Question

$7\hat{i}-4\hat{j}+7\hat{k}, \hat{i}-6\hat{j}+10\hat{k}, -\hat{i}-3\hat{j}+4\hat{k}, 5\hat{i}-\hat{j}+\hat{k}$ are the position vectors of the points A, B, C, D respectively. If $p\hat{i} + q\hat{j} + r\hat{k}$ is the position vector of the point of intersection of the diagonals of the quadrilateral ABCD, then $p+q+r=$

• A. 4
• B. 5
• C. 0
• D. 1

Hint:

Before finding the intersection of diagonals of a quadrilateral, it's always a good idea to check if it's a special type, like a parallelogram. If $\vec{A}, \vec{B}, \vec{C}, \vec{D}$ are vertices in order, check if $\vec{A}+\vec{C} = \vec{B}+\vec{D}$. If they are equal, it's a parallelogram, and the intersection point is simply $\frac{\vec{A}+\vec{C}}{2}$ or $\frac{\vec{B}+\vec{D}}{2}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The problem provides the position vectors of the four vertices of a quadrilateral ABCD. We need to find the position vector of the intersection of its diagonals. If the quadrilateral is a parallelogram, its diagonals bisect each other. The intersection point would then be the midpoint of either diagonal.
Step 2: Key Formula or Approach
1. First, determine if the quadrilateral ABCD is a parallelogram by checking if one pair of opposite sides are equal and parallel (i.e., if $\vec{AB} = \vec{DC}$). 2. If it is a parallelogram, the point of intersection of the diagonals is the midpoint of diagonal AC (or BD). The position vector of the midpoint of AC is given by $\frac{\vec{OA} + \vec{OC}}{2}$. 3. Once the position vector $p\hat{i} + q\hat{j} + r\hat{k}$ is found, calculate the sum $p+q+r$.
Step 3: Detailed Explanation
Let the position vectors of the vertices be: $\vec{A} = 7\hat{i}-4\hat{j}+7\hat{k}$ $\vec{B} = \hat{i}-6\hat{j}+10\hat{k}$ $\vec{C} = -\hat{i}-3\hat{j}+4\hat{k}$ $\vec{D} = 5\hat{i}-\hat{j}+\hat{k}$ 1. Check if ABCD is a parallelogram: Calculate the vector $\vec{AB}$: \[ \vec{AB} = \vec{B} - \vec{A} = (\hat{i}-6\hat{j}+10\hat{k}) - (7\hat{i}-4\hat{j}+7\hat{k}) = -6\hat{i}-2\hat{j}+3\hat{k} \] Calculate the vector $\vec{DC}$: \[ \vec{DC} = \vec{C} - \vec{D} = (-\hat{i}-3\hat{j}+4\hat{k}) - (5\hat{i}-\hat{j}+\hat{k}) = -6\hat{i}-2\hat{j}+3\hat{k} \] Since $\vec{AB} = \vec{DC}$, the quadrilateral ABCD is a parallelogram. 2. Find the point of intersection of diagonals: Since ABCD is a parallelogram, its diagonals AC and BD bisect each other. The point of intersection is the midpoint of either diagonal. Let's find the midpoint of diagonal AC. Let the intersection point be P. Its position vector is $\vec{P}$. \[ \vec{P} = \frac{\vec{A} + \vec{C}}{2} = \frac{(7\hat{i}-4\hat{j}+7\hat{k}) + (-\hat{i}-3\hat{j}+4\hat{k})}{2} \] \[ \vec{P} = \frac{(7-1)\hat{i} + (-4-3)\hat{j} + (7+4)\hat{k}}{2} = \frac{6\hat{i} - 7\hat{j} + 11\hat{k}}{2} \] \[ \vec{P} = 3\hat{i} - \frac{7}{2}\hat{j} + \frac{11}{2}\hat{k} \] We are given that the position vector is $p\hat{i} + q\hat{j} + r\hat{k}$. So, $p = 3$, $q = -\frac{7}{2}$, and $r = \frac{11}{2}$. 3. Calculate $p+q+r$: \[ p+q+r = 3 + \left(-\frac{7}{2}\right) + \frac{11}{2} = 3 + \frac{11-7}{2} = 3 + \frac{4}{2} = 3+2=5 \] Step 4: Final Answer
The sum of the components of the position vector of the intersection point is $p+q+r = 5$.