Find the Area of a Parallelogram Formed by Vectors :
Here we are going to see how to find the area of parallelogram formed by vectors.
Question 1 :
Find the area of the parallelogram whose two adjacent sides are determined by the vectors i vector + 2j vector + 3k vector and 3i vector − 2j vector + k vector.
Let a vector = i vector + 2j vector + 3k vector
b vector = 3i vector − 2j vector + k vector.
Vector area of parallelogram = a vector x b vector
= i[2+6] - j[1-9] + k[-2-6]
= 8i + 8j - 8k
= √82 + 82 + (-8)2
Question 2 :
Find the area of the triangle whose vertices are A(3, - 1, 2), B(1, - 1, - 3) and C(4, - 3, 1).
A vector = (3i - j + 2k) vector
B vector = (i - j - 3k) vector
C vector = (4i - 3j + k) vector
AB vector = -2i vector -5k vector
AC vector = i vector - 2j vector - k vector
Area of triangle = (1/2) |AB vector x AC vector|
= (1/2) √(-10)2 + 72 + (-4)2
= (1/2) √(100+49+16)
= (1/2) √165
Hence area of the given triangle is (1/2) √165.
Question 3 :
If a vector, b vector, c vector are position vectors of the vertices A, B, C of a triangle ABC, show that the area of the triangle ABC is (1/2) |a × b + b × c + c × a| vector. Also deduce the condition for collinearity of the points A, B, and C.
Let OA vector = a vector
OB vector = b vector
OC vector = c vector
Area of triangle ABC = (1/2) |AB vector x AC vector|
= (1/2) |(OB - OA) x (OC - OA)|
= (1/2) |(b - a) x (c - a)|
= (1/2) |(b x c - b x a - a x c + a x a)|
= (1/2) |(b x c + a x b + c x a + 0 vector)|
= (1/2) |a x b + b x c + c x a|
If the points A, B and C are collinear, then
Area of triangle ABC = 0
(1/2) |a x b + b x c + c x a| = 0
|a x b + b x c + c x a| = 0
a x b + b x c + c x a = 0
Hence the required condition is a x b + b x c + c x a = 0.
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