The planes $p_1$, $p_2$ and $p_3$ have equations $$p_1 : -x + 2y + 4z = 5,$$\[p_2 : 3x + ay -

Question

The planes $p_1$, $p_2$ and $p_3$ have equations $$p_1 : -x + 2y + 4z = 5,$$$$p_2 : 3x + ay - 6z = 6,$$$$p_3 : 2x - 4y - 8z = b,$$where $a$ and $b$ are real constants, $b > 0$. The line $l$ has equation $\frac{x - 5}{8} = \frac{3 - y}{6} = \frac{z - 1}{5}$. Given that the planes $p_1$ and $p_2$ intersect at $l$, show that $a = -1$.[3] Find the acute angle between $p_1$ and $p_2$.[2] The plane $p_4$ has equation $\mathbf{r} \cdot \begin{pmatrix} 1 \ -2 \ -4 \end{pmatrix} = 0$. Given that $p_1$ and $p_3$ are equidistant from $p_4$, find the value of $b$.[2] The point $A$ with coordinates $(-1, 0, 1)$ lies on $p_1$, and $F$ is the foot of the perpendicular from $A$ to $l$. The point $B$ is the reflection of $F$ in the plane $p_3$. Find the length of $AF$ and hence find the area of triangle $AFB$.[5]

Accepted Answer

(i) $a=-1$ (ii) $21.1^\circ$ (iii) $b=10$ (iv) $AF=\dfrac35\sqrt{105}$ units, area $=6\sqrt5$ units${}^2$

2018 HCI CT Q8

2018 HCI CT Q8

Video Solution:

Solution:

Solution locked