<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

Prove that a vector equation of the plane containing the point
${\bf{a}}={\bf{j}}+2{\bf{k}}$ and which contains the vectors
${\bf{b}}=2{\bf{i}}-4{\bf{j}}+{\bf{k}}$ and
${\bf{c}}=3{\bf{i}}+{\bf{j}}+5{\bf{k}}$ is given by ${\bf{r}}
\cdot (-3{\bf{i}}-{\bf{j}}+2{\bf{k}})=3$.

Find the position vector of the point of intersection of this
plane with the line

$${\bf{r}}={\bf{i}}+{\bf{j}}+{\bf{k}}+\lambda(-{\bf{i}}+{\bf{k}}).$$

What is the angle between this position vector and the normal to
the plane?

\medskip

{\bf Answer}

${}$

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${\bf{b}} and {\bf{c}}$ lie in a plane.

$d={\bf{a}} \cdot \hat{{\bf{n}}}$ where ${\bf{n}}$ is a unit
vector normal to the plane.

A vector normal to the plane is given by ${\bf{b}} \times
{\bf{c}}$.

\begin{eqnarray*} {\bf{b}} \times {\bf{c}} &amp; = &amp;
\left|\begin{array}{ccc} {\bf{i}} &amp; {\bf{j}} &amp; {\bf{k}}\\ 2 &amp; -4 &amp;
1\\ 3 &amp; 1 &amp; 5 \end{array} \right|\\ &amp; = &amp;
{\bf{i}}\left|\begin{array}{cc} -4 &amp; 1\\1 &amp; 5 \end{array}\right| -
{\bf{j}}\left|\begin{array}{cc} 2 &amp; 1\\3 &amp; 5 \end{array}\right| +
{\bf{k}}\left|\begin{array}{cc} 2 &amp; -4\\3 &amp; 1 \end{array}\right|\\
&amp; = &amp; -21{\bf{i}}-7{\bf{j}}+14{\bf{k}}. \end{eqnarray*}

$\hat{{\bf{n}}}=\ds\frac{-21{\bf{i}}-7{\bf{j}}+14{\bf{k}}}{\sqrt{(-21)^2+(-7)^2+(14)^2}}$

\begin{eqnarray*} d={\bf{k}} \cdot \hat{{\bf{k}}} &amp; = &amp; (0,+1,2) \cdot
\ds\frac{(-21,-7,14)}{\sqrt{21^2+7^2+14^2}}\\ &amp; = &amp;
\ds\frac{-7+28}{\sqrt{21^2+7^2+14^2}}\\ &amp; = &amp;
\ds\frac{21}{\sqrt{21^2+7^2+14^2}} \end{eqnarray*}

Therefore

${\bf{r}} \cdot
\ds\frac{(-21{\bf{i}}-7{\bf{j}}+14{\bf{k}})}{\sqrt{21^2+7^2+14^2}}
=\ds\frac{21}{\sqrt{21^2+7^2+14^2}}$

$\Rightarrow {\bf{k}} \cdot (-3{\bf{i}}-{\bf{j}}+2{\bf{k}})=3$

\newpage
Intersection of plane and line is given by

$[({\bf{i}}+{\bf{j}}+{\bf{k}})+\lambda(-{\bf{i}}+{\bf{k}})]
\cdot(-3{\bf{i}}-{\bf{j}}+2{\bf{k}})=3$

$((1-\lambda),1,1+\lambda) \cdot (-3,-1,2)=3$

$\Rightarrow -3+3\lambda-1+2+2\lambda=3$

$\Rightarrow -2+5\lambda=3$

$\Rightarrow \un{\lambda=1}$

Therefore
\begin{eqnarray*} {\bf{r}}_p &amp; = &amp;
{\bf{i}}+{\bf{j}}+{\bf{k}}+1(-{\bf{j}}+{\bf{k}})\\ &amp; = &amp;
{\bf{j}}+2{\bf{k}} \end{eqnarray*}

Angle between this and the normal vector is given by,$\theta$
where

$${\bf{r}}_p \cdot {\bf{n}}=|{\bf{r}}_p||{\bf{n}}|\cos\theta$$

${\bf{r}}_p \cdot {\bf{n}}=(0,1,2) \cdot (-3,-1,2)=3$

$|{\bf{r}}_p|=\sqrt{1^2+2^2}=\sqrt{5}$

$|{\bf{n}}|=\sqrt{3^2+1^2+2^2}=\sqrt{14}$

Therefore $\cos \theta=\ds\frac{3}{\sqrt{5}\sqrt{14}}$

$\Rightarrow
\theta=\arccos\left(\ds\frac{3}{\sqrt{5}\sqrt{14}}\right)=68.98^{\circ}
\approx \un{69^{\circ}}$


\end{document}
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