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2021 Paper 3 Q2
D: 1500.0 B: 1500.0
3x3 matrices determinant inequality algebraic manipulation matrix equation am-gm inequality

  1. Let \[ x = \frac{a}{b - c}, \qquad y = \frac{b}{c - a} \qquad \text{and} \qquad z = \frac{c}{a - b}, \] where \(a\), \(b\) and \(c\) are distinct real numbers. Show that \[ \begin{pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \] and use this result to deduce that \(yz + zx + xy = -1\). Hence show that \[ \frac{a^2}{(b-c)^2} + \frac{b^2}{(c-a)^2} + \frac{c^2}{(a-b)^2} \geqslant 2. \]
  2. Let \[ x = \frac{2a}{b+c}, \qquad y = \frac{2b}{c+a} \qquad \text{and} \qquad z = \frac{2c}{a+b}, \] where \(a\), \(b\) and \(c\) are positive real numbers. Using a suitable matrix, show that \(xyz + yz + zx + xy = 4\). Hence show that \[ (2a + b + c)(a + 2b + c)(a + b + 2c) > 5(b+c)(c+a)(a+b). \] Show further that \[ (2a + b + c)(a + 2b + c)(a + b + 2c) > 7(b+c)(c+a)(a+b). \]

Solution:

  1. \(\,\) \begin{align*} && \begin{pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} &= \begin{pmatrix} a-xb+xc \\ ay+b-yc \\ -za+zb+c \end{pmatrix} \\ &&&= \begin{pmatrix} a-x(b-c) \\ b-y(c-a) \\ c-z(a-b) \end{pmatrix} \\ &&&= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \end{align*} Notice since \(a,b\) and \(c\) are distinct real numbers the vector \(\langle a,b,c \rangle\) cannot be the zero vector, so the determinant of the matrix is zero, ie \(0= 1(1+yz)+x(y-yz)+x(yz+z) = 1 +yz+yx+zx\). Notice also then that \begin{align*} && \frac{a^2}{(b-c)^2} + \frac{b^2}{(c-a)^2} + \frac{c^2}{(a-b)^2} &= x^2+y^2+z^2 \\ &&&= (x+y+z)^2 - 2(xy+yz+zx) \\ &&&= 2 + (x+y+z)^2 \geq 2 \end{align*}
  2. \(\,\) \begin{align*} && \begin{pmatrix} 2 & -x & -x \\ -y & 2 & -y \\ -z & -z & 2 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} &= \begin{pmatrix} 2a-xb-xc \\ -ay+2b-yc \\ -za-zb+2c \end{pmatrix} \\ &&&= \begin{pmatrix} 2a-x(b+c) \\ 2b-y(c+a) \\ 2c-z(a+b) \end{pmatrix} \\ &&&= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\ \Rightarrow && 0 &= \det \begin{pmatrix} 2 & -x & -x \\ -y & 2 & -y \\ -z & -z & 2 \end{pmatrix} \\ &&&= 2(4 -yz)+x(-2y-yz)-x(yz+2z) \\ &&&= 8 - 2yz-2yx-2xyz-2zx\\ \Rightarrow && 4 &= xyz+xy+yz+zx \end{align*} \begin{align*} && (2a + b + c)(a + 2b + c)(a + b + 2c) &> 5(b+c)(c+a)(a+b) \\ \Leftrightarrow && \left ( \frac{2a}{b+c}+1 \right)\left ( \frac{2b}{c+a}+1 \right)\left ( \frac{2c}{a+b}+1 \right) &> 5 \\ \Leftrightarrow && \left ( x+1 \right)\left ( y+1 \right)\left ( x+1 \right) &> 5 \\ \Leftrightarrow && xyz+xy+yz+zx+x+y+z+1 &> 5 \\ \Leftrightarrow && 5+x+y+z&> 5 \\ \end{align*} Which is clearly true since if \(a,b,c\) are positve real numbers so are \(x,y,z\). This final inequality is equivalent to showing \(x+y+z > 2\) ie \begin{align*} && x+y+z &> 2 \\ \Leftrightarrow && \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} &> 1 \\ \\ && \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} & > \frac{a}{a+b+c} + \frac{b}{a+b+c} + \frac{c}{a+b+c} = 1 \end{align*} So we're done.

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2020 Paper 3 Q4
D: 1500.0 B: 1500.0
3x3 matrices reflection rotation plane geometry dot product unit vector linear transformations

The plane \(\Pi\) has equation \(\mathbf{r} \cdot \mathbf{n} = 0\) where \(\mathbf{n}\) is a unit vector. Let \(P\) be a point with position vector \(\mathbf{x}\) which does not lie on the plane \(\Pi\). Show that the point \(Q\) with position vector \(\mathbf{x} - (\mathbf{x} \cdot \mathbf{n})\mathbf{n}\) lies on \(\Pi\) and that \(PQ\) is perpendicular to \(\Pi\).

  1. Let transformation \(T\) be a reflection in the plane \(ax+by+cz=0\), where \(a^2+b^2+c^2=1\). Show that the image of \(\mathbf{i} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) under \(T\) is \(\begin{pmatrix} b^2+c^2-a^2 \\ -2ab \\ -2ac \end{pmatrix}\), and find the images of \(\mathbf{j}\) and \(\mathbf{k}\) under \(T\). Write down the matrix \(\mathbf{M}\) which represents transformation \(T\).
  2. The matrix \[ \begin{pmatrix} 0.64 & 0.48 & 0.6 \\ 0.48 & 0.36 & -0.8 \\ 0.6 & -0.8 & 0 \end{pmatrix} \] represents a reflection in a plane. Find the cartesian equation of the plane.
  3. The matrix \(\mathbf{N}\) represents a rotation through angle \(\pi\) about the line through the origin parallel to \(\begin{pmatrix} a \\ b \\ c \end{pmatrix}\), where \(a^2+b^2+c^2=1\). Find the matrix \(\mathbf{N}\).
  4. Identify the single transformation which is represented by the matrix \(\mathbf{NM}\).

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2003 Paper 2 Q1
D: 1600.0 B: 1516.0
3x3 matrices simultaneous equations linear systems Gaussian elimination consistency conditions parametric solution optimisation minimisation

Consider the equations \begin{alignat*}{2} ax-&y- \ z && =3 \;,\\ 2ax -&y -3z && = 7 \;,\\ 3ax-&y-5z && =b \;, \end{alignat*} where \(a\) and \(b\) are given constants.

  1. In the case \(a=0\,\), show that the equations have a solution if and only if \(b=11\,\).
  2. In the case \(a\ne0\) and \(b=11\,\) show that the equations have a solution with \(z=\lambda\) for any given number \(\lambda\,\).
  3. In the case \(a=2\) and \(b=11\,\) find the solution for which \(x^2+y^2+z^2\) is least.
  4. Find a value for \(a\) for which there is a solution such that \(x>10^6\) and \(y^2+z^2<1\,\).

Solution:

  1. If \(a = 0\), then then the LHS second equation is the average of the first and last equations, ie \(7 = \frac{b+3}{2}\) so \(b = 11\). This clearly has solutions, say \(x = 0, y = -1, z = -2\).
  2. If \(a \neq 0\) and \(b = 11\), it is still the case that the third equation a linear combination of the first two. Therefore we can consider the linear system: \begin{cases} ax - y &= 3 + \lambda \\ 2ax - y &= 7 + 3\lambda \end{cases} and since \(-a+2a = a \neq 0\) the solution has a unique solution for \(x\) and \(y\).
  3. \begin{align*} \begin{cases} 2x - y &= 3 + \lambda \\ 4x - y &= 7 + 3\lambda \end{cases} \Rightarrow x = 2 +\lambda, y = 1 + \lambda \\ x^2 + y^2 + z^2 &= (2 + \lambda)^2 + (1+\lambda)^2 + \lambda^2 \\ &= (4 + 1) + (4+2)\lambda + 3\lambda^2 \\ &= 5 + 3((\lambda+1)^2 - 1) \\ &= 3(\lambda + 1)^2 + 2 \end{align*} Therefore the solution is minimized when \(\lambda = -1, x = 1, y = 0, z = -1\)
  4. \begin{align*} \begin{cases} ax - y &= 3 + \lambda \\ 2ax - y &= 7 + 3\lambda \end{cases} \Rightarrow x = \frac{4 +2\lambda}{a}, y = 1 + \lambda \end{align*} We want say \(\lambda = -\frac12\) then we have \(y^2 + z^2 = \frac12\) and \(x = \frac{3}{a}\), so choose \(a < \frac{3}{10^6}\)

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1996 Paper 2 Q2
D: 1600.0 B: 1500.0
3x3 matrices simultaneous equations determinant substitution nonlinear equations Cramer's rule symmetric functions

Consider the system of equations \begin{alignat*}{1} 2yz+zx-5xy & =2\\ yz-zx+2xy & =1\\ yz-2zx+6xy & =3 \end{alignat*} Show that \[xyz=\pm 6\] and find the possible values of \(x\), \(y\) and \(z\).

Solution: Consider the linear \(3\times 3\) system in \(yz, zx, xy\), then \begin{align*} \left(\begin{array}{ccc|c} 2 & 1 & -5 & 2 \\ 1 & -1 & 2 & 1 \\ 1 & -2 & 6 & 3 \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 1 & -2 & 6 & 3 \\ 2 & 1 & -5 & 2 \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 0 & -1 & 4 & 2 \\ 0 & 3 & -9 & 0 \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & -1 & 2 & 1 \\ 0 & -1 & 4 & 2 \\ 0 & 0 & 3 & 6 \\ \end{array}\right) \\ \end{align*} Therefore \(yz = 2, zx = 6, xy = 3 \Rightarrow (xyz)^2 = 36 \Rightarrow xyz = \pm 6\). If \(xyz = 6, x = 3, y = 1, z = 2\), if \(xyz = -6, x = -3, y = -1, z = -2\)

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1996 Paper 3 Q2
D: 1700.0 B: 1516.0
3x3 matrices simultaneous equations linear systems parameter analysis Gaussian elimination consistency conditions

For all values of \(a\) and \(b,\) either solve the simultaneous equations \begin{alignat*}{1} x+y+az & =2\\ x+ay+z & =2\\ 2x+y+z & =2b \end{alignat*} or prove that they have no solution.

Solution: Consider the matrix system: \begin{align*} \left(\begin{array}{ccc|c} 1 & 1 & a & 2 \\ 1 & a & 1 & 2 \\ 2 & 1 & 1 & 2b \\ \end{array}\right) \\ \left(\begin{array}{ccc|c} 1 & 1 & a & 2 \\ 0 & a-1 & 1-a & 0 \\ 0 & -1 & 1-2a & 2b-4 \\ \end{array}\right)\\ \left(\begin{array}{ccc|c} 1 & 1 & a & 2 \\ 0 & a-1 & 1-a & 0 \\ 0 & 0 & -2a & 2b-4 \\ \end{array}\right) \\ \end{align*} Assuming that \(a \neq 1, 0\) all steps are fine and: \(z = \frac{2-b}{a}, y = \frac{2-b}{a}, x +(1+a)y = 2, x = 2 - \frac{(2-b)(1+a)}{a} = \frac{ab+b-2}{a}\) If \(a = 0\), \(y = z\) and \(\begin{cases} x + y &= 2 \\ 2x + 2y &= 2b \end{cases} \Rightarrow b= 2, x = t, y = 2-t, z = 2-t\) If \(a = 1\), \(x = 2b-2, y = t, z = 4-t-2b\), where \(t \in \mathbb{R}\)

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1995 Paper 3 Q1
D: 1700.0 B: 1500.0
3x3 matrices simultaneous equations linear equations parameter analysis Gaussian elimination cases

Find the simultaneous solutions of the three linear equations \begin{alignat*}{1} a^{2}x+ay+z & =a^{2}\\ ax+y+bz & =1\\ a^{2}bx+y+bz & =b \end{alignat*} for all possible real values of \(a\) and \(b\).

Solution: \begin{align*} && a^{2}x+ay+z & =a^{2} \tag{1}\\ && ax+y+bz & =1 \tag{2}\\ && a^{2}bx+y+bz & =b \tag{3} \\ \\ (1) - a(2): && (1-ba)z &= a^2-a \\ \Rightarrow && z &= \frac{a^2-a}{1-ab} \tag{if \(ab \neq 1\)} \\ \\ (2) - (3): && (a-a^2b)x &= b - 1 \\ \Rightarrow && x &= \frac{b-1}{a(1-ab)} \tag{if \(a \neq 0, ab \neq 1\)} \\ \\ b(1) - (3): && (ab-1)y &= a^2 - b^2 \\ \Rightarrow && y &= \frac{a^2-b^2}{ab-1} \end{align*} Let's consider the cases where \(a = 0\), then \begin{align*} && z &= 0 \\ && y + bz &= 1 \\ && y+bz &= b \\ \Rightarrow && y &= 1 = b \end{align*} So if \(a = 0\) then \(b = 1\) and \(x \in \mathbb{R}, y = 1, z = 0\). If \(a \neq 0, ab = 1\), then \begin{align*} && a^2 x + ay + z &= a^2 \\ && ax + y + \frac1{a}z &= a \\ && ax + y + \frac{1}{a}z &= b \\ \end{align*} The last two equations imply \(a = b = \pm 1\). \(a = 1 \Rightarrow x+y+z = 1\), so we have a lot of solutions. \(a = -1 \Rightarrow x -y +z = 1\) so again, lots of solutions. Conclusion: If \(ab \neq 1, a \neq 0\), we have: \[ (x,y,z) = \left (\frac{b-1}{a(1-ab)}, \frac{a^2-b^2}{ab-1}, \frac{a^2-a}{1-ab} \right)\] If \(a = 0\) then \(b = 1\) and we have: \((x,y,z) = (t, 1, 0)\). If \(ab = 1\) then \(a = 1\) or \(a = -1\). If \(a = 1\) then \((x,y,z) = (t, s, 1-t-s)\) If \(a = -1\) then \((x,y,z) = (t,s,1-t+s)\)

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1993 Paper 3 Q3
D: 1700.0 B: 1516.0
matrix 3x3 matrices matrix factorization LU decomposition matrix inverse simultaneous equations lower triangular matrix upper triangular matrix

The matrices \(\mathbf{A},\mathbf{B}\) and \(\mathbf{M}\) are given by \[ \mathbf{A}=\begin{pmatrix}a & 0 & 0\\ b & c & 0\\ d & e & f \end{pmatrix},\quad\mathbf{B}=\begin{pmatrix}1 & p & q\\ 0 & 1 & r\\ 0 & 0 & 1 \end{pmatrix},\quad\mathbf{M}=\begin{pmatrix}1 & 3 & 2\\ 4 & 13 & 5\\ 3 & 8 & 7 \end{pmatrix}, \] where \(a,b,\ldots,r\) are real numbers. Given that \(\mathbf{M=AB},\) show that \(a=1,b=4,c=1,d=3,e=1,f=-2,p=3,q=2\) and \(r=-3\) gives the unique solution for \(\mathbf{A}\) and \(\mathbf{B}.\) Evaluate \(\mathbf{A}^{-1}\) and \(\mathbf{B}^{-1},\) Hence, or otherwise, solve the simultaneous equations \begin{alignat*}{1} x+3y+2z & =7\\ 4x+13y+5z & =18\\ 3x+8y+7z & =25. \end{alignat*}

Solution: \begin{align*} && \begin{pmatrix}1 & 3 & 2\\ 4 & 13 & 5\\ 3 & 8 & 7 \end{pmatrix} &= \begin{pmatrix}a & 0 & 0\\ b & c & 0\\ d & e & f \end{pmatrix}\begin{pmatrix}1 & p & q\\ 0 & 1 & r\\ 0 & 0 & 1 \end{pmatrix} \\ &&&= \begin{pmatrix} a & ap & aq \\ b & pb + c & qb + cr\\ d & pd + e & qd + er +f \end{pmatrix} \\ \Rightarrow && a,b,d,p,q&=1,4,3,3,2\\ &&&= \begin{pmatrix} 1 & 3 & 2 \\ 4 & 12 + c & 8+ cr\\ 3 & 9 + e & 6 + er +f \end{pmatrix} \\ \Rightarrow && c, e&=1,-1\\ &&&= \begin{pmatrix} 1 & 3 & 2 \\ 4 & 13 & 8+ r\\ 3 & 8 & 6 -r +f \end{pmatrix} \\ \Rightarrow && r, f &= -3, -2 \end{align*} \begin{align*} \mathbf{A}^{-1} &= \begin{pmatrix} 1 & 0 & 0 \\ 4 & 1 & 0\\ 3 & -1 & -2 \end{pmatrix}^{-1} \\ &=\frac{1}{-2} \begin{pmatrix} -2 & 0 & 0 \\ 8 & -2 & 0\\ -7 & 1 & 1 \end{pmatrix} \\ \\ \mathbf{B}^{-1} &= \begin{pmatrix} 1 & 3 & 2 \\ 0 & 1 & -3\\ 0 & 0 & 1 \end{pmatrix}^{-1} \\ &= \begin{pmatrix} 1 & -3 & -11 \\ 0 & 1 & 3\\ 0 & 0 & 1 \end{pmatrix} \\ \end{align*} We want to solve \(\mathbf{M}\mathbf{v} = \begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix}\), ie \begin{align*} \mathbf{v} &= \mathbf{M}^{-1}\begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix} \\ &= \mathbf{B}^{-1} \mathbf{A}^{-1}\begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix} \\ &= \frac{1}{-2}\mathbf{B}^{-1} \begin{pmatrix} -2 & 0 & 0 \\ 8 & -2 & 0\\ -7 & 1 & 1 \end{pmatrix} \begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix} \\ &= \frac{1}{-2}\mathbf{B}^{-1} \begin{pmatrix} -14 \\ 20 \\ -6 \end{pmatrix} \\ &= \begin{pmatrix} 1 & -3 & -11 \\ 0 & 1 & 3\\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 7 \\ -10 \\ 3 \end{pmatrix} \\ &= \begin{pmatrix} 4\\ -1 \\ 3 \end{pmatrix} \end{align*} This algorithm is called the "LU-decomposition"

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1991 Paper 3 Q2
D: 1700.0 B: 1516.0
complex numbers similar triangles 3x3 matrices determinant Argand diagram geometric proof centroid equilateral triangle factorisation

The distinct points \(P_{1},P_{2},P_{3},Q_{1},Q_{2}\) and \(Q_{3}\) in the Argand diagram are represented by the complex numbers \(z_{1},z_{2},z_{3},w_{1},w_{2}\) and \(w_{3}\) respectively. Show that the triangles \(P_{1}P_{2}P_{3}\) and \(Q_{1}Q_{2}Q_{3}\) are similar, with \(P_{i}\) corresponding to \(Q_{i}\) (\(i=1,2,3\)) and the rotation from \(1\) to \(2\) to \(3\) being in the same sense for both triangles, if and only if \[ \frac{z_{1}-z_{2}}{z_{2}-z_{3}}=\frac{w_{1}-w_{2}}{w_{1}-w_{3}}. \] Verify that this condition may be written \[ \det\begin{pmatrix}z_{1} & z_{2} & z_{3}\\ w_{1} & w_{2} & w_{3}\\ 1 & 1 & 1 \end{pmatrix}=0. \]

  1. Show that if \(w_{i}=z_{i}^{2}\) (\(i=1,2,3\)) then triangle \(P_{1}P_{2}P_{3}\) is not similar to triangle \(Q_{1}Q_{2}Q_{3}.\)
  2. Show that if \(w_{i}=z_{i}^{3}\) (\(i=1,2,3\)) then triangle \(P_{1}P_{2}P_{3}\) is similar to triangle \(Q_{1}Q_{2}Q_{3}\) if and only if the centroid of triangle \(P_{1}P_{2}P_{3}\) is the origin. {[}The centroid of triangle \(P_{1}P_{2}P_{3}\) is represented by the complex number \(\frac{1}{3}(z_{1}+z_{2}+z_{3})\).{]}
  3. Show that the triangle \(P_{1}P_{2}P_{3}\) is equilateral if and only if \[ z_{2}z_{3}+z_{3}z_{1}+z_{1}z_{2}=z_{1}^{2}+z_{2}^{2}+z_{3}^{2}. \]

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1991 Paper 3 Q4
D: 1700.0 B: 1516.0
3x3 matrices vectors straight line in 3D lines of sight parametric simultaneous equations 3D geometry intersection

The point \(P\) moves on a straight line in three-dimensional space. The position of \(P\) is observed from the points \(O_{1}(0,0,0)\) and \(O_{2}(8a,0,0).\) At times \(t=t_{1}\) and \(t=t_{1}'\), the lines of sight from \(O_{1}\) are along the lines \[ \frac{x}{2}=\frac{z}{3},y=0\quad\mbox{ and }\quad x=0,\frac{y}{3}=\frac{z}{4} \] respectively. At times \(t=t_{2}\) and \(t=t_{2}'\), the lines of sight from \(O_{2}\) are \[ \frac{x-8a}{-3}=\frac{y}{1}=\frac{z}{3}\quad\mbox{ and }\quad\frac{x-8a}{-4}=\frac{y}{2}=\frac{z}{5} \] respectively. Find an equation or equations for the path of \(P\).

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1989 Paper 2 Q10
D: 1600.0 B: 1500.0
3x3 matrices line intersection planes vectors parametric simultaneous equations linear algebra geometric proof

State carefully the conditions which the fixed vectors \(\mathbf{a,b,u}\) and \(\mathbf{v}\) must satisfy in order to ensure that the line \(\mathbf{r=a+}\lambda\mathbf{u}\) intersects the line \(\mathbf{r=b+\mu}\mathbf{v}\) in exactly one point. Find the two values of the fixed scalar \(b\) for which the planes with equations \[ \left.\begin{array}{c} x+y+bz=b+2\\ bx+by+z=2b+1 \end{array}\right\} \tag{*} \] do not intersect in a line. For other values of \(b\), express the line of intersection of the two planes in the form \(\mathbf{r=a}+\lambda\mathbf{u},\) where \(\mathbf{a\cdot u}=0\). Find the conditions which \(b\) and the fixed scalars \(c\) and \(d\) must satisfy to ensure that there is exactly one point on the line \[ \mathbf{r=}\left(\begin{array}{c} 0\\ 0\\ c \end{array}\right)+\mu\left(\begin{array}{c} 1\\ d\\ 0 \end{array}\right) \] whose coordinates satisfy both equations \((*)\).

Solution: There are two requirements (assuming they are lines not fixed points): 1. They cannot be parallel, ie \(\mathbf{u} \neq \lambda \mathbf{v}\) for any \(\lambda\) 2. They must lie in the same plane, ie \((\mathbf{b}-\mathbf{a})\cdot (\mathbf{u} \times \mathbf{v}) = 0\) The planes will not intersect in a line if they are either parallel and separate or parallel and the same. If \(b = 1\) or \(b=-1\) the planes are parallel. \begin{align*} && (x+y) + b z &= b+ 2\\ &&b(x+y) + z &= 2b + 1 \\ \Rightarrow && (1-b^2)z &= 2b+1 - b^2 -2b \\ &&&= 1-b^2 \\ \Rightarrow && z &= 1 \\ && x+ y &= 2 \\ \end{align*} Therefore our line is \(\mathbf{r} = \begin{pmatrix} 1+t \\ 1-t \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \) We must have: \(d \neq -1\) to ensure that the lines aren't parallel. We must also have: \begin{align*} 0 &= \left ( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} -\begin{pmatrix} 0 \\ 0 \\ c \end{pmatrix}\right) \cdot \left ( \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ d \\ 0 \end{pmatrix} \right) \\ &= \begin{pmatrix} 1 \\ 1 \\ 1-c \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ d+1 \end{pmatrix} \\ &= (1-c)(d+1) \end{align*} So \(c =1\)

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