Question

The area of the parallelogram determined by A = 2i+j-3k and B 12j-2K is approximately:
a. 43 b. 56 c.38 d.74

Answer 1

Correct question:

To find the area of the parallelogram whose adjacent sides are determined by $\vec{A} = 2i + j -3k$ , and $\vec{B}= 12j -2k$ .

Theory :

if $\vec{A}$ and $\vec{B}$ represent the adjacent sides of a parallelogram , then its area is given by ;

Area of parallelogram = | A × B|

Solution :

★ Given :

$\vec{A} = 2i + j -3k$ and

$\vec{B}= 12j -2k$

$\vec{a} \: \times \vec{b}$

= i(-2+36) -j ( -4) +k (24)

= 34i +4j +24k

$| \vec{a} \: \times \vec{b} \: | = \sqrt{34 {}^{2} + 4 { }^{2} + 24 {}^{2} }$

$= \sqrt{1748}$

$= 41.80$

therefore , Area of parellogram = 43 square units ( approximately )

so the correct option is a) 43 sq units .

______________________

More formulas:

if diagonals of a parallelogram, d1 and d2 are given then ,

Area of parallelogram =$\frac{1}{2}$|d1 × d2|

Answer 2

Correct question:

The area of the parallelogram determined by $\vec{a}=2\hat{i}+\hat{j}-3\hat{k}$ and $\vec{b}=12\hat{j}-2\hat{k}$ is approximately:

• A. 43
• B. 56
• C. 38
• D. 74

Answer:

$\large\boxed{\sf{(A)\;43\;sq.\;units}}$

Explanation:

Given that there are sides of paralleolgram reprsented by vectors,

$\vec{a}=2\hat{i}+\hat{j}-3\hat{k}$

and

$\vec{b}=12\hat{j}-2\hat{k}$

Now, we have to find the area of paralleolgram.

But, we know that, area of paralleolgram is given by modulus of their cross product of sides in vector form, i.e., $\bold{|\vec{a}\times\vec{b}|}$

Now, we have to find the cross product of the given vectors.

Also, we know that, in cross product,

• $\hat{ i } \times \hat{ i } = \hat{ j } \times \hat{ j} = \hat{ k } \times \hat{ k } = 0$
• $\hat{i } \times \hat{ j } = - ( \hat{j } \times \hat{ i} )= \hat{ k }$
• $\hat{j } \times \hat{k } = - ( \hat{ k } \times\hat{j }) = \hat{ i }$
• $\hat{ k } \times \hat{ i} = - (\hat{ i } \times \hat{ k }) = \hat{ j }$

On calculation, we will get,

$\vec{ a } \times \vec{ b } = 34\hat{ i } + 4\hat{ j } + 24\hat{ k }$

$\red{Note:-}Refer\;to\;the\; attachment.$

Therefore, we will get,

$= > | \vec{a } \times \vec{ b} | = \sqrt{ {(34)}^{2} + {(4)}^{2} + {(24)}^{2} } \\ \\ = > | \vec{a } \times \vec{ b} | = \sqrt{1156 + 16 + 576} \\ \\ = > | \vec{a } \times \vec{ b} | = \sqrt{1748} \\ \\ = > | \vec{a } \times \vec{ b} | = 41.80$

Thus, we have required area nearest to 41.80.

Hence, the correct option is (A) 43 sq. units