Physics, asked by sarayuka, 11 months ago

If diagonals of a parallelogram are (i-j) and (i+j). then area of the parallelogram will be (1) 0.5 unit (3) 2 unit (2) 1 unit (4) 4 unit​

Answers

Answered by shadowsabers03
24

We know the magnitude of area of a parallelogram made by the vectors \vec{\sf{P}} and \vec{\sf{Q}} as the adjacent sides is,

\longrightarrow\textsf{A}=\left|\vec{\sf{P}}\times\vec{\sf{Q}}\right|\quad\quad\dots\sf{(1)}

Let \vec{\sf{D_1}} and \vec{\sf{D_2}} be the diagonal vectors of the parallelogram. Then,

\longrightarrow\vec{\sf{D_1}}=\vec{\sf{P}}+\vec{\sf{Q}}\quad\quad\dots\sf{(2)}

\longrightarrow\vec{\sf{D_2}}=\vec{\sf{P}}-\vec{\sf{Q}}\quad\quad\dots\sf{(3)}

On adding (2) and (3),

\longrightarrow\vec{\sf{P}}=\dfrac{\vec{\sf{D_1}}+\vec{\sf{D_2}}}{\sf{2}}

And on subtracting (3) from (2),

\longrightarrow\vec{\sf{Q}}=\dfrac{\vec{\sf{D_1}}-\vec{\sf{D_2}}}{\sf{2}}

Hence (1) becomes,

\longrightarrow\textsf{A}=\dfrac{\left|\left(\vec{\sf{D_1}}+\vec{\sf{D_2}}\right)\times\left(\vec{\sf{D_1}}-\vec{\sf{D_2}}\right)\right|}{\sf{4}}

\longrightarrow\textsf{A}=\dfrac{\left|\left(\vec{\sf{D_1}}\times\vec{\sf{D_1}}\right)-\left(\vec{\sf{D_1}}\times\vec{\sf{D_2}}\right)+\left(\vec{\sf{D_2}}\times\vec{\sf{D_1}}\right)-\left(\vec{\sf{D_2}}\times\vec{\sf{D_2}}\right)\right|}{\sf{4}}

Since \vec{\sf{a}}\times\vec{\sf{a}}=\vec{\sf{0}} for any arbitrary vector \vec{\sf{a}},

\longrightarrow\textsf{A}=\dfrac{\left|-\left(\vec{\sf{D_1}}\times\vec{\sf{D_2}}\right)+\left(\vec{\sf{D_2}}\times\vec{\sf{D_1}}\right)\right|}{\sf{4}}

Since \vec{\sf{a}}\times\vec{\sf{b}}=-\left(\vec{\sf{b}}\times\vec{\sf{a}}\right),

\longrightarrow\textsf{A}=\dfrac{\left|\left(\vec{\sf{D_2}}\times\vec{\sf{D_1}}\right)+\left(\vec{\sf{D_2}}\times\vec{\sf{D_1}}\right)\right|}{\sf{4}}

\longrightarrow\underline{\underline{\textsf{A}=\dfrac{\left|\vec{\sf{D_2}}\times\vec{\sf{D_1}}\right|}{\sf{2}}}}

Or,

\longrightarrow\underline{\underline{\textsf{A}=\dfrac{\left|\vec{\sf{D_1}}\times\vec{\sf{D_2}}\right|}{\sf{2}}}}

Here,

  • \vec{\sf{D_1}}=\sf{\hat i-\^j}

  • \vec{\sf{D_2}}=\sf{\hat i+\^j}

Then,

\longrightarrow\textsf{A}=\dfrac{\sf{\left|\left(\hat i-\hat j\right)\times\left(\hat i+\hat j\right)\right|}}{\sf{2}}

\longrightarrow\textsf{A}=\dfrac{\sf{\left|\left(\hat i\times\hat i\right)+\left(\hat i\times\hat j\right)-\left(\hat j\times\hat i\right)-\left(\hat j\times\hat j\right)\right|}}{\sf{2}}

\longrightarrow\textsf{A}=\dfrac{\sf{\left|\left(\hat i\times\hat j\right)-\left(\hat j\times\hat i\right)\right|}}{\sf{2}}

Since \sf{\hat i\times\^j=\^k} and \sf{\^j\times\hat i=-\^k,}

\longrightarrow\textsf{A}=\dfrac{\sf{\left|\left(\hat k\right)-\left(-\hat k\right)\right|}}{\sf{2}}

\longrightarrow\textsf{A}=\dfrac{\sf{\left|\hat k+\hat k\right|}}{\sf{2}}

\longrightarrow\textsf{A}=\dfrac{\sf{2\left|\^k\right|}}{\sf{2}}

\longrightarrow\textsf{A}=\sf{\left|\^k\right|}

\longrightarrow\underline{\underline{\sf{A=1}}}

Hence the answer is 1 unit.

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