a,b>.<3,-1>=0 \implies 3a-b = 0$

$3a = b$.

So all values which satisfy the above condition are solutions of the V.

Then unit vector will be, $\hat{V} = \frac{<V>}{||V||} = \frac{<a,b>}{\sqrt{a^2+b^2}}$

Since b = 3a, then,

$\hat{V}= \frac{<a,b>}{\sqrt{a^2+b^2}} = \frac{<a,3a>}{\sqrt{a^2+9a^2}} = \frac{<1,3>}{\sqrt{10}}$