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
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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
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Correct question:
To find the area of the parallelogram whose adjacent sides are determined by [tex]\vec{A} = 2i + j -3k [/tex] , and [tex]\vec{B}= 12j -2k [/tex] .
Theory :
if [tex]\vec{A}[/tex] and [tex]\vec{B}[/tex] represent the adjacent sides of a parallelogram , then its area is given by ;
Area of parallelogram = | A × B|
Solution :
★ Given :
[tex]\vec{A} = 2i + j -3k [/tex] and
[tex]\vec{B}= 12j -2k [/tex]
[tex] \vec{a} \: \times \vec{b} [/tex]
= i(-2+36) -j ( -4) +k (24)
= 34i +4j +24k
[tex] | \vec{a} \: \times \vec{b} \: | = \sqrt{34 {}^{2} + 4 { }^{2} + 24 {}^{2} } [/tex]
[tex] = \sqrt{1748} [/tex]
[tex] = 41.80[/tex]
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 =[tex]\frac{1}{2}[/tex]|d1 × d2|
Verified answer
Correct question:
The area of the parallelogram determined by [tex]\vec{a}=2\hat{i}+\hat{j}-3\hat{k}[/tex] and [tex]\vec{b}=12\hat{j}-2\hat{k}[/tex] is approximately:
Answer:
[tex]\large\boxed{\sf{(A)\;43\;sq.\;units}}[/tex]
Explanation:
Given that there are sides of paralleolgram reprsented by vectors,
[tex]\vec{a}=2\hat{i}+\hat{j}-3\hat{k}[/tex]
and
[tex]\vec{b}=12\hat{j}-2\hat{k}[/tex]
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., [tex]\bold{|\vec{a}\times\vec{b}|}[/tex]
Now, we have to find the cross product of the given vectors.
Also, we know that, in cross product,
On calculation, we will get,
[tex]\vec{ a } \times \vec{ b } = 34\hat{ i } + 4\hat{ j } + 24\hat{ k }[/tex]
[tex]\red{Note:-}Refer\;to\;the\; attachment.[/tex]
Therefore, we will get,
[tex] = > | \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[/tex]
Thus, we have required area nearest to 41.80.
Hence, the correct option is (A) 43 sq. units