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Question

Question: The unit vector perpendicular to![](https://cdn.pureessence.tech/canvas_232.png?top_left_x=1055&top_...

The unit vector perpendicular to and

is

A

B

3i+5j^+7k^83\frac { 3 \vec { i } + 5 \hat { j } + 7 \hat { k } } { \sqrt { 83 } }

C

5i+3j^7k^83\frac { 5 \vec { i } + 3 \hat { j } - 7 \hat { k } } { \sqrt { 83 } }

D

3i5j^+7k^83\frac { 3 \vec { i } - 5 \hat { j } + 7 \hat { k } } { \sqrt { 83 } }

Answer

3i+5j^+7k^83\frac { 3 \vec { i } + 5 \hat { j } + 7 \hat { k } } { \sqrt { 83 } }

Explanation

Solution

× B\vec { B } is a vector ⊥ to both and B\vec { B }

Now,× B\vec { B } = () × ( 3i+j^2k^3 \overrightarrow { \mathrm { i } } + \hat { \mathrm { j } } - 2 \hat { \mathrm { k } } ) =

3i+5j^+7k^3 \overrightarrow { \mathrm { i } } + 5 \hat { \mathrm { j } } + 7 \hat { \mathrm { k } }

Now, n^\hat { n } = A×BA×B\frac { \overrightarrow { \mathrm { A } } \times \overrightarrow { \mathrm { B } } } { | \overrightarrow { \mathrm { A } } \times \overrightarrow { \mathrm { B } } | } = 3i+5j^+7k^32+52+72=3i+5j^+7k^83\frac { 3 \overrightarrow { \mathrm { i } } + 5 \hat { \mathrm { j } } + 7 \hat { \mathrm { k } } } { \sqrt { 3 ^ { 2 } + 5 ^ { 2 } + 7 ^ { 2 } } } = \frac { 3 \overrightarrow { \mathrm { i } } + 5 \hat { \mathrm { j } } + 7 \hat { \mathrm { k } } } { \sqrt { 83 } }

Hence (2) is correct.