Key Concepts and Formulas

• Direction Vector of a Line Perpendicular to Two Other Lines: If a line's direction vector $\vec{d}$ is perpendicular to the direction vectors $\vec{d_1}$ and $\vec{d_2}$ of two other lines, then $\vec{d}$ can be found using the cross product: $\vec{d} = \vec{d_1} \times \vec{d_2}$.
• Vector Equation of a Line: A line passing through a point with position vector $\vec{a}$ and having a direction vector $\vec{d}$ can be represented as $\vec{r} = \vec{a} + \lambda \vec{d}$, where $\lambda$ is a scalar parameter.
• Perpendicularity of Vectors: Two vectors are perpendicular if their dot product is zero.
• Distance Formula in 3D: The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Step-by-Step Solution

Step 1: Determine the Direction Vector of the Required Line The problem states that the line we are interested in passes through point $P(5,-4,3)$ and is perpendicular to two given lines. Let the direction vectors of these two lines be $\vec{d_1}$ and $\vec{d_2}$.

The first line is given by $\vec{r}=(-3 \hat{i}+2 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+5 \hat{k})$. Its direction vector is $\vec{d_1} = 2 \hat{i}+3 \hat{j}+5 \hat{k}$.

The second line is given by $\vec{r}=(\hat{i}-2 \hat{j}+\hat{k})+\mu(-\hat{i}+3 \hat{j}+2 \hat{k})$. Its direction vector is $\vec{d_2} = -\hat{i}+3 \hat{j}+2 \hat{k}$.

Since our required line is perpendicular to both of these lines, its direction vector, let's call it $\vec{d}$, must be perpendicular to both $\vec{d_1}$ and $\vec{d_2}$. We can find such a vector by taking the cross product of $\vec{d_1}$ and $\vec{d_2}$.

$\vec{d} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ -1 & 3 & 2 \end{vmatrix}$ Expanding the determinant: $\vec{d} = \hat{i}((3)(2) - (5)(3)) - \hat{j}((2)(2) - (5)(-1)) + \hat{k}((2)(3) - (3)(-1))$ $\vec{d} = \hat{i}(6 - 15) - \hat{j}(4 + 5) + \hat{k}(6 + 3)$ $\vec{d} = -9 \hat{i} - 9 \hat{j} + 9 \hat{k}$ To simplify calculations, we can divide this vector by a common factor, $-9$. A simplified direction vector is: $\vec{d}_{simplified} = \frac{-9 \hat{i} - 9 \hat{j} + 9 \hat{k}}{-9} = \hat{i} + \hat{j} - \hat{k}$ We will use $\vec{d} = \hat{i} + \hat{j} - \hat{k}$ as the direction vector of the line.

Step 2: Write the Vector Equation of the Line The line passes through point $P(5,-4,3)$, so its position vector is $\vec{p} = 5\hat{i} - 4\hat{j} + 3\hat{k}$. The direction vector is $\vec{d} = \hat{i} + \hat{j} - \hat{k}$. The vector equation of the line is given by $\vec{r} = \vec{p} + \lambda \vec{d}$. $\vec{r} = (5\hat{i} - 4\hat{j} + 3\hat{k}) + \lambda (\hat{i} + \hat{j} - \hat{k})$ A general point $R$ on this line can be represented by its position vector $\vec{r_R}$: $\vec{r_R} = (5+\lambda)\hat{i} + (-4+\lambda)\hat{j} + (3-\lambda)\hat{k}$ Thus, the coordinates of any point $R$ on the line are $(5+\lambda, -4+\lambda, 3-\lambda)$.

Step 3: Find the Foot of the Perpendicular (R) from Q to the Line We need to find the distance of point $Q(0,2,-2)$ from the line. Let $R(5+\lambda, -4+\lambda, 3-\lambda)$ be the foot of the perpendicular from $Q$ to the line. The vector $\vec{QR}$ connects point $Q$ to point $R$. The position vector of $Q$ is $\vec{r_Q} = 0\hat{i} + 2\hat{j} - 2\hat{k}$.

$\vec{QR} = \vec{r_R} - \vec{r_Q}$ $\vec{QR} = ((5+\lambda) - 0)\hat{i} + ((-4+\lambda) - 2)\hat{j} + ((3-\lambda) - (-2))\hat{k}$ $\vec{QR} = (5+\lambda)\hat{i} + (-6+\lambda)\hat{j} + (5-\lambda)\hat{k}$ Since $R$ is the foot of the perpendicular from $Q$ to the line, the vector $\vec{QR}$ must be perpendicular to the direction vector of the line, $\vec{d} = \hat{i} + \hat{j} - \hat{k}$. Their dot product must be zero: $\vec{QR} \cdot \vec{d} = 0$.

$((5+\lambda)\hat{i} + (-6+\lambda)\hat{j} + (5-\lambda)\hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) = 0$ $(5+\lambda)(1) + (-6+\lambda)(1) + (5-\lambda)(-1) = 0$ $5+\lambda - 6+\lambda - 5+\lambda = 0$ $(5 - 6 - 5) + (\lambda + \lambda + \lambda) = 0$ $-6 + 3\lambda = 0$ $3\lambda = 6$ $\lambda = 2$ Substitute $\lambda=2$ back into the coordinates of $R$: $R = (5+2, -4+2, 3-2) = (7, -2, 1)$ So, the foot of the perpendicular is $R(7, -2, 1)$.

Step 4: Calculate the Distance QR Now we have the coordinates of point $Q(0,2,-2)$ and the foot of the perpendicular $R(7,-2,1)$. The distance between these two points is the shortest distance from $Q$ to the line. Using the distance formula:

$QR = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$ $QR = \sqrt{(7-0)^2 + (-2-2)^2 + (1-(-2))^2}$ $QR = \sqrt{(7)^2 + (-4)^2 + (3)^2}$ $QR = \sqrt{49 + 16 + 9}$ $QR = \sqrt{74}$

Common Mistakes & Tips

• Cross Product Calculation: Be very careful with the signs and arithmetic when calculating the cross product. A small error here will propagate through the rest of the solution.
• Dot Product Equation: Ensure you are correctly forming the vector $\vec{QR}$ and performing the dot product with the correct direction vector. Solving the linear equation for $\lambda$ should be straightforward.
• Simplifying Direction Vector: While not strictly necessary, simplifying the direction vector by dividing out common factors can make the dot product calculation easier.

Summary To find the distance of point $Q$ from the given line, we first determined the direction vector of the line by taking the cross product of the direction vectors of the two lines it is perpendicular to. We then wrote the vector equation of the line and found a general point $R$ on it. By setting the dot product of $\vec{QR}$ and the line's direction vector to zero, we found the specific value of the parameter $\lambda$ that corresponds to the foot of the perpendicular. Finally, we calculated the distance between point $Q$ and the foot of the perpendicular $R$ using the 3D distance formula.

The final answer is $\boxed{\sqrt{74}}$.