Plane intersections and orthogonal directions · source reference

Question

Planes $\Pi_1,\Pi_2$ are $\mathbf r\cdot(2,4,1)=10$ and $\mathbf r\cdot(1,3,1)=8$.

(i) Find their acute angle. [3]

(ii) Obtain a vector equation of intersection line $l_1$. [4]

Another line $l_2$ is $(x-2)/6=(7-z)/3$, $y=m$, m real.

(iii) If $\Pi_1$ and $l_2$ intersect at $(6,m,5)$, find m. [2]

(iv) Show $l_1,l_2$ are perpendicular for all m. [2]

[2010/DHS/Prelim/II/4]

Source working

(i) Let θ be the acute angle between planes.

$$\cos\theta=\frac{|(2,4,1)\cdot(1,3,1)|}{\sqrt{21}\sqrt{11}}=\frac{15}{\sqrt{21}\sqrt{11}},$$

so $\theta=9.3^\circ$ to nearest $0.1^\circ$.

(ii) Solve $2x+4y+z=10$, $x+3y+z=8$. Using GC: $x=-1+t/2$, $y=3-t/2$, $z=t$.

$$l_1:\mathbf r=(-1,3,0)+\alpha(1,-1,2),\quad\alpha=t/2\in\mathbb R.$$

Alternative: Direction $(2,4,1)\times(1,3,1)=(1,-1,2)$. Set $z=0$: $2x+4y=10$, $x+3y=8$ give $x=-1,y=3$, so $(-1,3,0)$ lies on both planes. The same vector equation follows.

(iii) $(6,m,5)\cdot(2,4,1)=10\Rightarrow12+4m+5=10\Rightarrow m=-7/4$.

(iv)

$$l_2:\mathbf r=(2,m,7)+\beta(2,0,-1),\quad\beta\in\mathbb R.$$ $$\mathbf d_1\cdot\mathbf d_2=(1,-1,2)\cdot(2,0,-1)=2-2=0,$$

independent of m. The printed conclusion calls the lines perpendicular for all real m.

Qualification

The directions are orthogonal for all m. The lines themselves intersect only when $m=-2/5$: equating coordinates gives $\alpha=17/5$, $\beta=1/5$. For other m they are skew with perpendicular directions, so “perpendicular lines” needs that directional interpretation.

Teacher notes

Define θ. Show the dot-product symbol clearly in the working.

GC simultaneous-equation instructions: press apps; choose 9↓PlySmlt2; choose 2: Simultaneous Eqn Solver; set Eqns: 2, Unknowns: 3; press NEXT; key in equations; press SOLVE. $x_1=x,x_2=y,x_3=z$.

Alternative construction: $l_1$ lies on both planes, hence is perpendicular to both normals. Its direction is $\mathbf n_1\times\mathbf n_2$. Put one variable to zero for a simple point on the line.

For (iii), substitute $(6,m,5)$ into $\Pi_1$ because it lies there.

Printed answer key

(i) $9.3^\circ$; (ii) $l_1:\mathbf r=(-1,3,0)+\alpha(1,-1,2)$, $\alpha\in\mathbb R$; (iii) $m=-7/4$. The short key prints the (iv) label without an extra numerical answer.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question document pages 3; solution document pages 12,13. Original page numbers are retained as text metadata.