Question

Find a vector $\overrightarrow{x}$ which is perpendicular to both $\overrightarrow{A}$ and$\overrightarrow{B}$ but has magnitude equal to that of $\overrightarrow{B}$

Rule : Inter change coeff. of$\widehat{i}$ and $\widehat{j}$and change sign of one of the vectors.

• A. $\frac{1}{\sqrt{10}}\left( \widehat{i} + 10\widehat{j} + 17\widehat{k} \right)$
• B. $\frac{1}{\sqrt{10}}\left( \widehat{i} - 10\widehat{j} + 17\widehat{k} \right)$
• C. $\frac{\sqrt{29}}{\sqrt{390}}\left( \widehat{i} + 10\widehat{j} + 17\widehat{k} \right)$
• D. $\frac{\sqrt{29}}{\sqrt{390}}\left( \widehat{i} - 10\widehat{j} + 17\widehat{k} \right)$

Answer 1

Answer: (D)

$\frac{\sqrt{29}}{\sqrt{390}}\left( \widehat{i} - 10\widehat{j} + 17\widehat{k} \right)$

Answer 2

$\overrightarrow{A} \times \overrightarrow{B} = \left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 3 & - 2 & 1 \\ 4 & 3 & - 2 \end{matrix} \right|$

$\hat { n } = \frac { \overrightarrow { \mathrm { A } } \times \overrightarrow { \mathrm { B } } } { | \overrightarrow { \mathrm { A } } \times \overrightarrow { \mathrm { B } } | } = \frac { \hat { \mathrm { i } } + 10 \hat { \mathrm { j } } + 17 \hat { \mathrm { k } } } { \sqrt { 390 } }$ $\overrightarrow{x}$ = |B|$\widehat{n}$ =

$\frac{\sqrt{29}\left( \widehat{i} + 10\widehat{j} + 17\widehat{k} \right)}{\sqrt{390}}$