Let \( \vec{a} \) and \( b \) be the vectors given by \( \vec{a}=\langle 2,0,-1\rangle \) and \( b=\langle 1,-1,3\rangle \). (1) Compute (i) \( \|\vec{a}\| \) and \( \|\vec{b}\| \) (ii) \( \vec{a} \cdot \vec{a}, \vec{a} \cdot \vec{b} \) in two different ways. (iii) \( \vec{a} \times \vec{b} \) and \( \vec{b} \times \vec{b} \) (2) Assume that \( \theta \) is the angle between the two vectors \( \vec{a} \) and \( \vec{b} \). Write down \( \cos \theta \) in terms of \( \vec{a} \cdot \vec{b} \) and \( \|\vec{a}\| \) and \( \|\vec{b}\| \). Deduce \( \cos \theta \) and find the angle \( \theta \) between the two vectors. (3) Find an equation of the plane \( P \) containing \( \vec{a} \) and \( \vec{b} \)

To tackle the tasks involving vectors \( \vec{a}=\langle 2,0,-1\rangle \) and \( \vec{b}=\langle 1,-1,3\rangle \), let's jump right in! ### (1) (i) The magnitudes \( \|\vec{a}\| \) and \( \|\vec{b}\| \) can be computed as follows: \[ \|\vec{a}\| = \sqrt{2^2 + 0^2 + (-1)^2} = \sqrt{4 + 0 + 1} = \sqrt{5}. \] \[ \|\vec{b}\| = \sqrt{1^2 + (-1)^2 + 3^2} = \sqrt{1 + 1 + 9} = \sqrt{11}. \] (ii) For the dot products: \[ \vec{a} \cdot \vec{a} = 2^2 + 0^2 + (-1)^2 = 4 + 0 + 1 = 5. \] \[ \vec{a} \cdot \vec{b} = 2 \cdot 1 + 0 \cdot (-1) + (-1) \cdot 3 = 2 + 0 - 3 = -1. \] (iii) The cross product \( \vec{a} \times \vec{b} \): \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & -1 \\ 1 & -1 & 3 \end{vmatrix} = \hat{i}(0 \cdot 3 - (-1) \cdot (-1)) - \hat{j}(2 \cdot 3 - (-1) \cdot 1) + \hat{k}(2 \cdot (-1) - 0 \cdot 1) = \hat{i}(0 - 1) - \hat{j}(6 + 1) + \hat{k}(-2) = -\hat{i} - 7\hat{j} - 2\hat{k}. \] Thus, \[ \vec{a} \times \vec{b} = \langle -1, -7, -2 \rangle. \] The cross product \( \vec{b} \times \vec{b} \) is zero since any vector crossed with itself is zero: \[ \vec{b} \times \vec{b} = \langle 0, 0, 0 \rangle. \] ### (2) For the angle \( \theta \) between the two vectors, we use the formula: \[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{\|\vec{a}\| \|\vec{b}\|}. \] Substituting the values we found: \[ \cos \theta = \frac{-1}{\sqrt{5} \cdot \sqrt{11}} = \frac{-1}{\sqrt{55}}. \] Now, to find \( \theta \): \[ \theta = \cos^{-1}\left(\frac{-1}{\sqrt{55}}\right). \] ### (3) To find the equation of the plane \( P \) containing the vectors \( \vec{a} \) and \( \vec{b} \), we can use the normal vector which is obtained from the cross product, \( \vec{n} = \vec{a} \times \vec{b} = \langle -1, -7, -2 \rangle \). The equation of the plane can be written as: \[ -1(x - x_0) - 7(y - y_0) - 2(z - z_0) = 0 \] Where \( (x_0, y_0, z_0) \) is a point on the plane. To use \( \vec{a} = (2,0,-1) \): \[ -1(x - 2) - 7(y - 0) - 2(z - (-1)) = 0. \] Thus, simplifying gives: \[ -1x + 2 - 7y - 2z - 2 = 0 \rightarrow x + 7y + 2z = 0. \] So the equation of the plane \( P \) is \( x + 7y + 2z = 0 \).