Question

If a,b,c are unit vectors such that a is perpendicular to plane of b and c amd the sngle between b and c is 60 the |a+b+c| is

Answer 1

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

★Given that

$\rm :\longmapsto\:\vec{a}, \: \vec{b}, \: \vec{c} \: are \: unit \: vectors.$

$\bf\implies \: |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 - - - (1)$

★Again given that,

$\rm :\longmapsto\:\vec{a} \: is \: \perp \: to \: plane \: of \: \vec{b} \: and \: \vec{c}$

$\bf\implies \:\vec{a} \: \parallel \: \vec{b} \times \vec{c}$

$\bf\implies \:\vec{a} \: \perp \: \vec{b} \: and \: \vec{a} \: \perp \: \vec{c}$

$\bf\implies \:\vec{a} \: . \: \vec{b} = 0 \: \: and \: \: \vec{b} \: . \: \vec{c} = 0 \: \: - - (2)$

★Again provided that

$\rm :\longmapsto\:Angle \: between \: \vec{b} \: and \: \vec{c} \: is \: 60 \degree$

Consider,

$\rm :\longmapsto\:\vec{b} \: . \: \vec{c}$

$\rm \: = \: \: \: |\vec{b}| |\vec{c}| cos60\degree$

$\rm \: = \: \: \:1 \times 1 \times \dfrac{1}{2}$

$\rm \: = \: \: \:\dfrac{1}{2}$

$\bf\implies \:\vec{b} \: . \: \vec{c} = \dfrac{1}{2} - - - (3)$

Now,

Consider,

$\bf :\longmapsto\: { |\vec{a} + \vec{b} + \vec{c}|}^{2}$

$\rm \: = \: \: \:(\vec{a} + \vec{b} + \vec{c}).(\vec{a} + \vec{b} + \vec{c})$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: { |\vec{a}| }^{2} = \vec{a}.\vec{a}\bigg \}}$

$\rm \: = \: \: \:\vec{a}.\vec{a} + \vec{a}.\vec{b} + \vec{a}.\vec{c} + \vec{b}.\vec{a} + \vec{b}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a} + \vec{c}.\vec{b} + \vec{c}.\vec{c}$

$\rm \: = \: \: \: { |\vec{a}| }^{2} + { |\vec{b}| }^{2} + { |\vec{c}| }^{2} + 2(\vec{a}.\vec{b}) + 2(\vec{b}.\vec{c}) + 2(\vec{c}.\vec{a})$

$\red{\bigg \{ \because \: \vec{a}.\vec{a} = { |\vec{a}| }^{2} \: and \: \vec{a}.\vec{b} = \vec{b}.\vec{a}\bigg \}}$

$\rm \: = \: \: \:1 + 1 + 1 + 2 \times 0 + 2 \times \dfrac{1}{2} + 2 \times 0$

$\red{\bigg \{ \because \:using \: equations \: (1),(2),(3) \: and \: (4) \bigg \}}$

$\rm \: = \: \: \:4$

$\rm \: = \: \: \: {2}^{2}$

$\bf\implies \: |\vec{a} + \vec{b} + \vec{c}| = 2$

Additional information :-

$\rm :\longmapsto\:\vec{a} \times \vec{b} = - \: \vec{b} \times \vec{a}$

$\rm :\longmapsto\:\vec{a} \times \vec{a} = 0$

$\rm :\longmapsto\:\vec{a}.\vec{a} = { |\vec{a}| }^{2}$

$\rm :\longmapsto\:\vec{a}.\vec{b} \: = \: \vec{b}.\vec{a}$

$\rm :\longmapsto\:\vec{a}.\vec{b} = 0 \: \rm \implies\: \: \vec{a} \: \perp \: \vec{b}$

$\rm :\longmapsto\:\vec{a} \: \perp \: \vec{b} \: \rm \implies\: \: \vec{a} \:.\: \vec{b} = 0$

$\rm :\longmapsto\:\vec{a} \times \vec{b} = 0 \: \rm \implies\: \: \vec{a} \: \parallel \: \vec{b}$

$\rm :\longmapsto\:\vec{a}.\vec{b} = |\vec{a}| |\vec{b}| \: cos \theta$

$\rm :\longmapsto\: |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| sin \theta$

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