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Mathematics20273 marksShort answerThree-dimensional GeometryTool: 3D geometry and vectors calculator
Solve the following.
The planes π₁ and π₂ have equations 2x + 6y - 2z = 5 and 3x + 9y + pz = -(51)/2 respectively…
The planes $\pi_1$ and $\pi_2$ have equations $2x + 6y - 2z = 5$ and $3x + 9y + pz = -\frac{51}{2}$ respectively.
(a)[1]
Verify that the point $P(2, \frac{1}{2}, 1)$ lies on the plane $\pi_1$.
(b)[1]
Determine the value of $p$ if $\pi_2$ is parallel to $\pi_1$.
(c)[1]
A line through P normal to $\pi_1$ meets $\pi_2$ at the point Q. Find the coordinates of Q.
Answer
Answer (a)
Official answer keyFor $P\left(2, \frac{1}{2}, 1\right)$: $2(2) + 6\left(\frac{1}{2}\right) - 2(1) = 4 + 3 - 2 = 5$, which equals the right-hand side, so $P$ lies on $\pi_1$.
Answer (b)
Official answer keyFor parallel planes the coefficients are proportional: $\frac{3}{2} = \frac{9}{6} = \frac{p}{-2}$, so $p = -3$.
Answer (c)
Official answer keyThe line through $P$ normal to $\pi_1$ has direction $(2, 6, -2) \parallel (1, 3, -1)$: $\frac{x-2}{1} = \frac{y-\frac{1}{2}}{3} = \frac{z-1}{-1} = \lambda$, so a general point is $\left(\lambda + 2, 3\lambda + \frac{1}{2}, 1 - \lambda\right)$.
It lies on $\pi_2: 3x + 9y - 3z = -\frac{51}{2}$: $3(\lambda+2) + 9\left(3\lambda + \frac{1}{2}\right) - 3(1-\lambda) = -\frac{51}{2}$, so $33\lambda + \frac{15}{2} = -\frac{51}{2}$ and $\lambda = -1$.
$Q = \left(1, -\frac{5}{2}, 2\right)$.