Question

Find unit vectors perpendicular to the plane of the vectors a=i-2j+k and b=2i-k.

Answer 1

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Answer 2

Answer:

$\widehat{n}=\frac{2i+3j+4k}{\sqrt{29}}$

Step-by-step explanation:

The given vectors are:

$a=i-2j+k$ and $b=2i-k$

Now, in order to find the unit vector of the given vectors, we have to find the value of $a{\times}b$ that is

$a{\times}b=\left[\begin{array}{ccc}i&j&k\\1&-2&1\\2&0&-1\end{array}\right]$

=$i(2-0)-j(-1-2)+k(0+4)$

=$2i+3j+4k$

Thus, the unit vector is given as:

$\widehat{n}=\frac{2i+3j+4k}{\sqrt{4+9+16}}$

$\widehat{n}=\frac{2i+3j+4k}{\sqrt{29}}$

which is the required unit vector.