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+6 votes

A unit vector perpendicular to both the vectors $a=2i-2j+k$ and $b=1+j-2k$ is:

  1. $\frac{1}{\sqrt{3}} (i+j+k)$
  2. $\frac{1}{3} (i+j-k)$
  3. $\frac{1}{3} (i-j-k)$
  4. $\frac{1}{\sqrt{3}} (i+j-k)$
asked in Linear Algebra by Veteran (59.6k points)
edited by | 694 views
Unit + Perpendicular. Assume some vector then use dot product to find unknowns. or directly use cross product to find desired vector.

1 Answer

+7 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 ans should be $A.$

to find the perpendicular unit vector to two vectors the procedure is as follows:
answered by Loyal (8.2k points)
edited by
a.(Option A) = 0 should also satisfy.
Also, option B and option C are are not unit vectors.
The dot product of vector a and option (A) is not equal to 0.I think option A is not correct or some Typo.
take 1 as "i" it's typo
wrong ans
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 ?

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)


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

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..
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...

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