Let the volume of a parallelepiped whose coterminous edges are given by \[\overrightarrow{u}=\widehat{i}+\widehat{j}+\lambda \widehat{k},\vec{v}=\widehat{i}+\widehat{j}+3\widehat{k}\text{ and }\vec{w}=2\widehat{i}+\widehat{j}+\widehat{k}\] be 1 cu. units. If $\theta $ be the angle between the edge $\vec{u}$ and $\vec{w}$ , then $\cos \theta $ can be :
A. $\dfrac{5}{7}$
B. $\dfrac{5}{3\sqrt{3}}$
C. $\dfrac{7}{6\sqrt{6}}$
D. $\dfrac{7}{6\sqrt{3}}$
Answer
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Hint: We are given that the volume of parallelepiped with given edges is 1 cu.unit. We can write it as Volume $=\left| \left( u\times \vec{v} \right).w \right|=\left[ \vec{u}\vec{v}\vec{w} \right]=\left| \begin{matrix}
1 & 1 & \lambda \\
1 & 1 & 3 \\
2 & 1 & 1 \\
\end{matrix} \right|=\pm 1$ . We will find the value of $\lambda $ for 1 and -1 by expanding the determinant. Then, we will substitute these values in $\vec{u}$ for each values of $\lambda $ and find angle between vectors $\text{\vec{u} and \vec{w}}$ for each case using \[\cos \theta =\dfrac{\vec{u}\cdot \vec{w}}{\left| {\vec{u}} \right|\left| {\vec{w}} \right|}\].
Complete step-by-step solution:
We are given that the edges of parallelepiped are \[\overrightarrow{u}=\widehat{i}+\widehat{j}+\lambda \widehat{k},\vec{v}=\widehat{i}+\widehat{j}+3\widehat{k}\text{ and }\vec{w}=2\widehat{i}+\widehat{j}+\widehat{k}\] and the volume is 1 cu.unit. The figure below shows the parallelepiped.
We know that for a parallelepiped of edges $\vec{a}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k},\text{ }\vec{b}={{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+{{b}_{3}}\widehat{k}\text{ }$ and $\vec{c}={{c}_{1}}\widehat{i}+{{c}_{2}}\widehat{j}+{{c}_{3}}\widehat{k}$ , volume is given by
$\left| \left( \vec{a}\times \vec{b} \right).\vec{c} \right|=\left[ \vec{a}\vec{b}\vec{c} \right]=\left| \begin{matrix}
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|$
Now, let’s write the volume of parallelepiped for given edges.
Volume $=\left| \begin{matrix}
1 & 1 & \lambda \\
1 & 1 & 3 \\
2 & 1 & 1 \\
\end{matrix} \right|$
We are given that volume is 1 cu. unit. We will take $\pm 1$Hence,
$\left| \begin{matrix}
1 & 1 & \lambda \\
1 & 1 & 3 \\
2 & 1 & 1 \\
\end{matrix} \right|=\pm 1$
Let us solve the determinant.
$1\left( 1-3 \right)-1\left( 1-6 \right)+\lambda \left( 1-2 \right)=\pm 1...\left( i \right)$
Let’s solve this to find the value of $\lambda $ for $+1$ .
\[\begin{align}
& 1\times -2-1\times -5-\lambda =1 \\
& \Rightarrow -2+5-\lambda =1 \\
& \Rightarrow 3-\lambda =1 \\
\end{align}\]
Let’s collect constants on RHS. We will get
\[\begin{align}
& \Rightarrow -\lambda =1-3 \\
& \Rightarrow -\lambda =-2 \\
& \Rightarrow \lambda =2 \\
\end{align}\]
Now, let us solve (i) to find the value of $\lambda $ for $-1$ .
\[\begin{align}
& 1\times -2-1\times -5-\lambda =-1 \\
& \Rightarrow -2+5-\lambda =-1 \\
& \Rightarrow 3-\lambda =-1 \\
\end{align}\]
Let’s collect constants on RHS. We will get
\[\begin{align}
& \Rightarrow -\lambda =-1-3 \\
& \Rightarrow -\lambda =-4 \\
& \Rightarrow \lambda =4 \\
\end{align}\]
Now, we have to find the angle between $\vec{u}$ and $\vec{w}$ for $\lambda =2$ .
We know that angle between any two vectors $\vec{a}\text{ and }\vec{b}$ is given by
\[\cos \theta =\dfrac{\vec{a}\cdot \vec{b}}{\left| {\vec{a}} \right|\left| {\vec{b}} \right|}\]
Hence, angle between vectors $\text{\vec{u} and \vec{w}}$ is given by
\[\cos \theta =\dfrac{\vec{u}\cdot \vec{w}}{\left| {\vec{u}} \right|\left| {\vec{w}} \right|}...\left( a \right)\]
Let us find \[\vec{u}\cdot \vec{w}\] .
We know that for two vectors $\vec{a}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k},\text{ }\vec{b}={{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+{{b}_{3}}\widehat{k}\text{ }$ , dot product is found as follows.
$\begin{align}
& \vec{a}\cdot \vec{b}=\left( {{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k} \right)\cdot \left( {{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+{{b}_{3}}\widehat{k}\text{ } \right) \\
& \Rightarrow \vec{a}\cdot \vec{b}=\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}} \right) \\
\end{align}$
Now, we can find \[\vec{u}\cdot \vec{w}\] where, \[\overrightarrow{u}=\widehat{i}+\widehat{j}+2\widehat{k}\text{ and }\vec{w}=2\widehat{i}+\widehat{j}+\widehat{k}\] .
\[\begin{align}
& \vec{u}\cdot \vec{w}=\left( 1\times 2+1\times 1+2\times 1 \right) \\
& \Rightarrow \vec{u}\cdot \vec{w}=2+1+2=5...(i) \\
\end{align}\]
Now, we have to find \[\left| {\vec{u}} \right|\] and \[\left| {\vec{w}} \right|\] .
We know that for a vector $\vec{a}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k}$ ,
\[\left| {\vec{a}} \right|=\sqrt{{{a}_{1}}^{2}+{{a}_{2}}^{2}+{{a}_{3}}^{2}}\]
\[\begin{align}
& \Rightarrow \left| {\vec{u}} \right|=\sqrt{{{1}^{2}}+{{1}^{2}}+{{2}^{2}}}=\sqrt{1+1+4}=\sqrt{6}...\left( ii \right) \\
& \left| {\vec{w}} \right|=\sqrt{{{2}^{2}}+{{1}^{2}}+{{1}^{2}}}=\sqrt{4+1+1}=\sqrt{6}...\left( iii \right) \\
\end{align}\]
Now, we can substitute (i), (ii) and (iii) in (a).
\[\Rightarrow \cos \theta =\dfrac{5}{\sqrt{6}\times \sqrt{6}}=\dfrac{5}{6}\]
Now, let us find the angle between $\vec{u}$ and $\vec{w}$ for $\lambda =4$
First, we have to find \[\vec{u}\cdot \vec{w}\] where, \[\overrightarrow{u}=\widehat{i}+\widehat{j}+4\widehat{k}\text{ and }\vec{w}=2\widehat{i}+\widehat{j}+\widehat{k}\]
\[\begin{align}
& \vec{u}\cdot \vec{w}=\left( 1\times 2+1\times 1+4\times 1 \right) \\
& \Rightarrow \vec{u}\cdot \vec{w}=2+1+4=7...(iv) \\
\end{align}\]
Now, we have to find \[\left| {\vec{u}} \right|\] and \[\left| {\vec{w}} \right|\] .
\[\begin{align}
& \Rightarrow \left| {\vec{u}} \right|=\sqrt{{{1}^{2}}+{{1}^{2}}+{{4}^{2}}}=\sqrt{18}=3\sqrt{2}...\left( v \right) \\
& \left| {\vec{w}} \right|=\sqrt{{{2}^{2}}+{{1}^{2}}+{{1}^{2}}}=\sqrt{4+1+1}=\sqrt{6}...\left( vi \right) \\
\end{align}\]
Now, we can substitute (iv), (v) and (vi) in (a).
\[\begin{align}
& \Rightarrow \cos \theta =\dfrac{7}{3\sqrt{2}\times \sqrt{6}}=\dfrac{7}{3\sqrt{12}} \\
& \Rightarrow \cos \theta =\dfrac{7}{3\times 2\sqrt{3}}=\dfrac{7}{6\sqrt{3}} \\
\end{align}\]
We got two values, that is, \[\cos \theta =\dfrac{5}{6},\dfrac{7}{6\sqrt{3}}\] .
Hence, the correct option is D.
Note: You may make a mistake by finding the value of $\lambda $ for only $+1$ . You may make mistake by writing the formula for \[\cos \theta \] as \[\dfrac{\vec{a}\times \vec{b}}{\left| {\vec{a}} \right|\left| {\vec{b}} \right|}\] . Also, errors can be made when finding the dot product as \[\vec{a}\cdot \vec{b}=\sqrt{{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}}\] and the magnitude of vectors as \[\left| {\vec{a}} \right|={{a}_{1}}^{2}+{{a}_{2}}^{2}+{{a}_{3}}^{2}\] . You must know how to solve the determinants in order to proceed further.
1 & 1 & \lambda \\
1 & 1 & 3 \\
2 & 1 & 1 \\
\end{matrix} \right|=\pm 1$ . We will find the value of $\lambda $ for 1 and -1 by expanding the determinant. Then, we will substitute these values in $\vec{u}$ for each values of $\lambda $ and find angle between vectors $\text{\vec{u} and \vec{w}}$ for each case using \[\cos \theta =\dfrac{\vec{u}\cdot \vec{w}}{\left| {\vec{u}} \right|\left| {\vec{w}} \right|}\].
Complete step-by-step solution:
We are given that the edges of parallelepiped are \[\overrightarrow{u}=\widehat{i}+\widehat{j}+\lambda \widehat{k},\vec{v}=\widehat{i}+\widehat{j}+3\widehat{k}\text{ and }\vec{w}=2\widehat{i}+\widehat{j}+\widehat{k}\] and the volume is 1 cu.unit. The figure below shows the parallelepiped.
We know that for a parallelepiped of edges $\vec{a}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k},\text{ }\vec{b}={{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+{{b}_{3}}\widehat{k}\text{ }$ and $\vec{c}={{c}_{1}}\widehat{i}+{{c}_{2}}\widehat{j}+{{c}_{3}}\widehat{k}$ , volume is given by
$\left| \left( \vec{a}\times \vec{b} \right).\vec{c} \right|=\left[ \vec{a}\vec{b}\vec{c} \right]=\left| \begin{matrix}
{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|$
Now, let’s write the volume of parallelepiped for given edges.
Volume $=\left| \begin{matrix}
1 & 1 & \lambda \\
1 & 1 & 3 \\
2 & 1 & 1 \\
\end{matrix} \right|$
We are given that volume is 1 cu. unit. We will take $\pm 1$Hence,
$\left| \begin{matrix}
1 & 1 & \lambda \\
1 & 1 & 3 \\
2 & 1 & 1 \\
\end{matrix} \right|=\pm 1$
Let us solve the determinant.
$1\left( 1-3 \right)-1\left( 1-6 \right)+\lambda \left( 1-2 \right)=\pm 1...\left( i \right)$
Let’s solve this to find the value of $\lambda $ for $+1$ .
\[\begin{align}
& 1\times -2-1\times -5-\lambda =1 \\
& \Rightarrow -2+5-\lambda =1 \\
& \Rightarrow 3-\lambda =1 \\
\end{align}\]
Let’s collect constants on RHS. We will get
\[\begin{align}
& \Rightarrow -\lambda =1-3 \\
& \Rightarrow -\lambda =-2 \\
& \Rightarrow \lambda =2 \\
\end{align}\]
Now, let us solve (i) to find the value of $\lambda $ for $-1$ .
\[\begin{align}
& 1\times -2-1\times -5-\lambda =-1 \\
& \Rightarrow -2+5-\lambda =-1 \\
& \Rightarrow 3-\lambda =-1 \\
\end{align}\]
Let’s collect constants on RHS. We will get
\[\begin{align}
& \Rightarrow -\lambda =-1-3 \\
& \Rightarrow -\lambda =-4 \\
& \Rightarrow \lambda =4 \\
\end{align}\]
Now, we have to find the angle between $\vec{u}$ and $\vec{w}$ for $\lambda =2$ .
We know that angle between any two vectors $\vec{a}\text{ and }\vec{b}$ is given by
\[\cos \theta =\dfrac{\vec{a}\cdot \vec{b}}{\left| {\vec{a}} \right|\left| {\vec{b}} \right|}\]
Hence, angle between vectors $\text{\vec{u} and \vec{w}}$ is given by
\[\cos \theta =\dfrac{\vec{u}\cdot \vec{w}}{\left| {\vec{u}} \right|\left| {\vec{w}} \right|}...\left( a \right)\]
Let us find \[\vec{u}\cdot \vec{w}\] .
We know that for two vectors $\vec{a}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k},\text{ }\vec{b}={{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+{{b}_{3}}\widehat{k}\text{ }$ , dot product is found as follows.
$\begin{align}
& \vec{a}\cdot \vec{b}=\left( {{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k} \right)\cdot \left( {{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+{{b}_{3}}\widehat{k}\text{ } \right) \\
& \Rightarrow \vec{a}\cdot \vec{b}=\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}} \right) \\
\end{align}$
Now, we can find \[\vec{u}\cdot \vec{w}\] where, \[\overrightarrow{u}=\widehat{i}+\widehat{j}+2\widehat{k}\text{ and }\vec{w}=2\widehat{i}+\widehat{j}+\widehat{k}\] .
\[\begin{align}
& \vec{u}\cdot \vec{w}=\left( 1\times 2+1\times 1+2\times 1 \right) \\
& \Rightarrow \vec{u}\cdot \vec{w}=2+1+2=5...(i) \\
\end{align}\]
Now, we have to find \[\left| {\vec{u}} \right|\] and \[\left| {\vec{w}} \right|\] .
We know that for a vector $\vec{a}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k}$ ,
\[\left| {\vec{a}} \right|=\sqrt{{{a}_{1}}^{2}+{{a}_{2}}^{2}+{{a}_{3}}^{2}}\]
\[\begin{align}
& \Rightarrow \left| {\vec{u}} \right|=\sqrt{{{1}^{2}}+{{1}^{2}}+{{2}^{2}}}=\sqrt{1+1+4}=\sqrt{6}...\left( ii \right) \\
& \left| {\vec{w}} \right|=\sqrt{{{2}^{2}}+{{1}^{2}}+{{1}^{2}}}=\sqrt{4+1+1}=\sqrt{6}...\left( iii \right) \\
\end{align}\]
Now, we can substitute (i), (ii) and (iii) in (a).
\[\Rightarrow \cos \theta =\dfrac{5}{\sqrt{6}\times \sqrt{6}}=\dfrac{5}{6}\]
Now, let us find the angle between $\vec{u}$ and $\vec{w}$ for $\lambda =4$
First, we have to find \[\vec{u}\cdot \vec{w}\] where, \[\overrightarrow{u}=\widehat{i}+\widehat{j}+4\widehat{k}\text{ and }\vec{w}=2\widehat{i}+\widehat{j}+\widehat{k}\]
\[\begin{align}
& \vec{u}\cdot \vec{w}=\left( 1\times 2+1\times 1+4\times 1 \right) \\
& \Rightarrow \vec{u}\cdot \vec{w}=2+1+4=7...(iv) \\
\end{align}\]
Now, we have to find \[\left| {\vec{u}} \right|\] and \[\left| {\vec{w}} \right|\] .
\[\begin{align}
& \Rightarrow \left| {\vec{u}} \right|=\sqrt{{{1}^{2}}+{{1}^{2}}+{{4}^{2}}}=\sqrt{18}=3\sqrt{2}...\left( v \right) \\
& \left| {\vec{w}} \right|=\sqrt{{{2}^{2}}+{{1}^{2}}+{{1}^{2}}}=\sqrt{4+1+1}=\sqrt{6}...\left( vi \right) \\
\end{align}\]
Now, we can substitute (iv), (v) and (vi) in (a).
\[\begin{align}
& \Rightarrow \cos \theta =\dfrac{7}{3\sqrt{2}\times \sqrt{6}}=\dfrac{7}{3\sqrt{12}} \\
& \Rightarrow \cos \theta =\dfrac{7}{3\times 2\sqrt{3}}=\dfrac{7}{6\sqrt{3}} \\
\end{align}\]
We got two values, that is, \[\cos \theta =\dfrac{5}{6},\dfrac{7}{6\sqrt{3}}\] .
Hence, the correct option is D.
Note: You may make a mistake by finding the value of $\lambda $ for only $+1$ . You may make mistake by writing the formula for \[\cos \theta \] as \[\dfrac{\vec{a}\times \vec{b}}{\left| {\vec{a}} \right|\left| {\vec{b}} \right|}\] . Also, errors can be made when finding the dot product as \[\vec{a}\cdot \vec{b}=\sqrt{{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}}\] and the magnitude of vectors as \[\left| {\vec{a}} \right|={{a}_{1}}^{2}+{{a}_{2}}^{2}+{{a}_{3}}^{2}\] . You must know how to solve the determinants in order to proceed further.
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