Question:
show that the points A,B,C with position vector 2i -j k, i -3j -5k and 3i -4j -4k respectively, are the vertices of a right angled triangle. hence find the area of the triangle.
Answer:
Let OA = 2i - j + k
OB = i - 3j - 5k
OC = 3i - 4j - 4k
and AB, BC and CA represent the sides of trinalge ABC
Now, AB = (1 - 2)i + (-3 - 1)j + (-5 - 1)k = -i - 2j - 6k
and |AB| = √{(-1)2 + (-2)2 + (-6)2 } = √(1 + 4 + 36) = √41
BC = (3 - 1)i + (-4 + 3)j + (-4 + 5)k = 2i - j + k
Now, |BC| = √{(2)2 + (-1)2 + 12 } = √(4 + 1 + 1) = √6
and CA = (2 - 3)i + (-1 + 4)j + (1 + 4)k = -i + 3j + 5k
Now, |CA| = √{(-1)2 + 32 + 52 } = √(1 + 9 + 25) = √35
Again, |BC|2 + |CA|2 = 6 + 35 = 41 = |AB|2
So, triangle ABC is a right angle triangle.
Hence, area of triangle ABC = (1/2) * base * altitude
= (1/2) * |BC| * |CA|
= (1/2) * √6 * √35
= (1/2) * √(6 * 35)
= (1/2) * √(210)
= √(210/4)
= √(105/2) square units
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