Cross Product - Free MCQ Practice Test with solutions, JEE Maths

As we know that,

Area of triangle = ½(a * b)
a = (1, 0, -2)   b = (2, 3, 1)
= i(0 + 6) + j(-4 - 1) + k(3 - 0)
= 6i - 5j + 3k
|a * b| = (36 + 25 + 9)^½
|a * b| = (70)^½
Area of triangle = ½(a * b)
= [(70)^½]/2

 i.(j * k) + j.(k * i) + k(i * j)
= i.(i) + j.(j) + k.(k)
= 1 + 1 + 1 
= 3

The area of a parallelogram formed by two vectors a and b is given by the magnitude of the cross product of the two vectors:

Area = |a × b|

Given:

a = i - 3j + 4k
b = 3i + j - 2k

Step 1: Compute the cross product a × b

Expanding the determinant:

Calculate the 2x2 determinants:

Step 2: Find the magnitude of the cross product

Now, calculate the magnitude of the vector a × b