Q71 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics

If $A, B, C$ and $D$ are four points in the plane such that $|\mathbf{A B}|^2+|\mathbf{C D}|^2=|\mathbf{B C}|^2+|\mathbf{D A}|^2=100$, then $\mathbf{A C} \cdot \mathbf{B D}=$

  • A
    10
  • Correct answer
    B
    0
  • C
    $\frac{1}{10}$
  • D
    1

Solution

Let $A, B, C$ and $D$ are vertices of a quadrilateral So, $$ |\mathbf{A C}|^2+|\mathbf{B D}|^2=|\mathbf{A B}|^2+|\mathbf{C D}|^2+2 \mathbf{B C} \cdot \mathbf{A D} $$ and $$ |\mathbf{A C}|^2+|\mathbf{B D}|^2=|\mathbf{A D}|^2+|\mathbf{B C}|^2+2 \mathbf{A B} \cdot \mathbf{D C} \text {. } $$ From Eqs. (i) and (ii), $$ \begin{aligned} & \mathbf{B C} \cdot \mathbf{A D}=\mathbf{A B} \cdot \mathbf{D C} \\ & \Rightarrow(\mathbf{O C}-\mathbf{O B}) \cdot(\mathbf{O D}-\mathbf{O A})=(\mathbf{O B}-\mathbf{O A}) \cdot(\mathbf{O C}-\mathbf{O D}) \\ & \Rightarrow \quad \text { OC } \cdot \text { OD }- \text { OC } \cdot \text { OA }- \text { OB } \cdot \text { OD }+ \text { OB } \cdot \text { OA } \\ & \quad=\mathbf{O B} \cdot \mathbf{O C}-\mathbf{O B} \cdot \mathbf{O D}-\mathbf{O A} \cdot \mathbf{O C}+\mathbf{O A} \cdot \mathbf{O D} \\ & \Rightarrow \mathbf{O C} \cdot \mathbf{O D}+\mathbf{O B} \cdot \mathbf{O A}=\mathbf{O B} \cdot \mathbf{O C}+\mathbf{O A} \cdot \mathbf{O D} \\ & \Rightarrow \mathbf{O C} \cdot(\mathbf{O D}-\mathbf{O B})+\mathbf{O A} \cdot(\mathbf{O B}-\mathbf{O D})=0 \\ & \Rightarrow \mathbf{O C} \cdot \mathbf{B D}+\mathbf{O A} \cdot \mathbf{D B}=0 \Rightarrow \mathbf{B D} \cdot(\mathbf{O C}-\mathbf{O A})=0 \\ & \Rightarrow \mathbf{B D} \cdot \mathbf{A C}=0 \end{aligned} $$

Hence, the correct answer is option: 'B'.

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