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This is the second of our many quizzes on Adding, subtracting, and multiplying vectors. Click on this to access more such competitive quizzes on Vectors.

Adding, subtracting, and multiplying vectors Quiz 2

mathematics

  If  $\theta $  is the angle between the two vectors  $\overrightarrow{p}=2\widehat{i}+\widehat{j}-\widehat{k}$  and  $\overrightarrow{q}=3\widehat{i}-\widehat{j}+\widehat{k}$. Then find the value of  $\tan \theta 

  Use the formula  $\text{cos}\theta \text{=}\dfrac{\overrightarrow{p}\cdot \overrightarrow{q}}{\left| \overrightarrow{p} \right|\left| \overrightarrow{q} \right|}$   to find the value of  $\text{cos}\theta $  and then use basic trigonometry to obtain the answer.

$ \text{We have , cos}\theta \text{=}\dfrac{\overrightarrow{p}\cdot \overrightarrow{q}}{\left| \overrightarrow{p} \right|\left| \overrightarrow{q} \right|} \\ =\dfrac{6-1-1}{\sqrt{{{2}^{2}}+{{1}^{2}}+{{1}^{2}}}\sqrt{{{3}^{2}}+{{1}^{2}}+{{1}^{2}}}} \\ =\dfrac{4}{\sqrt{66}} \\ \text{Hence hypotenuse=}\sqrt{66};\,\text{adjacent=4} \\ \Rightarrow \text{opposite=}\sqrt{66-16}=5\sqrt{2} \\ \therefore \tan \theta =\dfrac{5\sqrt{2}}{4} $

 $\overrightarrow{a}=\widehat{i}-3\widehat{j}+4\widehat{k},\,\overrightarrow{b}=2\widehat{i}+\widehat{j}-\widehat{k}$  and  $\overrightarrow{c}=\widehat{i}+\widehat{j}-\widehat{k}$  are such that  $\,\overrightarrow{b}-\lambda \overrightarrow{a}$ is perpendicular to  $\overrightarrow{c}$. Then find the value of  $\lambda $ 

 Take the dot product of two vectors which are perpendicular to each other and equate it to $0$ to find the value of  $\lambda $.

 $ \overrightarrow{b}-\lambda \overrightarrow{a}=\left( 2\widehat{i}+\widehat{j}-\widehat{k} \right)-\lambda \left( \widehat{i}-3\widehat{j}+4\widehat{k} \right) \\ =\left( 2-\lambda  \right)\widehat{i}+\left( 1+3\lambda  \right)\widehat{j}-\left( 1+4\lambda  \right)\widehat{k} \\ \text{If }\overrightarrow{b}-\lambda \overrightarrow{a}\text{ is perpendicular to}\,\,\overrightarrow{c},\text{ then} \\ \left( \overrightarrow{b}-\lambda \overrightarrow{a} \right).\overrightarrow{c}=0 \\ \Rightarrow \left[ \left( 2-\lambda  \right)\widehat{i}+\left( 1+3\lambda  \right)\widehat{j}-\left( 1+4\lambda  \right)\widehat{k} \right].\left[ \widehat{i}+\widehat{j}-\widehat{k} \right]=0 \\ \Rightarrow \left( 2-\lambda  \right)+\left( 1+3\lambda  \right)+\left( 1+4\lambda  \right)=0 \\ \Rightarrow 4+6\lambda =0 \\ \therefore \lambda =-\dfrac{2}{3}$

 Find  $\lambda ,\mu $  if  $\left( \widehat{i}+6\widehat{j}+3\widehat{k} \right).\left( 3\widehat{i}+\widehat{\lambda }+9\widehat{\mu } \right)=0$  and given that  $\lambda =3\mu $ 

 Take the dot product of the two vectors and equate it to 0 to find the value of  $\lambda $ and then use the relation  $\lambda =3\mu $   to find the value of  $\mu $.

 $  \left( \widehat{i}+6\widehat{j}+3\widehat{k} \right).\left( 3\widehat{i}+\widehat{\lambda }+9\widehat{\mu } \right)=0 \\ \Rightarrow 3+6\lambda +27\mu =0 \\ \Rightarrow 3+6\lambda +27\left( \dfrac{\lambda }{3} \right)=0 \\ \Rightarrow 3+15\lambda =0 \\ \therefore \lambda =-\dfrac{1}{5} \\ \therefore \mu =\dfrac{\lambda }{3}=-\dfrac{1}{15} $

 Find magnitude of  $\overrightarrow{x}$ , given that  $\left| \overrightarrow{y} \right|=\sqrt{2}$  and  $\left( \overrightarrow{x}-2\overrightarrow{y} \right).\left( \overrightarrow{x}+2\overrightarrow{y} \right)=8$ 

 Using the given relation   $\left( \overrightarrow{x}-2\overrightarrow{y} \right).\left( \overrightarrow{x}+2\overrightarrow{y} \right)=8$   form a relation between  $\left| \overrightarrow{x} \right|$  and  $\left| \overrightarrow{y} \right|$  and then substitute the value of  $\left| \overrightarrow{y} \right|$  to obtain the answer.

 $ \left( \overrightarrow{x}-2\overrightarrow{y} \right).\left( \overrightarrow{x}+2\overrightarrow{y} \right)=8 \\ \Rightarrow \overrightarrow{x}.\overrightarrow{x}+2\overrightarrow{x}\overrightarrow{.y}-2\overrightarrow{x}\overrightarrow{.y}-4\overrightarrow{y}\overrightarrow{.y}=8 \\ \Rightarrow {{\left| \overrightarrow{x} \right|}^{2}}-4{{\left| \overrightarrow{y} \right|}^{2}}=8 \\  \Rightarrow {{\left| \overrightarrow{x} \right|}^{2}}-4\left( 2 \right)=8 \\ \Rightarrow {{\left| \overrightarrow{x} \right|}^{2}}=16 \\ \therefore \left| \overrightarrow{x} \right|=4$

  If  $\vec{a},\vec{b},\vec{c}$  represents the sides of tetrahedron and  $\theta $  be the angle between  $\vec{a}$  and  $\vec{b}$ and  $\phi $  that between  $\vec{a}$  and  $\vec{c}$ ,  $\varphi $  between  $\vec{b}$  and  $\vec{c}$ , and Volume of tetrahedron is  $V$ , then 

 Consider a tetrahedron  $OABC$  with  $\overrightarrow{OA}=\vec{a},\overrightarrow{OB}=\vec{b},\overrightarrow{OC}=\vec{c}$ , and then use the formula  $V=\dfrac{1}{6}\left[ \vec{a}\cdot \left( \vec{b}\times \vec{c} \right) \right]$   to obtain the answer.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d5a50605-7b72-4282-81c1-f2c78d2f89b75864236403852541343.png"/> $ \\  \text{Let }OABC\text{ represent a tetrahedron, with }\overrightarrow{OA}=\vec{a},\overrightarrow{OB}=\vec{b},\overrightarrow{OC}=\vec{c} \\ \text{Volume of tetrahedron }V=\dfrac{1}{6}\left[ \vec{a}\cdot \left( \vec{b}\times \vec{c} \right) \right] \\ \Rightarrow {{V}^{2}}=\dfrac{1}{36}{{\left\{ \vec{a}\cdot \left( \vec{b}\times \vec{c} \right) \right\}}^{2}} \\ =\dfrac{1}{36}{{\left[ \vec{a}\text{  }\vec{b}\text{  }\vec{c} \right]}^{2}}=\dfrac{1}{36}\left| \begin{matrix}   \vec{a}\cdot \vec{a} & \vec{a}\cdot \vec{b} & \vec{a}\cdot \vec{c}  \\   \vec{b}\cdot \vec{a} & \vec{b}\cdot \vec{b} & \vec{b}\cdot \vec{c}  \\   \vec{c}\cdot \vec{a} & \vec{c}\cdot \vec{b} & \vec{c}\cdot \vec{c}  \\\end{matrix} \right| \\ =\dfrac{1}{36}\left| \begin{matrix}   {{a}^{2}} & ab\cos \theta  & ac\cos \phi   \\   ab\cos \theta  & {{b}^{2}} & bc\cos \varphi   \\   ac\cos \phi  & bc\cos \varphi  & {{c}^{2}}  \\\end{matrix} \right| \\  =\dfrac{{{a}^{2}}{{b}^{2}}{{c}^{2}}}{36}\left| \begin{matrix}   1 & \cos \theta  & \cos \phi   \\   \cos \theta  & 1 & \cos \varphi   \\   \cos \phi  & \cos \varphi  & 1  \\\end{matrix} \right| $