Question

pls answer the attachment with
explanation.....​

Answer 1

Answer:

b vector = i vector + 3j vector + 4k vector. required vector perpendicular to given vectors = ± μ [(a x b)/ |a x b|] = i[8-3] - j[4-1] + k[3-2] a x b = 5i - 3j + k |a x b| = √5 2 + (-3) 2 + 1 2 = √(25+9+1) = √35. Required vector = ± (10 √ 3/ √35) (5i - 3j + k) = ± (10 √ 3/ √35) (5i - 3j + k) Question 2 : Find the unit vectors perpendicular to each of the vectors a vector + b vector and a vector - b vector where a vector = i vector + j vector + …

Answer 2

${\huge{\underbrace{\rm{Answer\checkmark}}}}$

★ Given -

• $\mapsto \rm \: \vec{a} = 3 \hat{i} + \hat{j} + 2 \hat{k}$
• $\rm \mapsto \: \vec{b} = 2 \hat{i} - 2 \hat{j} + 4 \hat{k}$

★ To Find -

• The unit vector

★ Solution -

→ Let $\vec{c}$ be perpendicular to $\vec{a} \: and \: \vec{b}$

→ Then $\vec{c}$ can be taken as the Cross product of $\vec{a} \: and\: \vec{b}$

• Now , refer to the 1st Attachment .
• We've got the Cross product now , finding the final answer

$\boxed{ \rm\mapsto \vec{c} = 8 \hat{i} - 8 \hat{j} - 8 \hat{k}}.....(i)$

$\therefore \rm \: | \vec{c}| = \sqrt{ {8}^{2} + {( - 8)}^{2} + {( - 8)}^{2} }$

$\implies \sqrt{64 + 64 + 64} = \sqrt{192}$

$\boxed{ \mapsto\rm | \vec{c}| = 8\sqrt{3} }$

Now , Unit vector along $\vec{c}$ is :

$\longrightarrow \: \rm \hat{c} \: = \dfrac{ \vec{c}}{ | \vec{c}| }$

$\implies \rm \: \dfrac{8 \hat{i} - 8 \hat{j} - 8 \hat{k}}{8 \sqrt{3} }$

$\implies \rm \: \dfrac{ \cancel8( \hat{i} - \hat{j} - \hat{j})}{ \cancel8 \sqrt{3} }$

$\implies \boxed{ \rm\star \: \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}}$

This is ↑ the required answer !!!!!