NIMCET 2020 — Mathematics PYQ

Solution

1. Define the Vertices:

Let the position vectors of the vertices be $A,B,$ and $C$:

• $\vec{A}=(1,-2,2)$

• $\vec{B}=(2,1,-1)$

• $\vec{C}=(3,-1,2)$

2. Calculate the Side Vectors:

• $\overrightarrow{AB}=\vec{B}-\vec{A}=(2-1)\hat{i}+(1-(-2))\hat{j}+(-1-2)\hat{k}=\hat{i}+3\hat{j}-3\hat{k}$

• $\overrightarrow{BC}=\vec{C}-\vec{B}=(3-2)\hat{i}+(-1-1)\hat{j}+(2-(-1))\hat{k}=\hat{i}-2\hat{j}+3\hat{k}$

• $\overrightarrow{CA}=\vec{A}-\vec{C}=(1-3)\hat{i}+(-2-(-1))\hat{j}+(2-2)\hat{k}=-2\hat{i}-\hat{j}+0\hat{k}$

3. Calculate the Magnitudes (Side Lengths):

• $|\overrightarrow{AB}|=\sqrt{1^2+3^2+(-3{)}^2}=\sqrt{1+9+9}=\sqrt{19}$

• $|\overrightarrow{BC}|=\sqrt{1^2+(-2{)}^2+3^2}=\sqrt{1+4+9}=\sqrt{14}$

• $|\overrightarrow{CA}|=\sqrt{(-2{)}^2+(-1{)}^2+0^2}=\sqrt{4+1+0}=\sqrt{5}$

4. Check for Right-Angle Property:

Using the converse of Pythagoras theorem:

Since $|\overrightarrow{BC}{|}^2+|\overrightarrow{CA}{|}^2=|\overrightarrow{AB}{|}^2$, the triangle satisfies the Pythagoras theorem.

Final Answer:

The triangle is a Right-Angled Triangle.