Question

the area of parallelogram whose one diagonal is 2i+j-2k and one sine is 3i+j-k​

Answer 1

Step-by-step explanation:

Area of parallelogram= ✓18 sq. units

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Answer 2

The require answer is  $\sqrt{18}$ units .

parallelogram: A parallelogram is a two-dimensional geometric figure whose sides are parallel to each other. It is a type of polygon with four sides (also called a quadrilateral) where a pair of parallel sides have the same length. The sum of the adjacent angles of a parallelogram is equal to 180 degrees. In geometry you must have learned about many 2D shapes and sizes like circles, squares, rectangles, rhombuses etc. All these shapes have a different set of properties. Also, the area and perimeter formulas of these shapes differ from each other and are used to solve many problems. Here we will learn the definition, formulas and properties of a parallelogram.

Cross product: The cross product is a binary operation on two vectors in three-dimensional space. The result is again a vector that is perpendicular to both vectors. The cross product of two vectors is calculated according to the right-hand rule.

The right hand rule is nothing more than that the resultant of any two vectors is perpendicular to the other two vectors. Using the cross product, we can also find the magnitude of the resulting vector.

As per question we need to find the area of a parallelogram whose one diagonal is 2i+j-2k and one side is 3i+j-k​

now find the other side of the parallelogram

let a= 3i+j-k​

  d =2i+j-2k

let the other side of the parallelogram is b.

now a+b=d then b= d-b= 2i +j-2k - 3i -j+k= -i -k

we know that area of parallelogram is

|a x b|.

now we need to find the a x b

a x b =  $\left[\begin{array}{ccc}i&j&k\\3&1&-1\\-1&0&-1\end{array}\right]$

       = i(-1*1 -0*-1) -j(-1*3 -(-1)(-1)) + k(0*3- -1*-1)

        =-1i+4j-k

|a x b| = $\sqrt{1^{2}+4^{2}+(-1)^{2} }$

         = $\sqrt{18}$ units

Hence the require answer is  $\sqrt{18}$ units .

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https://brainly.com/question/29740341?

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