Question

The unit vector is perpendicular to the two vectors 3i + j + 2k and 2i - 2j + 4k is ............ .
(A) 1/â3 (i - j - k)
(B) 1/â3 (i + j + k)
(C) (i - j - k)
(D) â3(i - j - k)

Answer 1

Answer:

$\hat{n} = (-\frac{j}{\sqrt{2}}-\frac{k}{\sqrt{2}})$

Explanation:

$\vec{A}$ = 3i+j+2k

$\vec{B}$ = 2i-2j+4k

$\hat{n}$ = unit vector perpendicular to $\vec{A}$ and $\vec{B}$

Now according to definition of the cross-product of two vector, cross-product of two vectors is itself a vector but perpendicular to plane of original vectors.

$\vec{A}\times \vec{B} =(3i+j+2k)\times (2i-2j+4k)$

$\vec{A}\times \vec{B} =3i\times i-6i\times j+12i\times k+2j\times i-2j\times j+4j\times k+4k\times i-4k\times j+8k\times k$

$\vec{A}\times \vec{B} = 0-6k-12j-2k-0+4i+4j-4i+0$

$\vec{A}\times \vec{B} = -8j-8k$

So unit vector perpendicular to each $\vec{A}$ and $\vec{B}$ is evaluated as below :

$\hat{n} = \frac{\vec{A}\times \vec{B}}{\left | \vec{A}\times \vec{B} \right | }$  

$\hat{n} = \frac{(-8j-8k)}{\left | (-8j-8k) \right | }$  

$\hat{n} = \frac{(-8j-8k)}{\left | 8\sqrt{2} \right | }$

$\hat{n} = (-\frac{j}{\sqrt{2}}-\frac{k}{\sqrt{2}})$