The line $l$ has equation $\frac{x+1}{2}=\frac{y-a}{b}=\frac{z-4}{3}$. The plane$~p$ has equation $x

Question

The line $l$ has equation $\frac{x+1}{2}=\frac{y-a}{b}=\frac{z-4}{3}$. The plane$~p$ has equation $x+y-z-11=0$. If $l$ and $~p$ do not intersect, what can be said about the values of $a$ and $b$? [3] For the rest of the question, use $a=-2$ and $b=1$. Find the coordinates of the point $B$, the foot of perpendicular from the point $A\left( -1,a,4 \right)$ to $p$. [2] Hence, or otherwise, find a vector equation of the line of reflection of $l$ in $~p$. [2] Find the possible position vectors of the point $C$ on $l$ which is a distance of $\sqrt{164}$ from $B$. [3] Find the angle $BCA$. [2]

Accepted Answer

(i) a ≠ 16, b = 1; (ii) B(5, 4, -2); (iii) r = (11, 10, -8) + t(2, 1, 3); (iv) OC = (3, 0, 10) 	ext{ or } (-5, -4, -2); (v) 54.2°

2021 YIJC P2 Q4

2021 YIJC P2 Q4

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