Question:

Step 1: Write the dot product equation. Given $\vec{A} = 2\hat{i} + \hat{j} + 5\hat{k}$ and $\vec{B} = x\hat{i} + y\hat{j} + z\hat{k}$, the dot product is: $$ \vec{A}\cdot\vec{B} = (2)(x) + (1)(y) + (5)(z) = 2x + y + 5z $$ Given $\vec{A}\cdot\vec{B} = 11$, we have: $$ 2x + y + 5z = 11 \quad (*)$$ Step 2: Compute $\vec{A}\times\vec{B}$ and equate components. The cross product $\vec{A}\times\vec{B}$ is calculated as: $$ \vec{A}\times\vec{B} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&5\\x&y&z\end{vmatrix} = (1 \cdot z - 5 \cdot y)\hat{i} - (2 \cdot z - 5 \cdot x)\hat{j} + (2 \cdot y - 1 \cdot x)\hat{k} $$ $$ \vec{A}\times\vec{B} = (z-5y)\hat{i} - (2z-5x)\hat{j} + (2y-x)\hat{k} $$ Given $\vec{A}\times\vec{B} = -13\hat{i} - 9\hat{j} + 7\hat{k}$, we equate the corresponding components: 1. $z - 5y = -13$ 2. $-(2z-5x) = -9 \Rightarrow 2z-5x = 9$ 3. $2y - x = 7$ Step 3: Solve the system of equations for $x, y, z$. From equation (3), we express $x$ in terms of $y$: $$ x = 2y-7 \quad (a) $$ From equation (1), we express $z$ in terms of $y$: $$ z = 5y-13 \quad (b) $$ Substitute expressions $(a)$ and $(b)$ into the dot product equation $(*)$: $$ 2(2y-7) + y + 5(5y-13) = 11 $$ $$ 4y - 14 + y + 25y - 65 = 11 $$ $$ 30y - 79 = 11 $$ $$ 30y = 90 $$ $$ y = 3 $$ Now, substitute $y=3$ back into expressions $(a)$ and $(b)$ to find $x$ and $z$: $$ x = 2(3) - 7 = 6 - 7 = -1 $$ $$ z = 5(3) - 13 = 15 - 13 = 2 $$ Thus, the values are $x=-1, y=3, z=2$. Step 4: Evaluate the expressions. Using the determined values $x=-1, y=3, z=2$: 1. $x^2 + y^2 = (-1)^2 + (3)^2 = 1 + 9 = 10$ 2. $y^2 + z^2 = (3)^2 + (2)^2 = 9 + 4 = 13$ 3. $x+y+z = -1+3+2 = 4$ 4. $\dfrac{x+y}{z} = \dfrac{-1+3}{2} = \dfrac{2}{2} = 1$