15 Gegeven zijn de lijnen \( I:\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\lambda\left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right), \quad m:\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}0 \\ 3 \\ -3\end{array}\right)+\mu\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right) \) en \( n:\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}3 \\ 0 \\ 3\end{array}\right)+v\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right) \) a. Geef een vectorvoorstelling van de lijn \( k \), die door het punt \( P(2,2,2) \) gaat en die de lijnen \( l \) en \( m \) snijdt. b. Bereken \( d(k, n) \)

We are given the three lines \[ I:\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \lambda \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \quad m:\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ -3 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \quad n:\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ 3 \end{pmatrix} + v \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}. \] In part (a) we need to find a vector equation for the line \( k \) that passes through the point \( P(2,2,2) \) and that intersects both the line \( I \) and the line \( m \). (In the original statement the letter “l” is mentioned but we assume that it should be \( I \).) We let - \( Q=(0,0,\lambda) \) be the point where \( k \) meets \( I \), - \( R=(\mu,3,-3) \) be the point where \( k \) meets \( m \). Since \( k \) goes through \( P \), \( Q \), and \( R \) they must be collinear. This means the vectors \( \vec{PQ} \) and \( \vec{PR} \) are proportional. 1. Write the vectors: - \( \vec{PQ} = Q - P = \begin{pmatrix} 0-2 \\ 0-2 \\ \lambda-2 \end{pmatrix} = \begin{pmatrix} -2 \\ -2 \\ \lambda-2 \end{pmatrix} \). - \( \vec{PR} = R - P = \begin{pmatrix} \mu-2 \\ 3-2 \\ -3-2 \end{pmatrix} = \begin{pmatrix} \mu-2 \\ 1 \\ -5 \end{pmatrix} \). 2. Since \( \vec{PQ} \) and \( \vec{PR} \) are collinear, there exists a scalar \( \alpha \) such that \[ \begin{pmatrix} -2 \\ -2 \\ \lambda-2 \end{pmatrix} = \alpha \begin{pmatrix} \mu-2 \\ 1 \\ -5 \end{pmatrix}. \] Equate components: \[ \begin{cases} -2 = \alpha (\mu-2), \\ -2 = \alpha (1), \\ \lambda-2 = \alpha(-5). \end{cases} \] 3. From the second equation: \[ \alpha = -2. \] 4. Substitute \(\alpha=-2\) into the first equation: \[ -2 = -2(\mu-2) \quad \Longrightarrow \quad 1 = \mu-2 \quad \Longrightarrow \quad \mu = 3. \] 5. Now substitute \(\alpha=-2\) into the third equation: \[ \lambda-2 = -2(-5) = 10 \quad \Longrightarrow \quad \lambda = 12. \] Thus, the intersection points are \[ Q=(0,0,12) \quad \text{and} \quad R=(3,3,-3). \] A convenient choice for the line \( k \) is to use the point \( P(2,2,2) \) with direction vector given by \[ \vec{d}_k = R-P = \begin{pmatrix} 3-2 \\ 3-2 \\ -3-2 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ -5 \end{pmatrix}. \] Thus, the vector equation of the line \( k \) is: \[ k:\quad \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} + t \begin{pmatrix} 1 \\ 1 \\ -5 \end{pmatrix},\quad t\in\mathbb{R}. \] --- For part (b), we need to compute the distance \( d(k,n) \) between the line \( k \) and the line \( n \). Recall that the distance between two skew (or non-intersecting) lines given by \[ k: \vec{r} = \vec{a} + t \vec{d}_k \quad \text{and} \quad n: \vec{r} = \vec{b} + v \vec{d}_n, \] is \[ d(k,n)=\frac{\left| (\vec{b}-\vec{a})\cdot (\vec{d}_k\times \vec{d}_n)\right|}{\|\vec{d}_k\times\vec{d}_n\|}. \] Let \[ \vec{a} = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} \quad \text{(a point on \( k \))}, \qquad \vec{d}_k = \begin{pmatrix} 1 \\ 1 \\ -5 \end{pmatrix}, \] \[ \vec{b} = \begin{pmatrix} 3 \\ 0 \\ 3 \end{pmatrix} \quad \text{(a point on \( n \))}, \qquad \vec{d}_n = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}. \] 1. Compute \(\vec{b}-\vec{a}\): \[ \vec{b} - \vec{a} = \begin{pmatrix} 3-2 \\ 0-2 \\ 3-2 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}. \] 2. Compute the cross product \(\vec{d}_k \times \vec{d}_n\): \[ \vec{d}_k \times \vec{d}_n = \begin{pmatrix} 1 \\ 1 \\ -5 \end{pmatrix} \times \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}. \] Using the formula for the cross product, we get: - \( x\)-component: \[ (1)(0)-(-5)(1)= 0+5 = 5, \] - \( y\)-component: \[ -\left[(1)(0)-(-5)(0)\right]= -\left[0-0\right]= 0, \] - \( z\)-component: \[ (1)(1) - (1)(0)= 1-0= 1. \] Thus, \[ \vec{d}_k \times \vec{d}_n = \begin{pmatrix} 5 \\ 0 \\ 1 \end{pmatrix}. \] 3. Compute the norm of the cross product: \[ \|\vec{d}_k \times \vec{d}_n\| = \sqrt{5^2+0^2+1^2} = \sqrt{25+1} = \sqrt{26}. \] 4. Now, compute the dot product \((\vec{b}-\vec{a})\cdot (\vec{d}_k \times \vec{d}_n)\): \[ \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 0 \\ 1 \end{pmatrix} = 1\cdot5 + (-2)\cdot0 + 1\cdot1 = 5+0+1 = 6. \] 5. Hence, the distance is \[ d(k,n)=\frac{|6|}{\sqrt{26}}=\frac{6}{\sqrt{26}}. \] --- Final answers: a. \( k:\quad \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} + t \begin{pmatrix} 1 \\ 1 \\ -5 \end{pmatrix},\quad t\in\mathbb{R}.\) b. \( d(k,n)=\frac{6}{\sqrt{26}}. \)