Question

The value of m for which the two vectors 2i-j+2k and 3i+mj+k are perpendicular is​

Answer 1

$\large\underline{\sf{Solution-}}$

Let we assume that

$\red{\rm :\longmapsto\: \vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k}}$

and

$\red{\rm :\longmapsto\: \vec{b} = 3 \hat{i} + m\hat{j} + \hat{k}}$

Now, it is given that,

$\rm :\longmapsto\: \vec{a} \: \perp \: \vec{b}$

$\bf\implies \: \vec{a} \: . \: \vec{b} \: = \: 0$

$\rm :\longmapsto\:(2 \hat{i} - \hat{j} + 2 \hat{k}).(3 \hat{i} + m\hat{j} + \hat{k}) = 0$

$\rm :\longmapsto\:6 - m + 2 = 0$

$\rm :\longmapsto\:8- m = 0$

$\bf\implies \:m \: = \: 8$

Additional Information :-

$\boxed{ \bf{ \: \vec{a}. \vec{b} = | \vec{a}| \: | \vec{b}| \: cos \theta}}$

$\boxed{ \bf{ \: \vec{a} \times \vec{b} = | \vec{a}| \: | \vec{b}| \: sin \theta \: \hat{n}}}$

$\boxed{ \bf{ \: {( \vec{a}. \vec{b})}^{2} + { | \vec{a} \times \vec{b}| }^{2} = { | \vec{a}| }^{2} { | \vec{b}| }^{2}}}$

$\boxed{ \bf{ \: \vec{a}. \vec{b} = \vec{b}. \vec{a}}}$

$\boxed{ \bf{ \: \vec{a} \times \vec{b} \: = - \: \vec{b} \times \vec{a}}}$

$\boxed{ \bf{ \:cos\theta = \frac{ \vec{a}. \vec{b}}{ | \vec{a}| \: | \vec{b}| }}}$