Question

Unit vector perpendicular to vector a=-3i-2j-3k and b=2i+4j+6k

Answer 1

Answer:

The unit vectors perpendicular to given vectors are ±$\frac{-3\vec{j}+2\vec{k}}{\sqrt{13}}$

Step-by-step explanation:

Unit vector perpendicular to vector a=-3i-2j-3k and b=2i+4j+6k

Formula used:

The unit vectors perpendicular to both $\vec{a}\:and\:\vec{b}$ are

$\frac{\vec{a}*\vec{b}}{|\vec{a}*\vec{b}|}$

$\vec{a}=-3\vec{i}-2\vec{j}-3\vec{k}$

$\vec{b}=2\vec{i}+4\vec{j}+6\vec{k}$

$\vec{a}*\vec{b}=\left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\3&-2&-3\\2&4&6\end{array}\right|$

$\vec{a}*\vec{b}=\vec{i}(-12+12)-\vec{j}(18+6)+\vec{k}(12+4)$

$\vec{a}*\vec{b}=-24\vec{j}+16\vec{k}$

$\implies\:\vec{a}*\vec{b}=8(-3\vec{j}+2\vec{k})$

$|\vec{a}*\vec{b}|=8\sqrt{(-3)^2+2^2}$

$|\vec{a}*\vec{b}|=8\sqrt{9+4}$

$\implies\:|\vec{a}*\vec{b}|=8\sqrt{13}$

Now,

The unit vectors perpendicular to both $\vec{a}\:and\:\vec{b}$ are

±$\frac{\vec{a}*\vec{b}}{|\vec{a}*\vec{b}|}$

=±$\frac{8(-3\vec{j}+2\vec{k}}{8\sqrt{13})}$

=±$\frac{-3\vec{j}+2\vec{k}}{\sqrt{13}}$

Answer 2

The dot and cross product has been used here
So the answer is here