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Q2   Find a unit vector perpendicular to each of the vector \vec a + \vec b \: \: and\: \: \vec a - \vec b , where                                                        \vec a = 3 \hat i + 2 \hat j + 2 \hat k \: \:and \: \: \vec b = \hat i + 2 \hat j - 2 \hat k
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\vec a = 3 \hat i + 2 \hat j + 2 \hat k \: \:and \: \: \vec b = \hat i + 2 \hat j - 2 \hat k

\vec a + \vec b =3\hat i +2\hat j+2\hat k+\hat i +2\hat j-2\hat k=4\hat i +4\hat j

\vec a - \vec b =3\hat i +2\hat j+2\hat k-\hat i -2\hat j+2\hat k=2\hat i +4\hat k

Now , A vector which perpendicular to both \vec a + \vec b \: \: and\: \: \vec a - \vec b is  (\vec a + \vec b) \times (\vec a - \vec b)

(\vec a + \vec b) \times (\vec a - \vec b)=\begin{vmatrix} \hat i &\hat j &\hat k \\ 4&4 &0 \\ 2& 0& 4 \end{vmatrix}

(\vec a + \vec b) \times (\vec a - \vec b)= \hat i (16-0)-\hat j(16-0)+\hat k(0-8)

(\vec a + \vec b) \times (\vec a - \vec b)= 16\hat i -16\hat j-8\hat k

And a unit vector in this direction :

\vec u =\frac{16\hat i-16\hat j-8\hat k}{|16\hat i-16\hat j-8\hat k|}=\frac{16\hat i-16\hat j-8\hat k}{\sqrt{16^2+(-16)^2+(-8)^2}}

\vec u =\frac{16\hat i-16\hat j-8\hat k}{24}=\frac{2}{3}\hat i-\frac{2}{3}\hat j-\frac{1}{3}\hat k

Hence Unit vector perpendicular to each of the vector \vec a + \vec b \: \: and\: \: \vec a - \vec b is \frac{2}{3}\hat i-\frac{2}{3}\hat j-\frac{1}{3}\hat k.

Posted by

Pankaj Sanodiya

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