Question

Consider a parallelogram whose adjacent sides are
given by a = 2m + ñ and b=m+ 2n, where m
and ñ are unit vectors inclined at an angle of 60°.
Based on this information, answer the following
questions.
One diagonal of the parallelogram is represented
by the vector
(a) m+n (b) 3m+ň (c) 3m-ñ (d) -ń+ 2n
The other diagonal of the parallelogram is
represented by the vector
(a) n-3m (b) m +3n (c) -m - 3n (d) m-n
The area of the parallelogram is
(a) 2/3 sq. units (b) 573 sq. units
(c) 13 sq. units
(d) 3
213​

Answer 1

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Answer 2

The value of one diagonal is 3(m+n).

The value of the other diagonal is m-n.

The area of the parallelogram is (5√3)/2 square units.

Given:

a = 2m + n

b=m+ 2n

m and n are unit vectors inclined at an angle(θ) of 60°.

To Find:

1) The value of one diagonal of the parallelogram is represented in a vector.

2) The value of another diagonal of the parallelogram is

represented by the vector,

3) The value of the area of the parallelogram.

Solution:

We know in the parallelogram of given two sides one diagonal is given as the sum of two sides and the other diagonal is given as the difference between the two sides.

a = 2m+n, b = m+2n are the sides of parallelogram

One diagonal = a+b

= 2m + n + m+ 2n

= 3m +3n

= 3(m+n)

Therefore, The value of one diagonal is 3(m+n).

Other diagonal = a-b

= 2m+n - m-2n

= m-n

Therefore, The value of the other diagonal is m-n.

The area of the parallelogram is equal to the cross product of two sides of the parallelogram

From the above information

Area, a×b = |a|×|b|sin θ -----(1)

|a| = √2²+1²

|a| = √5

|b| = √2²+1²

|b| = √5

Substitute the values of |a| and |b| in equation(1)

Area, a×b = √5×√5×sin 60°

a×b = (5√3)/2

Therefore, The area of the parallelogram is (5√3)/2 square units.

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