Relative to the origin $O$, a point $A$ has position vector $6\mathbf{i}+t\mathbf{j}+10\mathbf{k}$,

Question

Relative to the origin $O$, a point $A$ has position vector $6\mathbf{i}+t\mathbf{j}+10\mathbf{k}$, where $t$ is a real and negative constant. Given that the direction cosine of $A$ with respect to the $x$-axis is $\frac{3}{\sqrt{55}}$, find the value of $t$. Deduce the angle between $\overrightarrow{OA}$ and the $yz$-plane.[3] The planes ${{p}_{1}}$ and ${{p}_{2}}$ have equations $\mathbf{r}\cdot \left( \begin{matrix} 2 \ -2 \ 1 \ \end{matrix} ight)=8$ and $\mathbf{r}\cdot \left( \begin{matrix} -2 \ 2 \ -1 \ \end{matrix} ight)=k$ respectively, where $k$ is a real constant. It is given that the plane $p$ is equidistant from ${{p}_{1}}$ and ${{p}_{2}}$. Find, in terms of $k$, a vector equation of $p$ in the scalar product form.[3] Given $k=13$, find the coordinates of the point on line $l$ with equation $\frac{x-1}{3}=\frac{2y-1}{18}=\frac{3-z}{1}$ that is equidistant from ${{p}_{1}}$ and ${{p}_{2}}$.[3]

Accepted Answer

t = -2, θ = 30.5°; p: r·(2,-2,1) = (1/2)(8-k); Intersection point (2.5, 5, 2.5)

2019 NJC P1 Q6

2019 NJC P1 Q6

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