Equations of Lines — Question 4

PDF ↗

Question 4

The lines L1:r→=⟨1,2,−1⟩+s⟨2,−1,2⟩,L2:r→=⟨5,0,3⟩+t⟨1,2,−2⟩L_1:\vec r=\left\langle 1,2,-1\right\rangle+s\left\langle 2,-1,2\right\rangle,\qquad L_2:\vec r=\left\langle 5,0,3\right\rangle+t\left\langle 1,2,-2\right\rangle intersect. Find their intersection and the acute angle between them.

Original worksheet page 1: question and worked solution for 6-2-004
Show solutionHide solution

Question 4 – Solution

Strategy Solve the coordinate equations for the intersection; use direction vectors for the angle.

See the diagram in the original worksheet below.

Intersection Equating coordinates gives 1+2s=5+t1+2s=5+t, 2−s=2t2-s=2t, and −1+2s=3−2t-1+2s=3-2t. The solution is s=2,t=0s=2,t=0, so Q=(5,0,3)\boxed{Q=(5,0,3)}.

Angle With d→1=⟨2,−1,2⟩\vec d_1=\left\langle 2,-1,2\right\rangle and d→2=⟨1,2,−2⟩\vec d_2=\left\langle 1,2,-2\right\rangle, |d→1⋅d→2|=4,∥d→1∥=3,∥d→2∥=3.|\vec d_1\cdot\vec d_2|=4,\quad \|\vec d_1\|=3,\quad\|\vec d_2\|=3. Hence θ=cos⁡−1(4/9)≈63.6∘\boxed{\theta=\cos^{-1}(4/9)\approx 63.6^\circ}.

Verification Substituting s=2s=2 and t=0t=0 gives the same point on both lines.

Original worksheet page 2: question and worked solution for 6-2-004

Original worksheet layout. Use Enlarge or open the PDF for a closer view.