Solveeit Logo
Login

Question

Question: Let the volume of a parallelepiped whose coterminous edges are given by \[\overrightarrow{u}=\wideha...

Let the volume of a parallelepiped whose coterminous edges are given by u→=i^+j^+λk^,v⃗=i^+j^+3k^ and w⃗=2i^+j^+k^\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 θ\theta be the angle between the edge u⃗\vec{u} and w⃗\vec{w} , then cos⁡θ\cos \theta can be :
A. 57\dfrac{5}{7}
B. 533\dfrac{5}{3\sqrt{3}}
C. 766\dfrac{7}{6\sqrt{6}}
D. 763\dfrac{7}{6\sqrt{3}}

Explanation

Solution

We are given that the volume of parallelepiped with given edges is 1 cu.unit. We can write it as Volume =∣(u×v⃗).w∣=[u⃗v⃗w⃗]=∣11λ 113 211 ∣=±1=\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 λ\lambda for 1 and -1 by expanding the determinant. Then, we will substitute these values in u⃗\vec{u} for each values of λ\lambda and find angle between vectors \text{\vec{u} and \vec{w}} for each case using cos⁡θ=u⃗⋅w⃗∣u⃗∣∣w⃗∣\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 u→=i^+j^+λk^,v⃗=i^+j^+3k^ and w⃗=2i^+j^+k^\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 a⃗=a1i^+a2j^+a3k^, b⃗=b1i^+b2j^+b3k^ \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 c⃗=c1i^+c2j^+c3k^\vec{c}={{c}_{1}}\widehat{i}+{{c}_{2}}\widehat{j}+{{c}_{3}}\widehat{k} , volume is given by
∣(a⃗×b⃗).c⃗∣=[a⃗b⃗c⃗]=∣a1a2a3 b1b2b3 c1c2c3 ∣\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 =∣11λ 113 211 ∣=\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 ±1\pm 1Hence,
∣11λ 113 211 ∣=±1\left| \begin{matrix} 1 & 1 & \lambda \\\ 1 & 1 & 3 \\\ 2 & 1 & 1 \\\ \end{matrix} \right|=\pm 1
Let us solve the determinant.
1(1−3)−1(1−6)+λ(1−2)=±1...(i)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 λ\lambda for +1+1 .

& 1\times -2-1\times -5-\lambda =1 \\\ & \Rightarrow -2+5-\lambda =1 \\\ & \Rightarrow 3-\lambda =1 \\\ \end{aligned}$$ Let’s collect constants on RHS. We will get $$\begin{aligned} & \Rightarrow -\lambda =1-3 \\\ & \Rightarrow -\lambda =-2 \\\ & \Rightarrow \lambda =2 \\\ \end{aligned}$$ Now, let us solve (i) to find the value of $\lambda $ for $-1$ . $$\begin{aligned} & 1\times -2-1\times -5-\lambda =-1 \\\ & \Rightarrow -2+5-\lambda =-1 \\\ & \Rightarrow 3-\lambda =-1 \\\ \end{aligned}$$ Let’s collect constants on RHS. We will get $$\begin{aligned} & \Rightarrow -\lambda =-1-3 \\\ & \Rightarrow -\lambda =-4 \\\ & \Rightarrow \lambda =4 \\\ \end{aligned}$$ Now, we have to find the angle between $\vec{u}$ and $\vec{w}$ for $\lambda =2$ . We know that angle between any two vectors $\vec{a}\text{ and }\vec{b}$ is given by $$\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}$$ . We know that for two vectors $\vec{a}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k},\text{ }\vec{b}={{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+{{b}_{3}}\widehat{k}\text{ }$ , dot product is found as follows. $\begin{aligned} & \vec{a}\cdot \vec{b}=\left( {{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k} \right)\cdot \left( {{b}_{1}}\widehat{i}+{{b}_{2}}\widehat{j}+{{b}_{3}}\widehat{k}\text{ } \right) \\\ & \Rightarrow \vec{a}\cdot \vec{b}=\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}} \right) \\\ \end{aligned}$ Now, we can find $$\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{aligned} & \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{aligned}$$ Now, we have to find $$\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{aligned} & \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{aligned}$$ Now, we can substitute (i), (ii) and (iii) in (a). $$\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{aligned} & \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{aligned}$$ Now, we have to find $$\left| {\vec{u}} \right|$$ and $$\left| {\vec{w}} \right|$$ . $$\begin{aligned} & \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{aligned}$$ Now, we can substitute (iv), (v) and (vi) in (a). $$\begin{aligned} & \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{aligned}$$ We got two values, that is, $$\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.