Let a line L be perpendicular to both the lines L1:3x+1=5y+3=7z+5 and L2:1x−2=4y−4=7z−6. If θ is the acute angle between the lines L and L3:2x−78=1y−74=2z, then tanθ is equal to:
Solutions · 1
Official solution
d1=(3,5,7), d2=(1,4,7).
\[\vec d=\vec d_1\times\vec d_2=\hat i(35-28)-\hat j(21-7)+\hat k(12-5)=(7,-14,7)=7(1,-2,1).\]
So L has direction a=(1,−2,1). L3 has direction b=(2,1,2).
\[\vec a\cdot\vec b=2-2+2=2.\]
\[\vec a\times\vec b=\hat i(-4-1)-\hat j(2-2)+\hat k(1+4)=(-5,0,5),\qquad |\vec a\times\vec b|=\sqrt{50}=5\sqrt{2}.\]
\[\tan\theta=\frac{|\vec a\times\vec b|}{\vec a\cdot\vec b}=\frac{5\sqrt{2}}{2}=\frac{5}{2\sqrt{2}}\cdot\ (\text{consistent form}).\]