Planes containing a prescribed line · source reference

Printed question

The equations of the line $l_1$ and the plane $p_1$ are respectively, $\mathbf r=(1,0,0)+\mu(-1,1,1)$ and $\mathbf r=s(1,1,1)+t(2,-1,1)$, where $\mu,s,t\in\mathbb R$.

(i) Find the acute angle between $l_1,p_1$. [3]

(ii) A second plane $p_2$ has equation $\mathbf r\cdot(\alpha,1,\beta)=1$. Given that the two planes $p_1,p_2$ intersect at the line $l_2:\frac{4x-15}{-2}=y,z=\frac52$, find the values of $\alpha,\beta$. [3]

(iii) The plane $p_3$ with equation $2x+by+z=1$ is parallel to $l_2$. Find the value of $b$. Hence find the distance between $l_2,p_3$. [4]

[2010/CJC/Prelim/II/2]

Printed working

(i) A normal to $p_1=(1,1,1)\times(2,-1,1)=(2,1,-3)$.

A direction vector of $l_1=(-1,1,1)$.

Acute angle between $l_1$ and normal of $p_1$

$=\cos^{-1}\frac{|(2,1,-3)\cdot(-1,1,1)|}{\sqrt{(4+1+9)(1+1+1)}}=51.9^\circ$.

Hence acute angle between $l_1,p_1=90^\circ-51.9^\circ=38.1^\circ$ (to 1 d.p.).

(ii) $p_2:\mathbf r\cdot(\alpha,1,\beta)=1$.

$l_2:\mathbf r=(15/4,0,5/2)+a(-1,2,0)$ lies on $p_1,p_2$.

Thus $l_2$ is perpendicular to normal of $p_2$ and $(15/4,0,5/2)$ lies in $p_2$.

$\Rightarrow(-1,2,0)\cdot(\alpha,1,\beta)=0\Rightarrow-\alpha+2=0\Rightarrow\alpha=2$.

and $(15/4,0,5/2)\cdot(\alpha,1,\beta)=1\Rightarrow\frac{15}4\alpha+\frac52\beta=1\Rightarrow\beta=-\frac{13}5$.

Printed alternative solution

$(15/4,0,5/2)$ and $(4,-1/2,5/2)$ (using $a=-\frac14$) are two points on line $l_2$, and thus are points on $p_2$.

Substituting each point in equation of plane,

$3.75\alpha+2.5\beta=1$ — (1).

$3.25\alpha+2.5\beta=0$ — (2).

Using GC, $\alpha=2,\beta=-2.6$.

(iii) $p_3$ is parallel to $l_2\Rightarrow$ direction vector of $l_2$ is perpendicular to normal of $p_3$.

$(2,b,1)\cdot(-1,2,0)=0\Rightarrow b=1$.

$\therefore$ Equation of plane $p_3:\mathbf r\cdot(2,1,1)=1$.

Observe that $B(0,1,0)$ is a point on $p_3$.

$\overrightarrow{AB}=(0,1,0)-(15/4,0,5/2)=(-15/4,1,-5/2)$.

Thus, distance between $l_2,p_3=$ perpendicular distance from $A$ to $p_3$

$=\frac{|\overrightarrow{AB}\cdot(2,1,1)|}{|(2,1,1)|}=\frac1{\sqrt{4+1+1}}|(-15/4,1,-5/2)\cdot(2,1,1)|=\frac{|-15/2+1-5/2|}{\sqrt6}=\frac9{\sqrt6}=\frac{3\sqrt6}2$.

Teacher notes

Let $\theta$ be the acute angle between a line and a plane $l_1:\mathbf r=\mathbf a+\lambda\mathbf d$, $p_1:\mathbf r\cdot\mathbf n=p$.

$\cos\phi=\frac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}$; $\theta=90^\circ-\phi$.

$B(0,1,0)$ is a point on $p_3$ since it satisfies equation of $p_3$.

Printed answer key

(i) $38.1^\circ$ (to 1 d.p.). (ii) $\alpha=2,\beta=-\frac{13}5$. (iii) $b=1$; $\frac{3\sqrt6}2$.

Editorial qualification

The printed alternative point $(4,-1/2,5/2)$ correctly uses $a=-1/4$. Its second scalar equation is a source typo: substituting this point gives $4\alpha+2.5\beta=3/2$, rather than the printed $3.25\alpha+2.5\beta=0$. Together with the first equation, the corrected system still gives $\alpha=2,\beta=-13/5$.

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.