Finding a plane parallel to one given line and contains other given line

Question

For some reason I can't figure out how to Find a plane which contains L1:\begin{cases} 3x+3y+5z=8 & (1) \\ 5x+5y -3z=2 & (2)\end{cases}

and parallel to the direction of L2: (34,34,4)

I know that the cross product of L1 gives me the plane vector -34i + 34j +0k, but i'm getting stuck there. Would really appreciate some help.

Answer 1

Let $\vec{l_1}(a,b,c)$ be a direction of $l_1$.

Thus, $$3a+3b+5c=0$$ and $$5a+5b-3c=0,$$ which gives $$\vec{l_1}(1,-1,0).$$ Now, let $\vec{n}(a,b,c)$ be a normal of the plane.

Thus, $$17a+17b+2c=0$$ and $$a-b=0,$$ which gives $$\vec{n}(1,1,-17)$$ and since a point $(1,0,1)$ is placed on the plane, we obtain the following equation: $$1(x-1)+1(y-0)-17(z-1)=0$$ or $$x+y-17z+16=0.$$

$(1,0,1)$ is placed on the plane because $$3\cdot1+3\cdot0+5\cdot1=8$$ and $$5\cdot1+5\cdot0-3\cdot1=2.$$

Answer 2

Let $$P_1: 3x+3y+5z=8,~~P_2: 5x+5y-3z=2$$ Eq. of family of plane through the intersection of $P_1$ and $P_2$ is $$P_3: P_1+tP_2=0 \implies (3+5t)x+(3+5t)y+(5-3t)z-(8-2t)=0~~~~(1)$$ since $P_3$ contains a line with drs $\vec L\{34,34,4\}$ it is perpendicular to the normar to the plane $\vec N=\{3+5t, 3+5t,5-3t\}$ So $$\vec N . \vec L=0 \implies 68(3+5t)+4(5-3t)=0 \implies t=-\frac{28}{41} $$ Putting this value of $t$ in (1) we get the required plane.