Question

3
NAVNEET 21 M. L.Q. SETS : MATHEMATICS & STATISTICS-STD. XII (Science)
(ii) If a and b are unit vectors, then what is angle between a and b for
3a-b to be a unit vector?
(a) 30°
(b) 45°
(c) 60°
(d) 90°​

Answer 1

Appropriate Question

$\red{ \sf \: If \: \vec{a} \: and \: \vec{b} \: are \: unit \: vectors, then \: what \: is \: the \: angle} \\ \red{ \sf \: between \: \vec{a} \: and \: \vec{b} \: so \: that \: \sqrt{3}\vec{a} - \vec{b} \: is \: unit \: vector. \: }$

Basic Identities Used :-

$\boxed{ \sf \:\vec{a} \: is \: unit \: vector \implies \: |\vec{a}| = 1}$

$\boxed{ \sf \:\vec{a}.\vec{a} = { |\vec{a}| }^{2}}$

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

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

$\green{\large\underline{\bf{Solution-}}}$

Given that

${ \sf \:\vec{a} \: is \: unit \: vector \implies \: |\vec{a}| = 1}$

${ \sf \:\vec{b} \: is \: unit \: vector \implies \: |\vec{b}| = 1}$

Also given that

$\rm :\longmapsto\:{ \sf \: \sqrt{3} \vec{a} \: - \: \vec{b} \: is \: unit \: vector }$

$\rm :\longmapsto\: | \sqrt{3} \vec{a} - \vec{b}| = 1$

On squaring both sides, we get

$\rm :\longmapsto\: | \sqrt{3} \vec{a} - \vec{b}| ^{2} = 1$

$\rm :\longmapsto\:( \sqrt{3} \vec{a} - \vec{b}).( \sqrt{3} \vec{a} - \vec{b}) = 1$

$\rm :\longmapsto\:3\vec{a}.\vec{a} - \sqrt{3} \vec{a}.\vec{b} - \sqrt{3} \vec{b}.\vec{a} + \vec{b}.\vec{b} = 1$

$\rm :\longmapsto\:3 |\vec{a}|^{2} - \sqrt{3} \vec{a}.\vec{b} - \sqrt{3} \vec{a}.\vec{b} + { |\vec{b}| }^{2} = 1$

$\rm :\longmapsto\:3 - 2 \sqrt{3} \: \vec{a}.\vec{b} + 1 = 1$

$\rm :\longmapsto\:2 \sqrt{3} \: \vec{a}.\vec{b} = 3$

$\rm :\longmapsto\:2 \sqrt{3} \: |\vec{a}| |\vec{b}| cos\theta = \sqrt{3} \times \sqrt{3}$

$\rm :\longmapsto\:2 \: \times 1 \times 1 \times cos\theta = \sqrt{3}$

$\rm :\longmapsto\:cos\theta = \dfrac{ \sqrt{3} }{2}$

$\bf\implies \:\theta = 30 \degree$

Hence,

• Option (a) is correct.

Additional Information :-

$\boxed{ \sf \:\vec{a}.\vec{b} = 0 \implies \: \vec{a} \: \perp \: \vec{b}}$

$\boxed{ \sf \:\vec{a} \: \perp \: \vec{b} \implies \: \vec{a}.\vec{b} = 0}$

$\boxed{ \sf \:If \: \vec{a} \times \vec{b} = \vec{0}, then \: \vec{a} \: \parallel \: \vec{b}}$

$\boxed{ \sf \:If \: \vec{a} \: \parallel \: \vec{b}, \: then \: \vec{a} \times \vec{b} = \vec{0}}$

$\boxed{ \sf \: |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| sin\theta }$

$\boxed{ \sf \:\vec{a} \times \vec{a} = 0}$

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

Answer 2

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