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A unit vector perpendicular to both the vectors $a=2i-3j+k$ and $b=i+j-2k$ is:

1. $\frac{1}{\sqrt{3}} (i+j+k)$
2. $\frac{1}{3} (i+j-k)$
3. $\frac{1}{3} (i-j-k)$
4. $\frac{1}{\sqrt{3}} (i+j-k)$

the question has mistake in GO pdf

okay bro. Thank you for letting me know.
Let the unit vector be

X= (x,y,z)^T

X= (1/sqrt(x,y,j)) (xi+yj+zk) ……….1

a=2i−3j+k

and b=i+j−2k

a.X = b.X = 0

2x-3y+z=0=x+y-2z

It is a Homogeneous equation

Infinite Solutions are there

By solution y=z=k,(k=non zero constant)

x=k

So, X=(1,1,1)

So, replacing at eq1

we get optio A

The Cross Product of $a$ and $b$ is a unit vector perpendicular to both $a$ and $b$

The cross product of $a=<a_1,a_2,a_3>$ and $b=<b_1,b_2,b_3>$ is

$a \times b = \begin{vmatrix} i &j &k \\ a_1&a_2 &a_3 \\ b_1&b_2 &b_3 \end{vmatrix}$

From question , $a_1 =2, a_2=-3, a_3=1, b_1=1,b_2=1,b_3=-2$

$a \times b = \begin{vmatrix} i &j &k \\ 2&-3 &1 \\ 1&1 &-2 \end{vmatrix}$

$=i\begin{vmatrix} -3 &1 \\ 1& -2 \end{vmatrix} -j\begin{vmatrix} 2 &1 \\ 1& -2 \end{vmatrix}+k\begin{vmatrix} 2 &-3 \\ 1& 1 \end{vmatrix}$

$=5i+5j+5k$

Hence the vector $a \times b=5i+5j+5k=<5,5,5>$ is a vector perpendicular to $a=2i−3j+k$ and $b=i+j−2k$

The unit length of the vector $a \times b = \sqrt { 5^2+5^2+5^2} = \sqrt {5*5*3} = 5 \sqrt 3$

$\therefore$ A unit vector perpendicular to both $a=2i−3j+k$ and $b=i+j−2k$  is

$\frac{<5,5,5>}{5\sqrt 3} = \frac{<1,1,1>}{\sqrt 3} = \frac{1}{\sqrt 3} i+j+k$

Hence Option $A.$ is correct answer.

by

yes , that question is on differnt concept.
You are awesome ;-)
This should be the best answer.

Answer should be $A$.

Dot product of two perpendicular vector is $0.$
Vector given in option A gives $0$ dotproduct with vector $b$ while any other vector is not giving $0$ in dotproduct. Therefore answer should be $A.$

To find the perpendicular unit vector to two vectors please see:

by

There was misprint in question , now updated.
edited

@toxicdesire

@jayendra has written answer in his own style. Its better if I don't edit it.

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