Question

Prove by vector method that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Answer 1

Taking $\overrightarrow a$and $\overrightarrow b$to be the position vectors of AB and BC. We know that $\overrightarrow a .\overrightarrow b = 0$ since AB is perpendicular to BC. By using the triangle law of vector addition we get $AC = \overrightarrow b - \overrightarrow a$ squaring which we get the required proof.

Complete step by step solution:
Let ABC be a right angled triangle right angled at B.
The side opposite to the right angle is known as the hypotenuse.
So here AC is the hypotenuse.
Hence we need to prove $A{C^2} = A{B^2} + B{C^2}$
Now, let and be the position vectors of AB and BC and since AB is perpendicular to BC we have their dot product to be zero.
Hence, $\overrightarrow a .\overrightarrow b = 0$…….(1)
By triangle law of vector addition we have
$\Rightarrow AC = \overrightarrow b - \overrightarrow a$
Squaring on both sides we get

$$\Rightarrow {\left( {AC} \right)^2} = {\left( {\overrightarrow b - \overrightarrow a } \right)^2} \\\ \Rightarrow {\left( {AC} \right)^2} = {\overrightarrow a ^2} + {\overrightarrow b ^2} - 2\overrightarrow a .\overrightarrow b \\\$$

Using equation (1) we get

$$\Rightarrow {\left( {AC} \right)^2} = {\overrightarrow a ^2} + {\overrightarrow b ^2} - 0 \\\ \Rightarrow {\left( {AC} \right)^2} = {\overrightarrow a ^2} + {\overrightarrow b ^2} \\\$$

Now , since and be the position vectors of AB and BC
$\Rightarrow {\left( {AC} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2}$
Hence the proof.

Note:
When two vectors $\overrightarrow a$ and $\overrightarrow b$ are perpendicular then their dot product is 0.
This is obtained by
$\Rightarrow \overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta$
And since they are perpendicular then the angle between them is ${90}^ \circ$.
Hence
$\Rightarrow \overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos {90^ \circ } = 0$
This is the proof of the property used.