Project a point onto an intersection line · source reference

Printed question

Relative to the origin $O$, the point $A$ has position vector $\overrightarrow{OA}=(0,1,0)$ and $A$ lies on the plane $\Pi_1$ with equation defined by $\mathbf r\cdot(1,3,2)=3$. Another plane $\Pi_2$ has equation $y=x$. The planes $\Pi_1,\Pi_2$ intersect at line $l$.

(i) Find a vector equation of the line $l$. [1]

(ii) Show that the cosine of the acute angle between the planes $\Pi_1,\Pi_2$ is $\frac{\sqrt7}{7}$. [2]

(iii) Find the position vector of the foot of perpendicular, $\overrightarrow{OF}$, from point $A$ to the line $l$. Hence, find the exact length of projection of $AF$ onto the plane $\Pi_2$. [5]

(iv) Another plane $\Pi_3$ has equation $px+qy=1$, where $p,q$ are real constants. Find the condition which $p,q$ must satisfy such that $l$ intersects $\Pi_3$ at exactly one point. [2]

[2010/NJC/Prelim/I/9]

Printed working

(i) $\Pi_1:\mathbf r\cdot(1,3,2)=3\Rightarrow x+3y+2z=3$; $\Pi_2:-x+y+0z=0$.

Using GC, $x=\frac34-\frac12t$, $y=\frac34-\frac12t$, $z=t$.

$l:\mathbf r=(3/4,3/4,0)+\lambda(-1/2,-1/2,1)$, $\lambda\in\mathbb R$ or $\mathbf r=(3/4,3/4,0)+\lambda(1,1,-2)$, $\lambda\in\mathbb R$.

(ii) $\Pi_2:-x+y+0z=0\Rightarrow\mathbf r\cdot(-1,1,0)=0$.

Let $\theta$ be acute angle between the planes $\Pi_1,\Pi_2$.

Then $\cos\theta=\frac{|(1,3,2)\cdot(-1,1,0)|}{\sqrt{14}\sqrt2}=\frac{|-1+3|}{\sqrt{28}}=\frac2{2\sqrt7}=\frac{\sqrt7}{7}$ (Shown).

(iii) Since point $F$ lies on line $l$, $\overrightarrow{OF}=(3/4,3/4,0)+\lambda(1,1,-2)$ for some real $\lambda$.

$\overrightarrow{AF}=(3/4,3/4,0)+\lambda(1,1,-2)-(0,1,0)=(3/4,-1/4,0)+\lambda(1,1,-2)$.

$AF\perp l\Rightarrow[(3/4,-1/4,0)+\lambda(1,1,-2)]\cdot(1,1,-2)=0$.

$\Rightarrow(\frac34-\frac14)+(1+1+4)\lambda=0\Rightarrow\lambda=-\frac2{4(6)}=-\frac1{12}$.

$\therefore\overrightarrow{OF}=(3/4,3/4,0)-\frac1{12}(1,1,-2)=\frac1{12}(9-1,9-1,0+2)=\frac1{12}(8,8,2)=(2/3,2/3,1/6)$.

$\overrightarrow{AF}=(3/4,-1/4,0)-\frac1{12}(1,1,-2)=\frac1{12}(9-1,-3-1,0+2)=\frac16(4,-2,1)$.

and $|\overrightarrow{AF}|=\frac16\sqrt{16+4+1}=\frac{\sqrt{21}}6$.

Hence, exact length of projection of $\overrightarrow{AF}$ onto the plane $\Pi_2$

$=|\overrightarrow{AF}|\cos\theta=\frac{\sqrt{21}}6(\frac{\sqrt7}{7})=\frac{7\sqrt3}{6(7)}=\frac{\sqrt3}6$.

(iv) $l:\mathbf r=(3/4,3/4,0)+\lambda(1,1,-2)$ and $\Pi_3:px+qy=1$, i.e. $\mathbf r\cdot(p,q,0)=1$.

$l$ intersects $\Pi_3$ at exactly one point $\Rightarrow l$ is not perpendicular to normal of $\Pi_3$.

$\Rightarrow(1,1,-2)\cdot(p,q,0)\ne0\Rightarrow p+q\ne0\therefore p\ne-q$.

Teacher notes

Using GC: 1) press apps; 2) → 9:PlySmlt2; 3) → 2:Simultaneous Eqn Solver; 4) Put Eqns:2, Unknowns:3; 5) Press “NEXT”; 6) Key in the eqns; 7) Press “SOLVE”. Note: $x_1=x,x_2=y,x_3=z$.

For vector equation of line, must write “$\mathbf r=\ldots,\lambda\in\mathbb R$”.

Angle between the 2 planes is angle between their normals. For acute angle, put in the modulus sign for the dot product of the 2 vectors in the formula for $\cos\theta$.

$l$ intersects $\Pi_3$ at exactly one pt $\Rightarrow l$ is not parallel or lies on $\Pi_3$ $\Rightarrow l$ is not perpendicular to normal of $\Pi_3$.

Printed answer key

(i) $\mathbf r=(3/4,3/4,0)+\lambda(1,1,-2)$, $\lambda\in\mathbb R$. (iii) $(2/3,2/3,1/6)$; $\frac{\sqrt3}6$. (iv) $p\ne-q$. Short key omits (ii).

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question booklet PDF pages 4–12 (printed 80–88); solution booklet PDF pages 17–44. Column vectors are typeset as ordered coordinate triples here; the entries, products, intermediate equations and original parameter/domain restrictions are retained.