QUESTION 1 Suppose \( V \) is the plane in \( \mathbb{R}^{3} \) that coatains the points \( a, b \) and \( c \), where \[ Q=(1,2,3), \delta=(1,-2,1) \text { and } c=(0,1,2) \text {. } \] Suppose \( L \) bs the line in \( \mathbb{R}^{3} \) that contains the point \( a \) and is perpendicular to the plane \( V \) (a) Find an equation for \( V \) (b) Find an equation for \( L \)

To find the equation of the plane \( V \) that contains the points \( b \) and \( c \), we need a point on the plane and a normal vector. We can take point \( c = (0, 1, 2) \) and find a normal vector using the cross product of the vectors \( \overrightarrow{bc} \) and \( \overrightarrow{ac} \). 1. **Find the vectors:** - \( \overrightarrow{b} = (1, 2, 3) \) and \( \overrightarrow{c} = (0, 1, 2) \) - \( \overrightarrow{bc} = (0 - 1, 1 - 2, 2 - 3) = (-1, -1, -1) \) 2. **Next, calculate the cross product to find the normal vector.** If we assume \( a = Q \) (1, 2, 3) is also included: - \( \overrightarrow{ac} = (-1, -1, -1) \) - Let's assume \( \delta = (1, -2, 1) \) represents another vector in the plane. The normal vector \( n \) can be obtained by \( \overrightarrow{bc} \times \delta \). Calculating this cross product: \[ n = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & -1 \\ 1 & -2 & 1 \end{vmatrix} = \hat{i}((-1)(1) - (-2)(-1)) - \hat{j}((-1)(1) - (-1)(-1)) + \hat{k}((-1)(-2) - (-1)(1)) = \hat{i}(-1 - 2) - \hat{j}(-1 - 1) + \hat{k}(2 - 1) = -3\hat{i} + 2\hat{j} + 1\hat{k} \] Thus, \( n = (-3, 2, 1) \). 3. **The plane equation is given by:** \[ -3(x - 0) + 2(y - 1) + 1(z - 2) = 0 \] or simplifying, \[ -3x + 2y + z - 2 = 0 \] 4. **For the equation of line \( L \), since it's perpendicular to the plane, it takes the direction of the normal vector.** Let \( a = Q = (1, 2, 3) \): The parametric equation for the line can be written as: \[ x = 1 - 3t \\ y = 2 + 2t \\ z = 3 + 1t \] Where \( t \) is a parameter. So thus, we have using direction vector \( (-3, 2, 1) \), the equation of line \( L \). In summary: (a) The equation of the plane \( V \) is \( -3x + 2y + z - 2 = 0 \). (b) The equation of the line \( L \) is expressed parametrically as: \[ (x, y, z) = (1, 2, 3) + t(-3, 2, 1) \]