Question

can anyone explain ​

Answer 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,

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

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

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

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

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