Suppose u⎯⎯⎯=⟨−1,1⟩u¯=⟨−1,1⟩ and v⎯⎯⎯=⟨0,−2⟩v¯=⟨0,−2⟩ are two vectors that form the sides of a parallelogram. then the lengths of the two diagonals of the parallelogram are

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Answer:

The length of the two diagonals of a parallelogram are:

                 [tex]\sqrt{2}\ \text{units\ and}\ \sqrt{10}\ \text{units}[/tex]

Step-by-step explanation:

We know that if two vectors form the sides of a parallelogram then the two diagonals of the parallelogram are:

sum of the two vectors and difference of two vectors.

i.e. if u and v are two vectors such that they form the side of  a parallelogram the,

u+v and u-v form the diagonal of a parallelogram.

Here we have:

u=<-1,1>

and v=<0,-2>

Hence,

u+v=<-1,1>+<0,-2>

i.e.

u+v=<-1+0,1-2>

i.e.

u+v=<-1,-1>

Also, the length of the diagonal will be:

||u+v||=||<-1,-1>||

[tex]||<-1,-1>||=\sqrt{(-1)^2+(-1)^2}\\\\i.e.\\\\||<-1,-1>||=\sqrt{2}[/tex]

Length of one diagonal is: √2 units

and the other diagonal is formed by the vector:

u-v

i.e.

u-v=<-1,1>-<0,-2>

i.e.

u-v=<-1-0,1-(-2)>

i.e.

u-v=<-1,3>

i.e. the length of the other diagonals is:

||u-v||=||<-1,3>||

Hence,

[tex]||<-1,3>||=\sqrt{(-1)^2+(3)^2}\\\\||<-1,3>||=\sqrt{1+9}\\\\i.e.\\\\||<-1,3>||=\sqrt{10}[/tex]

Hence, length of other diagonal is: √10 units.

The lengths of the two diagonals of the parallelogram are [tex]\sqrt{2}[/tex] and [tex]\sqrt{10}[/tex], respectively.

Vectorially speaking, the diagonals of the parallelogram can be found by means of the following vectorial formulas:

[tex]\vec {d}_{1} = \vec u + \vec v[/tex] (1)

[tex]\vec d_{2} = \vec u - \vec v[/tex] (2)

If we know that [tex]\vec u = \langle -1, 1 \rangle[/tex] and [tex]\vec v = \langle 0, -2 \rangle[/tex], then the vectors diagonal are:

[tex]\vec {d_{1}} = \langle -1, 1 \rangle + \langle 0, -2 \rangle[/tex]

[tex]\vec d_{1} = \langle -1, -1 \rangle[/tex]

[tex]\vec {d_{2}} = \langle -1, 1 \rangle - \langle 0, -2 \rangle[/tex]

[tex]\vec d_{2} = \langle -1, 3 \rangle[/tex]

And the lengths of the diagonals are calculated by Pythagorean theorem:

[tex]\|\vec {d}_{1}\| = \sqrt{(-1)^{2}+(-1)^{2}}[/tex]

[tex]\|\vec {d}_{1}\| = \sqrt{2}[/tex]

[tex]\|\vec {d}_{2}\| = \sqrt{(-1)^{2}+3^{2}}[/tex]

[tex]\|\vec {d}_2\| = \sqrt{10}[/tex]

The lengths of the two diagonals of the parallelogram are [tex]\sqrt{2}[/tex] and [tex]\sqrt{10}[/tex], respectively.

We kindly invite to check this question on parallelograms: https://brainly.com/question/1563728

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