Question

Find the vector area and the area of the parallelogram having a = i + 2j - k and b = 2i - j + 2k as adjacent sides.

Answer 1

Step-by-step explanation:

Given a=I+2j-k

b=2i-j+2k

if the area of parallelogram having a and b as a adjacent sides is axb and their vector area is |axb|

Now,axb= I j k

1 2 -1

2 -1 2

=I(4-1)-j(2+2)+k(-1-4)

=3i-4j-5k

Therefore, the area of parallelogram having a and b as adjacent sides is axb=3i-4j-5k

Now vector area=|axb|

=√9+16+25

=√50

=√25×√2

=5√2 sq.units

Therefore, the area of parallelogram having a and b as adjacent sides is axb=3i-4j-5k and their vector area is |axb|=5√2 sq.units

Answer 2

Given:

Parallelogram having a = i + 2j - k and b = 2i - j + 2k as adjacent sides.

To Find:

The vector area of the parallelogram.

Solution:

The area of a parallelogram in vector form using the adjacent sides is,

|a×b|

where,

a and b are vectors representing two adjacent sides

we know that,

• i×i=j×j=k×k=0
• i×j=k; j×k=i; k×i=j
• j×i=−k; k×j=−i; i×k=−j

so,

Area of parallelogramm = |a×b|

a×b = (i + 2j - k) ×  (2i - j + 2k)

⇒ i × (2i - j + 2k) + 2j × (2i - j + 2k) - k × (2i - j + 2k)

⇒ - k - 2j - 4k + 4i - 2j - i

⇒ 3i - 4j - 5k

|a×b|  = √3² + 5²+ 4²

|a×b|  = √9+ 16 +25

|a×b|  = √50 = 5√2

Hence, the area of parallelogram is 5√2 units.