The plane ${{p}_{1}}$ has equation $x-2y+3z=5$. The plane ${{p}_{2}}$ contains the points \(A\

Question

The plane ${{p}_{1}}$ has equation $x-2y+3z=5$. The plane ${{p}_{2}}$ contains the points $A$ and $B$ with coordinates $\left( 1,-3,1 \right)$ and $\left( 4,3,-2 \right)$ respectively and is perpendicular to ${{p}_{1}}$. Find the cartesian equation of ${{p}_{2}}$.[3] Find the exact perpendicular distance from $A$ to ${{p}_{1}}$.[2] Find the vector equation of the line $l$ where ${{p}_{1}}$and ${{p}_{2}}$ meet.[2] Given that $C$ is a general point on $l$ find an expression for the square of the distance $BC$. Hence, or otherwise, find the position vector of the point on $l$ which is nearest to $B$.[5]

Accepted Answer

(i) $x-y-z=3$ (ii) $\frac5{\sqrt{14}}$ units (iii) $\mathbf r=\begin{pmatrix}1\-2\0\end{pmatrix}+\lambda\begin{pmatrix}5\4\1\end{pmatrix},\ \lambda\in\mathbb R$ (iv) $BC^2=42\lambda^2-66\lambda+38$, $\overrightarrow{OC}=\frac1{14}\begin{pmatrix}69\16\11\end{pmatrix}$

2015 IJC P1 Q11

2015 IJC P1 Q11

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