The line $l$ has equation $1-x=\frac{z-\alpha }{2}$, $y=1$, where $\alpha $ is a consta

Question

The line $l$ has equation $1-x=\frac{z-\alpha }{2}$, $y=1$, where $\alpha $ is a constant. Given that the point of intersection between the line $l$ and the $yz$-plane is $\left( 0,\,1,\,5 \right)$, show that $\alpha =3$.[2] Referred to the origin, the points $A$ and $B$ have position vectors $-\mathbf{j}+3\mathbf{k}$ and $12\mathbf{i}+5\mathbf{j}-\mathbf{k}$ respectively. Find the cartesian equation of the plane ${{\pi }_{1}}$ which contains the point $A$ and the line $l$.[3] Find the acute angle between the line $AB$ and ${{\pi }_{1}}$.[3] The plane ${{\pi }_{2}}$ has cartesian equation $x+2z=5$. Find the cartesian equations of the planes such that the perpendicular distance from each plane to ${{\pi }_{2}}$ is $2\sqrt{5}$.[4]

Accepted Answer

(i) $\alpha =3$ (ii) $2x-y+z=4$ or $-2x+y-z=-4$ (iii) ${{24.1}^{\circ }}$ (iv) $x+2z=15$ or $x+2z=-5$

2022 NYJC Promo Q9

2022 NYJC Promo Q9

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