91Ó°ÊÓ

Since we are given vector \(\langle 0,1,1 \rangle\), we can find a vector that is orthogonal to it by assuming two arbitrary values and finding the third value using the dot product formula: \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos{\theta}\). Since the vectors are orthogonal, the angle between them is \(90^\circ\), and therefore, the dot product is zero. Let's assume our orthogonal vector to be \(\langle x, y, z \rangle\). We can write the dot product as \(0 \cdot x + 1 \cdot y + 1 \cdot z = 0\) We can choose arbitrary values for x and y, and solve for z. Let x = 1 and y = 1. Then the equation becomes \(1 + z = 0\) Solving for z, we get z = -1. Therefore, the vector orthogonal to \(\langle 0,1,1 \rangle\) is \(\langle 1,1,-1 \rangle\).

Now that we have the first orthogonal vector \(\langle 1,1,-1 \rangle\), we can use the cross product to find the second orthogonal vector. The cross product is given by the following formula: \(\vec{c} = \vec{a} \times \vec{b}\) Where \(\vec{a}\) is the given vector, \(\vec{b}\) is the first orthogonal vector we found, and \(\vec{c}\) is the second orthogonal vector we are seeking. The cross product of two 3-dimensional vectors is calculated as follows: \(\vec{c} = \langle c_x, c_y, c_z \rangle\) \(c_x = a_y \cdot b_z - a_z \cdot b_y\) \(c_y = a_z \cdot b_x - a_x \cdot b_z\) \(c_z = a_x \cdot b_y - a_y \cdot b_x\) Given \(\vec{a} = \langle 0,1,1 \rangle\) and \(\vec{b} = \langle 1,1,-1 \rangle\), we can compute the elements of \(\vec{c}\). \(c_x = 1 \cdot (-1) - 1 \cdot 1 = -2\) \(c_y = 1 \cdot 1 - 0 \cdot (-1) = 1\) \(c_z = 0 \cdot 1 - 1 \cdot 1 = -1\) Thus, the second orthogonal vector is \(\vec{c} = \langle -2, 1, -1 \rangle\). So, the two vectors orthogonal to \(\langle 0,1,1 \rangle\) and to each other are \(\langle 1,1,-1 \rangle\) and \(\langle -2, 1, -1 \rangle\).