Question

The vectors from origin to the points A and B. A = 3i-6j+2k and B= 2i+j-2k resp. The area of ∆AOB be

Answer 1

Area of ∆AOB =1/2| A × B|
=1/2| (3i-6j+2k)×(2i+j-2k) |
=1/2| i(12-2)-j(-6-4)+k(3+12) |
=1/2| 10i+10j+15k |
=1/2√(100+100+225)
=1/2√(425)
=1/2(20.6)
=10.3 unit square

Answer 2

The area of triangle AOB formed would be 10.3 units

1) The area of such a triangle would be given be 1/2 * | A x B |

2) A x B = $\left[\begin{array}{ccc}i&j&k\\3&-6&2\\2&1&-2\end{array}\right]$

3) A x B = i( 12 -2) -j( -6-4) + k( 3+12)

4) A x B = 10 i + 10 j + 15 k

5) | A x B | = √10² +10²+15² = √425 = 5√17

6) The area of triangle would be hence, 5/2* √17

7) Substituting the value of sqrt part gives:

Area of AOB = 10.3 units