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Question: Find the value of \({{\lambda }}\) in unit vector \({{0}}{{.4}}\;{{\hat i}}\;{{ + 0}}{{.8}}\;{{\hat ...

Find the value of λ{{\lambda }} in unit vector 0.4  i^  +0.8  j^  +  λ  k^{{0}}{{.4}}\;{{\hat i}}\;{{ + 0}}{{.8}}\;{{\hat j}}\;{{ + }}\;{{\lambda }}\;{{\hat k}}

Explanation

Solution

A unit vector is that vector which have magnitude 1.1. so we will put the magnitude of this vector equal to 11 and then calculate value of λ{{\lambda }} , A unit vector can be in any direction. Unit vectors are helpful to determine the base form of a vector space. Every vector in a given space can be expressed as a linear combination of unit vectors.

Formula used:
∣A⃗∣  =Ax2+Ay2+Az2\left| {{{\vec A}}} \right|\;{{ = }}\sqrt {{{{A}}_{{x}}}^{{2}}{{ + }}{{{A}}_{{y}}}^{{2}}{{ + }}{{{A}}_{{z}}}^{{2}}}
Where ∣A⃗∣\left| {{{\vec A}}} \right| is any vector and Ax  ,  Ay  ,Az{{{A}}_{{{x}}\;{{,}}\;}}{{{A}}_{{{y}}\;{{,}}}}{{{A}}_{{z}}} are its components along x, y, z directions respectively and ∣A⃗∣\left| {{{\vec A}}} \right| is the magnitude of this vector.

Complete step by step solution:
The given vector is 410i^  + 810j^  +  λk^\dfrac{{{4}}}{{{{10}}}}{{\hat i}}\;{{ + }}\,\dfrac{{{8}}}{{{{10}}}}{{\hat j}}\;{{ + }}\;{{\lambda \hat k}}
As, the vector is unit vector so, its magnitude will be equal to 11 i.e.
(410)2+(810)2+λ2=1\sqrt {{{\left( {\dfrac{{{4}}}{{{{10}}}}} \right)}^{{2}}}{{ + }}{{\left( {\dfrac{{{8}}}{{{{10}}}}} \right)}^{{2}}}{{ + }}{{{\lambda }}^{{2}}}} {{ = 1}}
⇒  16100+64100+λ2  =  1\Rightarrow \;\dfrac{{{{16}}}}{{{{100}}}}{{ + }}\dfrac{{{{64}}}}{{{{100}}}}{{ + }}{{{\lambda }}^{{2}}}\;{{ = }}\;{{1}}
⇒100100+λ2  =  1\Rightarrow \dfrac{{{{100}}}}{{{{100}}}}{{ + }}{{{\lambda }}^{{2}}}\;{{ = }}\;{{1}}
⇒λ  =  0\Rightarrow {{\lambda }}\;{{ = }}\;{{0}}
To find a unit vector with the same direction as a given vector, we divide the vector by its magnitude.
Any vector can be converted into a unit vector by dividing it by the magnitude of the given vector. The dot product for any two unit vectors is a scalar quantity whereas the cross product of any two arbitrary unit vectors results in a third vector orthogonal to both of them.

Note: For such a question always put the magnitude equal to 1.1. Normal vector is a vector which is perpendicular to the surface at a given point. They are also called “normal,” to a surface is a vector. When normals are estimated on any closed surfaces, the normal pointing towards the interior of the surface and normal pointing outward are usually discovered.