Using only vectors.
(1) The altitudes of a triangle are concurrent.
(2) The angle subtended on semicircle is right angle.
Using only vectors.
(1) The altitudes of a triangle are concurrent.
(2) The angle subtended on semicircle is right angle.
(1) To prove the altitudes of a triangle are concurrent by using vectors:
Let the given triangle be ΔABC
with vertices A, B, C of position vectors respectively.
Let AD, BE be the altitudes of the given triangle.
Let O be the orthocentre of the given triangle of position vectors at origin.
We know that orthocentre is the point of intersection of the altitudes. Thus, O is the point of intersection of AD and BE.
The vector representation of OA=→a
The vector representation of OB=→b
The vector representation of OC=→c
We know that AD is perpendicular to BC, by the definition of an altitude of a triangle.
Thus, we get that OA is perpendicular to BC.
The vector representation of the above statement is
→a.(→c−→b)=0
Since we know that the dot product of perpendicular vectors is 0.
By solving the above equation, we get
→a.→c−→a.→b=0 …… (1)
We know that BE is perpendicular to CA, by the definition of an altitude of a triangle.
Thus, we get that OB is perpendicular to CA.
The vector representation of the above statement is
→b.(→a−→c)=0
→a.→b−→c.→b=0
By adding the above equation and equation (1), we get
→a.→b−→c.→b+→a.→c−→a.→b=0+0
→a.→c−→c.→b=0
→c.(→a−→b)=0
Thus, we get OC is perpendicular to BA.
This implies OC is an altitude of the given triangle.
Thus, the altitudes of a triangle are concurrent.
(2) The angle subtended on semicircle is right angle using vectors:
Consider a circle with centre c and radius r as follows
Where AB is a diameter and →AC=→CB=→a and →CP=→r is a position vector of a point P on the circle.
Since, →AP=→AC+→CP [From vector addition
→AP=→a+→r.......(i)
Similar way, →BP=→BC+→CP
=−→CB+→CP
∴→BP=−→a+→r......(ii)
Consider →AP.→BP=(→r+→a).(→r−→a)
=→r.→r−→r.→a+→a.→r−→a.→a
=|r|2−|a|2
∴→AP.→BP=0 [As →r=→a]
Thus, →AP⊥→BP=0
∴∠APB=90∘
Hence, the angle subtended on a semicircle is the right angle.
Therefore , the correct option is A (True)
(1) To prove the altitudes of a triangle are concurrent by using vectors:
Let the given triangle be ΔABC
with vertices A, B, C of position vectors respectively.
Let AD, BE be the altitudes of the given triangle.
Let O be the orthocentre of the given triangle of position vectors at origin.
We know that orthocentre is the point of intersection of the altitudes. Thus, O is the point of intersection of AD and BE.
The vector representation of OA=→a
The vector representation of OB=→b
The vector representation of OC=→c
We know that AD is perpendicular to BC, by the definition of an altitude of a triangle.
Thus, we get that OA is perpendicular to BC.
The vector representation of the above statement is
→a.(→c−→b)=0
Since we know that the dot product of perpendicular vectors is 0.
By solving the above equation, we get
→a.→c−→a.→b=0 …… (1)
We know that BE is perpendicular to CA, by the definition of an altitude of a triangle.
Thus, we get that OB is perpendicular to CA.
The vector representation of the above statement is
→b.(→a−→c)=0
→a.→b−→c.→b=0
By adding the above equation and equation (1), we get
→a.→b−→c.→b+→a.→c−→a.→b=0+0
→a.→c−→c.→b=0
→c.(→a−→b)=0
Thus, we get OC is perpendicular to BA.
This implies OC is an altitude of the given triangle.
Thus, the altitudes of a triangle are concurrent.
(2) The angle subtended on semicircle is right angle using vectors:
Consider a circle with centre c and radius r as follows
Where AB is a diameter and →AC=→CB=→a and →CP=→r is a position vector of a point P on the circle.
Since, →AP=→AC+→CP [From vector addition
→AP=→a+→r.......(i)
Similar way, →BP=→BC+→CP
=−→CB+→CP
∴→BP=−→a+→r......(ii)
Consider →AP.→BP=(→r+→a).(→r−→a)
=→r.→r−→r.→a+→a.→r−→a.→a
=|r|2−|a|2
∴→AP.→BP=0 [As →r=→a]
Thus, →AP⊥→BP=0
∴∠APB=90∘
Hence, the angle subtended on a semicircle is the right angle.
Therefore , the correct option is A (True)
