Separate plane angles, intersections and orthogonal directions

The normals have dot product 15 and magnitudes√21,√11. Thus $\alpha=\cos^{-1}[15/(\sqrt{21}\sqrt{11})]=0.162$ rad(3s.f.), or 9.3°(1d.p.). This acute normal angle is the acute angle between planes.

$2x+4y+z=10$ and $x+3y+z=8$ give $x=-1+z/2$, $y=3-z/2$. Taking z as a free parameter yields $l_1:\mathbf r=(-1,3,0)+\lambda(1/2,-1/2,1)$, λ real.

$2(6)+4m+5=10$ gives m=−7/4. The point also satisfies l2 since $(6-2)/6=(7-5)/3=2/3$.

Writing the common ratio of l2 as μ gives $l_2:\mathbf r=(2,m,7)+\mu(6,0,-3)$. The direction dot product is $(1/2,-1/2,1)\cdot(6,0,-3)=3-3=0$, independent of m. Thus their directions are orthogonal for all m, as intended by the printed proof.

The two lines intersect only when μ=1/15, λ=34/5 and m=−2/5, found by matching x,z then y. For other m they are skew, despite the orthogonal directions. Therefore “perpendicular for all m” must be read as the angle between directions, not as intersecting at a right angle for every m.