Question

is a bar is equals to 3 ICAP + J cap + 2 k cap and b bar is equals to two ICAP - 2 J cap + 4K cab then unit vector perpendicular to the plane containing a bar and Bieber is.

Answer 1

Answer:

The required vectors are

$\vec{n}=\pm\frac{(\vec{i}-\vec{j}-\vec{k})}{\sqrt{3}}$

Explanation:

Formula used:

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

$\vec{n}=\pm\frac{(\vec{a}\times\vec{b})}{|\vec{a}\times\vec{b}|}$

Given:

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

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

$\vec{a}\times\vec{b}=\left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\3&1&2\\2&-2&4}\end{array}\right|$

$\vec{a}\times\vec{b}=\vec{i}(4+4)-\vec{j}(12-4)+\vec{k}(-6-2)$

$\vec{a}\times\vec{b}=8\vec{i}-8\vec{j}-8\vec{k}$

$\implies\:\vec{a}\times\vec{b}=8(\vec{i}-\vec{j}-\vec{k})$

$|\vec{a}\times\vec{b}|=8\sqrt{1+1+1}$

$|\vec{a}\times\vec{b}|=8\sqrt{3}$

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

$=\pm\frac{(\vec{a}\times\vec{b})}{|\vec{a}\times\vec{b}|}$

$=\pm\frac{8(\vec{i}-\vec{j}-\vec{k})}{8\sqrt{3}}$

$=\pm\frac{(\vec{i}-\vec{j}-\vec{k})}{\sqrt{3}}$

Answer 2

Answer:

Answer:

The required vectors are

\vec{n}=\pm\frac{(\vec{i}-\vec{j}-\vec{k})}{\sqrt{3}}n=±3(i−j−k)

Explanation:

Formula used:

The unit vector perpendicular to both \vec{a}\:and\:\vec{b}aandb are

\vec{n}=\pm\frac{(\vec{a}\times\vec{b})}{|\vec{a}\times\vec{b}|}n=±∣a×b∣(a×b)

Given:

\vec{a}=3\vec{i}+\vec{j}+2\vec{k}a=3i+j+2k

\vec{b}=2\vec{i}-2\vec{j}+4\vec{k}b=2i−2j+4k



\vec{a}\times\vec{b}=\vec{i}(4+4)-\vec{j}(12-4)+\vec{k}(-6-2)a×b=i(4+4)−j(12−4)+k(−6−2)

\vec{a}\times\vec{b}=8\vec{i}-8\vec{j}-8\vec{k}a×b=8i−8j−8k

\implies\:\vec{a}\times\vec{b}=8(\vec{i}-\vec{j}-\vec{k})⟹a×b=8(i−j−k)

|\vec{a}\times\vec{b}|=8\sqrt{1+1+1}∣a×b∣=81+1+1

|\vec{a}\times\vec{b}|=8\sqrt{3}∣a×b∣=83

The unit vector perpendicular to both \vec{a}\:and\:\vec{b}aandb

=\pm\frac{(\vec{a}\times\vec{b})}{|\vec{a}\times\vec{b}|}=±∣a×b∣(a×b)

=\pm\frac{8(\vec{i}-\vec{j}-\vec{k})}{8\sqrt{3}}=±838(i−j−k)

=\pm\frac{(\vec{i}-\vec{j}-\vec{k})}{\sqrt{3}}=±3(i−j−

............$iiiij