Planes $p$ and $q$ are perpendicular. Plane $p$ has equation $x+2y-3z=12$. Plane $q$ conta

Question

Planes $p$ and $q$ are perpendicular. Plane $p$ has equation $x+2y-3z=12$. Plane $q$ contains the line $l$ with equation $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{4}$. The point $A$ on $l$ has coordinates $\left( 1,-1,3 \right)$. Find a cartesian equation of $q$.[4] Find a vector equation of the line $m$ where $p$ and $q$ meet.[4] $B$ is a general point on $m$. Find an expression for the square of the distance $AB$. Hence, or otherwise, find the coordinates of the point on $m$ which is nearest to $A$.[5]

Accepted Answer

(i) $-x+2y+z=0$ (ii) $\mathbf{r}=\left( \begin{matrix} \begin{matrix} 6 \ 3 \ 0 \ \end{matrix} \ \end{matrix} \right)+\mu \left( \begin{matrix} \begin{matrix} 4 \ 1 \ 2 \ \end{matrix} \ \end{matrix} \right)$, $\mu \in \mathbb{R}$ (iii) $21{{\mu }^{2}}+36\mu +50$; $\left( \frac{18}{7},\frac{15}{7},-\frac{12}{7} \right)$

RI Vectors III Planes Tutorial Q3

RI Vectors III Planes Tutorial Q3

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