The plane ${{p}_{1}}$ has equation $2x-3y+z=8$ and it meets the $x$- and $z$- axes at the po

Question

The plane ${{p}_{1}}$ has equation $2x-3y+z=8$ and it meets the $x$- and $z$- axes at the points $A$ and $B$ respectively. State the position vectors of $A$ and $B$, relative to the origin $O$, and show that a Cartesian equation of the line $AB$ is $4-x=\frac{z}{2}$, $y=0$.[4] The point $C$ lies on $AB$ such that $\frac{AC}{CB}=\frac{1}{3}$. The plane ${{p}_{2}}$ passes through $C$ and is parallel to the vector $\mathbf{i}-2\mathbf{j}$. Given that ${{p}_{2}}$ is also perpendicular to ${{p}_{1}}$, find the cartesian equation of ${{p}_{2}}$,[5] the perpendicular distance from $A$ to ${{p}_{2}}$.[3]

Accepted Answer

$A=(4,0,0)$, $B=(0,0,8)$, $AB:4-x=\dfrac z2$, $y=0$ (i) $2x+y-z=4$ (ii) $\dfrac4{\sqrt6}$ units

2013 IJC P1 Q12

2013 IJC P1 Q12

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