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  • Let the volume of a parallelopiped whose coterminous edges are given by

Let the volume of a parallelopiped whose coterminous edges are given by \vec{u}=\widehat{i}+\widehat{j}+\lambda \widehat{k},\vec{v}=\widehat{i}+\widehat{j}+3\widehat{k} and \vec{w}=2\widehat{i}+\widehat{j}+\widehat{k} be 1 cu. unit. If \theta be the angle between the edges \vec{u} and \vec{w}, then \cos \theta can be :
Option: 1 \frac{7}{6\sqrt{6}}
Option: 2 \frac{5}{7}
Option: 3 \frac{7}{6\sqrt{3}}
Option: 4 \frac{5}{3\sqrt{3}}
 

Answers (1)

best_answer

 

 

Dot (Scalar) Product in Terms of Components -


Angle between two vectors

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\vec{\mathbf a} \cdot \vec{\mathbf b}=|\vec{\mathbf a}||\vec{\mathbf b}| \cos \theta\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;}\cos \theta=\frac{\vec{\mathbf a} \cdot \vec{\mathbf b}}{|\vec{\mathbf a}||\vec{\mathbf b}| }\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \theta=\cos^{-1}\left (\frac{\vec{\mathbf a} \cdot \vec{\mathbf b}}{|\vec{\mathbf a}||\vec{\mathbf b}| } \right )\\\\\text{If }\;\;\vec {\mathbf a}=a_{1} \hat{\mathbf{i}}+a_{2} \hat{\mathbf{j}}+a_{3} \hat{\mathbf{k}}\;\;\text{and }\;\;\vec{\mathbf b}=b_{1} \hat{\mathbf{i}}+b_{2} \hat{\mathbf{j}}+b_{3} \hat{\mathbf{k}}\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \theta=\cos ^{-1}\left(\frac{a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}} \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}}\right)

 

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Geometrical Interpretation of Scalar Triple Product -

Let vectors \vec {\mathbf a},\;\;\vec {\mathbf b} and \vec {\mathbf c} represented the sides of a parallelepiped OA, OB and OC respectively. Then, \vec {\mathbf b}\times \vec {\mathbf c} is a vector perpendicular to the plane of \vec {\mathbf b} and \vec {\mathbf c}. Let ? be the angle between vectors \vec{\mathbf b} and \vec {\mathbf c} and α be the angle between \vec {\mathbf a} and \vec {\mathbf b}\times\vec {\mathbf c}.

If \hat {\mathbf n} is a unit vector along \vec {\mathbf b}\times\vec {\mathbf c}, then α is the angle between \hat {\mathbf n} and \vec {\mathbf a}.

\\\left [ \vec{\mathbf a}\;\;\vec{\mathbf b}\;\;\vec{\mathbf c} \right ]= \vec{\mathbf c}\cdot \left ( \vec{\mathbf b}\times \vec{\mathbf c} \right )\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\vec{\mathbf a} \cdot(\mathbf {b} \mathbf {c} \sin \theta\; \hat{\mathbf n})\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=(\mathbf {b} \mathbf {c} \sin \theta)(\vec{\mathbf a} \cdot \hat{\mathbf n})\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=(\mathbf {b} \mathbf {c} \sin \theta)({\mathbf a} \cdot {\mathbf 1}\cdot \cos \alpha)\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=({\mathbf a} \cdot \cos \alpha)(\mathbf {b} \mathbf {c} \sin \theta)\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=(\text{Height} )\cdot \text{(Area of Base)}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\text{Volume of parallelepiped}

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\\\pm 1=\left|\begin{array}{lll}{1} & {1} & {\lambda} \\ {1} & {1} & {3} \\ {2} & {1} & {1}\end{array}\right| \Rightarrow=-\lambda+3=\pm 1 \Rightarrow \lambda=2 \text { or } \lambda=4 \\ {\text { For } \lambda=4 \frac{1}{90}} \\\\ {\cos \theta=\frac{2+1+4}{\sqrt{6} \sqrt{18}}=\frac{7}{6 \sqrt{3}}}

Correct Option (3)

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Kuldeep Maurya

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