Question

5) If vectors -3i+4j+lambdavec k and mubar(i)+bar(j)+3bar(k) are collinear then find lambda and mu​

Answer 1

Basic Concept Used :-

Let us consider two vectors,

$\rm :\longmapsto\:\vec{a} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$

$\rm :\longmapsto\:\vec{b} = a_2\hat{i} + b_2\hat{j} + c_2\hat{k}$

then ,

$\rm :\longmapsto\:\vec{a} \: and \: \vec{b} \: are \: collinear \: iff \: \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

Let's solve the problem now!!!!

Given that,

$\rm :\longmapsto\: - 3\hat{i} + 4\hat{j} + \lambda\hat{k} \: and \: \mu \: \hat{i} + \hat{j} + 3\hat{k} \: are \: collinear.$

$\rm :\implies\:\dfrac{ - 3}{ \mu} = \dfrac{4}{1} = \dfrac{ \lambda}{3}$

Taking first and second, we get

$\rm :\longmapsto\:\dfrac{ - 3}{ \mu} = \dfrac{4}{1}$

$\bf\implies \: \mu \: = \: - \: \dfrac{3}{4}$

Taking second and third, we get

$\rm :\longmapsto\: \dfrac{4}{1} = \dfrac{ \lambda}{3}$

$\bf\implies \: \lambda \: = \: 12$

Additional Information :-

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

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

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

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

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

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

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

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