Given the vectors $\overrightarrow{v_{1}}=\hat{\imath}+3 \hat{\jmath} ; \vec{v}_{2}=2 \hat{\imath}-4 \hat{\jmath}+3 \hat{k},$ the vector $\overrightarrow{v_{3}}$ that is perpendicular to both $\overrightarrow{v_{1}}$
and $\overrightarrow{v_{2}}$ is given by:
(A) $\overrightarrow{v_{3}}=\overrightarrow{v_{1}}-\left(\overrightarrow{v_{1}} \cdot \overrightarrow{v_{2}}\right) \frac{\overrightarrow{v_{2}}}{\left|\overrightarrow{v_{2}}\right|}$
(B) $\overrightarrow{v_{3}}=\hat{k}$
(C) $\overrightarrow{v_{3}}=\overrightarrow{v_{2}}-\left(\overrightarrow{v_{1}} \cdot \overrightarrow{v_{2}}\right) \frac{\overrightarrow{v_{1}}}{\mid \overrightarrow{v_{1} \mid}}$
(D) $\overrightarrow{v_{3}}=\frac{\overrightarrow{v_{1}} \times \overrightarrow{v_{2}}}{\left|\overrightarrow{v_{1}} \times \overrightarrow{v_{2}}\right|}$
Given the vectors $\overrightarrow{v_{1}}=\hat{\imath}+3 \hat{\jmath} ; \vec{v}_{2}=2 \hat{\imath}-4 \hat{\jmath}+3 \hat{k},$ the vector $\overrightarrow{v_{3}}$ that is perpendicular to both $\overrightarrow{v_{1}}$
and $\overrightarrow{v_{2}}$ is given by:
(A) $\overrightarrow{v_{3}}=\overrightarrow{v_{1}}-\left(\overrightarrow{v_{1}} \cdot \overrightarrow{v_{2}}\right) \frac{\overrightarrow{v_{2}}}{\left|\overrightarrow{v_{2}}\right|}$
(B) $\overrightarrow{v_{3}}=\hat{k}$
(C) $\overrightarrow{v_{3}}=\overrightarrow{v_{2}}-\left(\overrightarrow{v_{1}} \cdot \overrightarrow{v_{2}}\right) \frac{\overrightarrow{v_{1}}}{\mid \overrightarrow{v_{1} \mid}}$
(D) $\overrightarrow{v_{3}}=\frac{\overrightarrow{v_{1}} \times \overrightarrow{v_{2}}}{\left|\overrightarrow{v_{1}} \times \overrightarrow{v_{2}}\right|}$
(D) $\overrightarrow{v_{3}}=\frac{\overrightarrow{v_{1}} \times \overrightarrow{v_{2}}}{\left|\overrightarrow{v_{1}} \times \overrightarrow{v_{2}}\right|}$