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JEE Main 2023
Vector Algebra
Vector Algebra
Hard

Question

Let ABCD\mathrm{ABCD} be a quadrilateral. If E\mathrm{E} and F\mathrm{F} are the mid points of the diagonals AC\mathrm{AC} and BD\mathrm{BD} respectively and (AB→−BC→)+(AD→−DC→)=kFE→(\overrightarrow{A B}-\overrightarrow{B C})+(\overrightarrow{A D}-\overrightarrow{D C})=k \overrightarrow{F E}, then kk is equal to :

Options

Solution

  1. Key Concepts and Formulas
  • Position Vector: The position vector of a point PP with respect to an origin OO is denoted by OP→\overrightarrow{OP} or p⃗\vec{p}.
  • Displacement Vector: The vector from point PP to point QQ is given by PQ→=OQ→−OP→=q⃗−p⃗\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \vec{q} - \vec{p}.
  • Midpoint Formula: If MM is the midpoint of the line segment joining points PP and QQ with position vectors p⃗\vec{p} and q⃗\vec{q} respectively, then the position vector of MM is m⃗=p⃗+q⃗2\vec{m} = \frac{\vec{p} + \vec{q}}{2}.
  1. Step-by-Step Solution

Step 1: Assign Position Vectors to Vertices and Midpoints Let the position vectors of the vertices A,B,C,DA, B, C, D of the quadrilateral be a⃗,b⃗,c⃗,d⃗\vec{a}, \vec{b}, \vec{c}, \vec{d} respectively, with respect to an arbitrary origin OO. Since EE is the midpoint of the diagonal ACAC, its position vector e⃗\vec{e} is given by the midpoint formula: e⃗=a⃗+c⃗2\vec{e} = \frac{\vec{a} + \vec{c}}{2} Similarly, since FF is the midpoint of the diagonal BDBD, its position vector f⃗\vec{f} is: f⃗=b⃗+d⃗2\vec{f} = \frac{\vec{b} + \vec{d}}{2}

Step 2: Simplify the Left Hand Side (LHS) of the Given Equation The given equation is (AB→−BC→)+(AD→−DC→)=kFE→(\overrightarrow{AB} - \overrightarrow{BC}) + (\overrightarrow{AD} - \overrightarrow{DC}) = k \overrightarrow{FE}. We will first simplify the LHS using the displacement vector formula PQ→=q⃗−p⃗\overrightarrow{PQ} = \vec{q} - \vec{p}. AB→=b⃗−a⃗\overrightarrow{AB} = \vec{b} - \vec{a} BC→=c⃗−b⃗\overrightarrow{BC} = \vec{c} - \vec{b} AD→=d⃗−a⃗\overrightarrow{AD} = \vec{d} - \vec{a} DC→=c⃗−d⃗\overrightarrow{DC} = \vec{c} - \vec{d} Substitute these into the LHS: LHS=(AB→−BC→)+(AD→−DC→)\text{LHS} = (\overrightarrow{AB} - \overrightarrow{BC}) + (\overrightarrow{AD} - \overrightarrow{DC}) LHS=((b⃗−a⃗)−(c⃗−b⃗))+((d⃗−a⃗)−(c⃗−d⃗))\text{LHS} = ((\vec{b} - \vec{a}) - (\vec{c} - \vec{b})) + ((\vec{d} - \vec{a}) - (\vec{c} - \vec{d})) Expand the terms carefully, paying attention to the signs: LHS=(b⃗−a⃗−c⃗+b⃗)+(d⃗−a⃗−c⃗+d⃗)\text{LHS} = (\vec{b} - \vec{a} - \vec{c} + \vec{b}) + (\vec{d} - \vec{a} - \vec{c} + \vec{d}) Combine like terms within each parenthesis: LHS=(2b⃗−a⃗−c⃗)+(2d⃗−a⃗−c⃗)\text{LHS} = (2\vec{b} - \vec{a} - \vec{c}) + (2\vec{d} - \vec{a} - \vec{c}) Now, combine all terms: LHS=2b⃗+2d⃗−a⃗−a⃗−c⃗−c⃗\text{LHS} = 2\vec{b} + 2\vec{d} - \vec{a} - \vec{a} - \vec{c} - \vec{c} LHS=2b⃗+2d⃗−2a⃗−2c⃗\text{LHS} = 2\vec{b} + 2\vec{d} - 2\vec{a} - 2\vec{c} Factor out a 2: LHS=2(b⃗+d⃗−a⃗−c⃗)\text{LHS} = 2(\vec{b} + \vec{d} - \vec{a} - \vec{c})

Step 3: Simplify the Right Hand Side (RHS) of the Given Equation The RHS is kFE→k \overrightarrow{FE}. First, express FE→\overrightarrow{FE} using position vectors: FE→=e⃗−f⃗\overrightarrow{FE} = \vec{e} - \vec{f} Substitute the expressions for e⃗\vec{e} and f⃗\vec{f} from Step 1: FE→=(a⃗+c⃗2)−(b⃗+d⃗2)\overrightarrow{FE} = \left(\frac{\vec{a} + \vec{c}}{2}\right) - \left(\frac{\vec{b} + \vec{d}}{2}\right) Combine the terms over a common denominator: FE→=a⃗+c⃗−b⃗−d⃗2\overrightarrow{FE} = \frac{\vec{a} + \vec{c} - \vec{b} - \vec{d}}{2} Therefore, the RHS is: RHS=k(a⃗+c⃗−b⃗−d⃗2)\text{RHS} = k \left(\frac{\vec{a} + \vec{c} - \vec{b} - \vec{d}}{2}\right) RHS=k2(a⃗+c⃗−b⃗−d⃗)\text{RHS} = \frac{k}{2}(\vec{a} + \vec{c} - \vec{b} - \vec{d})

Step 4: Equate LHS and RHS and Solve for kk Now, we equate the simplified LHS and RHS: 2(b⃗+d⃗−a⃗−c⃗)=k2(a⃗+c⃗−b⃗−d⃗)2(\vec{b} + \vec{d} - \vec{a} - \vec{c}) = \frac{k}{2}(\vec{a} + \vec{c} - \vec{b} - \vec{d}) Observe that the vector term (b⃗+d⃗−a⃗−c⃗)(\vec{b} + \vec{d} - \vec{a} - \vec{c}) on the LHS is the negative of the vector term (a⃗+c⃗−b⃗−d⃗)(\vec{a} + \vec{c} - \vec{b} - \vec{d}) on the RHS. Let V⃗=a⃗+c⃗−b⃗−d⃗\vec{V} = \vec{a} + \vec{c} - \vec{b} - \vec{d}. Then (b⃗+d⃗−a⃗−c⃗)=−(a⃗+c⃗−b⃗−d⃗)=−V⃗(\vec{b} + \vec{d} - \vec{a} - \vec{c}) = -(\vec{a} + \vec{c} - \vec{b} - \vec{d}) = -\vec{V}. Substitute this into the equation: 2(−V⃗)=k2(V⃗)2(-\vec{V}) = \frac{k}{2}(\vec{V}) −2V⃗=k2V⃗-2\vec{V} = \frac{k}{2}\vec{V} Since V⃗\vec{V} is generally not the zero vector for a general quadrilateral, we can equate the scalar coefficients: −2=k2-2 = \frac{k}{2} Multiply both sides by 2 to solve for kk: k=−2×2k = -2 \times 2 k=−4k = -4

  1. Common Mistakes & Tips
  • Sign Errors: Be meticulously careful with signs when subtracting vectors or expanding expressions. For example, −(c⃗−b⃗)-(\vec{c} - \vec{b}) should be expanded to −c⃗+b⃗-\vec{c} + \vec{b}.
  • Vector Direction: Always remember that PQ→\overrightarrow{PQ} represents the displacement from PP to QQ, so PQ→=q⃗−p⃗\overrightarrow{PQ} = \vec{q} - \vec{p}.
  • Midpoint Formula Application: Ensure the midpoint formula is applied correctly to the correct diagonals (ACAC for EE, BDBD for FF).
  1. Summary

The problem involves simplifying a vector equation related to a quadrilateral. By representing the vertices and midpoints using position vectors and applying the definitions of displacement vectors and the midpoint formula, the LHS and RHS of the given equation were systematically simplified. The resulting algebraic equation allowed for the determination of the scalar value kk. The key was to recognize the relationship between the vector terms on both sides of the equation.

The final answer is −4\boxed{-4}, which corresponds to option (C).

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