Physics, asked by premsingh60, 9 months ago

can anyone explain ​

Attachments:

Answers

Answered by shadowsabers03
1

\longrightarrow\sf{\overrightarrow{\sf{A}}\cdot\overrightarrow{\sf{B}}=\left|\overrightarrow{\sf{A}}\times\overrightarrow{\sf{B}}\right|}

\longrightarrow\sf{AB\cos\theta=AB\sin\theta}

\longrightarrow\sf{\sin\theta=\cos\theta}

\Longrightarrow\sf{\theta=45^o\quad\Longleftarrow\quad\tan\theta=1}

Then,

\longrightarrow\sf{C=\left|\overrightarrow{\sf{A}}+\overrightarrow{\sf{B}}\right|}

\longrightarrow\sf{C=\sqrt{A^2+B^2+2AB\cos45^{\circ}}}

\longrightarrow\sf{C=\sqrt{A^2+B^2+2AB\cdot\dfrac{1}{\sqrt2}}}

\longrightarrow\sf{\underline{\underline{C=\sqrt{A^2+B^2+\sqrt2\ AB}}}}

======================================

The vector \sf{\overrightarrow{\sf{A}}} is,

\longrightarrow\sf{\overrightarrow{\sf{A}}=1(-\hat i)}

\longrightarrow\sf{\overrightarrow{\sf{A}}=-\hat i}

And the vector \sf{\overrightarrow{\sf{B}}} is,

\longrightarrow\sf{\overrightarrow{\sf{B}}=2\sqrt3\cos30^o\ \hat i+2\sqrt3\sin30^o\ \hat j}

\longrightarrow\sf{\overrightarrow{\sf{B}}=2\sqrt3\cdot\dfrac{\sqrt3}{2}\ \hat i+2\sqrt3\cdot\dfrac{1}{2}\ \hat j}

\longrightarrow\sf{\overrightarrow{\sf{B}}=3\ \hat i+\sqrt3\ \hat j}

Now the vector difference \overrightarrow{\sf{A}}-\overrightarrow{\sf{B}} is,

\longrightarrow\sf{\overrightarrow{\sf{A}}-\overrightarrow{\sf{B}}=-\hat i-(3\ \hat i+\sqrt3\ \hat j)}

\longrightarrow\sf{\overrightarrow{\sf{A}}-\overrightarrow{\sf{B}}=-\hat i-3\ \hat i-\sqrt3\ \hat j}

\longrightarrow\sf{\underline{\underline{\overrightarrow{\sf{A}}-\overrightarrow{\sf{B}}=-4\ \hat i-\sqrt3\ \hat j}}}

Alternate method:-

\overrightarrow{\sf{B}} is in first quadrant so -\overrightarrow{\sf{B}} should be in third quadrant (negative vector is opposite to the original vector).

\overrightarrow{\sf{A}} is along X axis, so surely \overrightarrow{\sf{A}}-\overrightarrow{\sf{B}} should be in third quadrant, because \overrightarrow{\sf{A}}-\overrightarrow{\sf{B}} comes between \overrightarrow{\sf{A}} and -\overrightarrow{\sf{B}}.

  • \sf{-4\ \hat i-\sqrt3\ \hat j} lies in third quadrant.

  • \sf{4\ \hat i-\sqrt3\ \hat j} lies in fourth quadrant.

  • \sf{\sqrt3\ \hat i-4\ \hat j} also lies in fourth quadrant.

  • \sf{-\hat i+2\sqrt3\ \hat j} lies in second quadrant.

Hence possible chance for the answer is \sf{\underline{\underline{\overrightarrow{\sf{A}}-\overrightarrow{\sf{B}}=-4\ \hat i-\sqrt3\ \hat j.}}}

Similar questions