2008 H2 Mathematics Paper 1 Question 11

Vectors II: Lines and Planes

Answers

(−411,−411,711){\displaystyle \left( - \frac{4}{11}, - \frac{4}{11}, \frac{7}{11} \right)}

(i)

l:r=(−1−10)+λ(111),λ∈R{l: \mathbf{r} = \begin{pmatrix} - 1 \\ - 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \lambda \in \mathbb{R}}

(ii)

λ=−22,  μ=17{\lambda = - 22, \; \mu = 17}

(iii)

λ=−22,  μ≠17,μ∈R{\lambda = - 22, \; \mu \neq 17, \mu \in \mathbb{R}}

(iv)

−3x+y+2z=2{- 3 x + y + 2 z = 2}

Full solutions

2x−5y+3z=33x+2y−5z=−55x−20.9y+17z=16.6\begin{align} \qquad 2x - 5y + 3z &= 3 \\ \qquad 3x + 2y - 5z &= -5 \\ \qquad 5x - 20.9y + 17z &= 16.6 \end{align}

From GC, coordinates of point of intersection: (−411,−411,711){\displaystyle \left( - \frac{4}{11}, - \frac{4}{11}, \frac{7}{11} \right)}

(i)

Solving (1) and (2) with a GC,
l:r=(−1−10)+λ(111),λ∈Rl: \mathbf{r} = \begin{pmatrix} - 1 \\ - 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \lambda \in \mathbb{R}

(ii)

Since all three planes meet in l,l{l, l} lies in p3{p_3}
Hence n3⋅d=0{\mathbf{n_3} \cdot \mathbf{d} = 0}
(5λ17)⋅(111)=0{\begin{pmatrix}5 \\ \lambda \\ 17\end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 0}
22+λ=0{22 + \lambda = 0}
λ=−22{\lambda = - 22}
Since all three planes meet in l{l}, the point in l{l} lies in p3{p_3}
(−1−10)⋅(5−2217)=μ{\begin{pmatrix} - 1 \\ - 1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ - 22 \\ 17 \end{pmatrix} = \mu}
−5+22+0=μ{- 5 + 22 + 0 = \mu}
μ=17{\mu = 17}

(iii)

Since all three planes have no point in common, l{l} is parallel to p3{p_3}
Hence n3⋅d=0{\mathbf{n_3} \cdot \mathbf{d} = 0}
λ=−22{\lambda = - 22}
Since all three planes have no point in common, the point in l{l} does not lie in p3{p_3}
(−1−10)⋅(5−2217)≠μ{\begin{pmatrix} - 1 \\ - 1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ - 22 \\ 17 \end{pmatrix} \neq \mu}
μ≠17{\mu \neq 17}

(iv)

We get a direction vector of the plane by joining the point on l{l} to (1,−1,3),{(1,-1,3),}
d1=(1−13)−(−1−10)=(203)\begin{align*}\mathbf{d_1} &= \begin{pmatrix} 1 \\ - 1 \\ 3 \end{pmatrix} - \begin{pmatrix} - 1 \\ - 1 \\ 0 \end{pmatrix} \\ &= \begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix}\end{align*}
The direction vector of l{l} is also a direction vector of the plane
n=(203)×(111)=(−312)\begin{align*}\mathbf{n} &= \begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \\ &= \begin{pmatrix} - 3 \\ 1 \\ 2 \end{pmatrix}\end{align*}
a⋅n=(1−13)⋅(−312)=2\begin{align*}\mathbf{a}\cdot\mathbf{n} &= \begin{pmatrix} 1 \\ - 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} - 3 \\ 1 \\ 2 \end{pmatrix} \\ &= 2 \end{align*}
p:r⋅(−312)=2{p: \mathbf{r} \cdot \begin{pmatrix} - 3 \\ 1 \\ 2 \end{pmatrix} = 2}
Cartesian equation of plane: −3x+y+2z=2{- 3 x + y + 2 z = 2}