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: Check the Nature of Quadrilateral ABCD: Let the position vectors be \( \vec{a}, \vec{b}, \vec{c}, \vec{d} \). Midpoint of diagonal AC: \[ M_{AC} = \frac{\vec{a} + \vec{c}}{2} = \frac{(7-1)\bar{i} + (-4-3)\bar{j} + (7+4)\bar{k}}{2} = \frac{6\bar{i} - 7\bar{j} + 11\bar{k}}{2} \] Midpoint of diagonal BD: \[ M_{BD} = \frac{\vec{b} + \vec{d}}{2} = \frac{(1+5)\bar{i} + (-6-1)\bar{j} + (10+1)\bar{k}}{2} = \frac{6\bar{i} - 7\bar{j} + 11\bar{k}}{2} \] Since the midpoints of the diagonals coincide, ABCD is a parallelogram. The point of intersection of the diagonals is this common midpoint.
Step 2: Identify p, q, r: Position vector of intersection point \( = 3\bar{i} - 3.5\bar{j} + 5.5\bar{k} \). So, \( p = 3 \), \( q = -3.5 \), \( r = 5.5 \).
Step 3: Calculate Sum: \[ p + q + r = 3 + (-3.5) + 5.5 = 3 + 2 = 5 \]