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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. $\frac{1}{\sqrt{3}} (i+j-k)$
in Linear Algebra by Veteran
edited by | 1.2k views
0
Unit + Perpendicular. Assume some vector then use dot product to find unknowns. or directly use cross product to find desired vector.
+2
There is a misprint in this question. The original question is:

 

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

A. $\frac{1}{\sqrt{3}}(i+j+k)$

B. $\frac{1}{3}(i+j-k)$

C. $\frac{1}{3}(i-j-k)$

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

Please edit the question if possible.

2 Answers

+10 votes
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:
http://www.leadinglesson.com/problem-on-finding-a-vector-perpendicular-to-two-vectors

by Loyal
edited by
+2
a.(Option A) = 0 should also satisfy.
+1
Also, option B and option C are are not unit vectors.
+1
The dot product of vector a and option (A) is not equal to 0.I think option A is not correct or some Typo.
0
take 1 as "i" it's typo
0
wrong ans
0
Can anyone please explain how option (A) is correct. I have tried both dot product and cross product methods but option (A) is not matched.

1) using dot product :-

$\vec{a} = 2i - 2j + k$ and $\vec{b} = i + j -2k$ and let, option (A) = $\vec{c} = \frac{(i + j +k) }{\sqrt{3}}$

Now , $\vec{a}. \vec{c} $  = $\frac{2-2+1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \neq 0$

2) by using cross product of $\vec{a}\;\;and\;\;  \vec{b} $ to get the a vector which is perpendicular to the plane in which vectors $\vec{a}\;\; and\;\; \vec{b}$ lies :-

$\vec{n}$= $\begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 2&-2 &1 \\ 1&1 &-2 \end{vmatrix} = \hat{i} \begin{vmatrix} -2 &1 \\ 1& -2 \end{vmatrix} -\hat{j}\begin{vmatrix} 2 &1 \\ 1& -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 &-2 \\ 1& 1 \end{vmatrix} = 3\hat{i} + 5\hat{j} + 4\hat{k}$

Now , unit length vector in the direction of vector $\vec{n}$ is :-

$\hat{n}$ = $\frac{\vec{n}}{|\vec{n}|}$ = $\frac{3\hat{i} + 5\hat{j} + 4\hat{k}}{\sqrt{9 +25+16}}$ = $\frac{3\hat{i} + 5\hat{j} + 4\hat{k}}{5\sqrt{2}}$

It is not matching with any given options. So , Please tell me , Am I missing something or given options are wrong ?
0

you are doing dot product for a & b.

You should do dot product for a with all the options ( A ,B,C,D)

and b with all the options ( A ,B,C,D)  --> Selected Answer does this.

Your method fails... as a & b need not be perpendicular to each other. (it is not mentioned in the question also)

0

@MIRIYALA , Have I done the dot product of a & b ?

0
sorry ... I did not  read properly..

but taking the dot product with a is not giving 0 ( for any of the options)

Do the dot product of b with all options ( option A will give you 0 ).

Even I too didn't understand why it is the case..
+3
yes, if we consider it as a correct question then none of the options are matched here..I think something is missing in the question...
0
vector given in option A gives 0 on dotproduct with vector b, but not with vector a. Why option A is chosen as correct answer
0
There was misprint in question , now updated.
0

@Satbir will you please edit the answer too?

0

@toxicdesire

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

You can see my answer.

+3 votes

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.


Reference :- http://www.leadinglesson.com/problem-on-finding-a-vector-perpendicular-to-two-vectors

by Boss
+1

Two non-zero vectors $\vec{u}$ and $\vec{v}$ are orthogonal(perpendicular) if and only if $\vec{u}\cdot \vec{v} = 0.$

For given vector $\vec{C},$ the unit vector $\hat{C} = \dfrac{\vec{C}}{\left \|\vec{C} \right \|}$ 

Ref:https://brilliant.org/wiki/unit-vectors/

0

@Satbir yes right approach.

But that procedure only for unit vector , right?

Means cannot applicable here https://gateoverflow.in/118311/gate2017-1-30

 

0
yes , that question is on differnt concept.
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