Question #211907

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.


1
Expert's answer
2021-07-04T16:31:56-0400

Q1. Since we need to find the plane which is parallel to the given plane x+3y2z=6-x+3y-2z=6 . I can write it as, r.(i^+3j^2k^)=6\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+3y2z=c-x+3y-2z=c where c is constant.

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

So, the equation will be, x+3y2z=0-x+3y-2z=0




Q2. Distance of the point (x1,x2,x3)(x_1,x_2,x_3) from the plane Ax+By+Cz+D=0Ax+By+Cz+D = 0 is given by, d=Ax1+Bx2+Cx3+DA2+B2+C2d = \frac{|Ax_1+Bx_2+Cx_3+D|}{\sqrt{A^2+B^2+C^2}}

Simply putting values, we get, d=(1×3)+2+(4×0)+292+(1)2+42=3+2+214=114d = \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>=0V.<3,-1>=0

Let V=<a,b>V = <a,b>

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

3a=b3a = b.

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

Then unit vector will be, V^=<V>V=<a,b>a2+b2\hat{V} = \frac{<V>}{||V||} = \frac{<a,b>}{\sqrt{a^2+b^2}}

Since b = 3a, then,

V^=<a,b>a2+b2=<a,3a>a2+9a2=<1,3>10\hat{V}= \frac{<a,b>}{\sqrt{a^2+b^2}} = \frac{<a,3a>}{\sqrt{a^2+9a^2}} = \frac{<1,3>}{\sqrt{10}}




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