Let the volume of a parallelopiped whose coterminous edges are given by $$\vec{u} = \hat{i} + \hat{j} + \lambda \hat{k}, \vec{v} = \hat{i} + \hat{j} + 3\hat{k}$$ and $$\\ \vec{w} = 2\hat{i} + \hat{j} + \hat{k}$$ be $$1 cu$$. units. If $$\theta$$ be the angle between the edge $$\vec{u}$$ and $$\vec{w}$$, then $$\cos \theta$$ can be:
Given, $$\vec{u} = \hat{i} + \hat{j} + \lambda \hat{k}$$, $$\vec{v} = \hat{i} + \hat{j} + 3\hat{k}$$ $$\vec{w} = 2\hat{i} + \hat{j} + \hat{k}$$ Volume of parallelopied $$= u.(\vec{v} \times \vec{w}) = [\vec{u} \,\vec{v}\, \vec{w}]$$ $$= \begin{vmatrix} 1&1 &\lambda \\ 1& 1&\lambda \\2 & 1 & 1\end{vmatrix}$$ $$= 1(1-3)-1(1-6)+\lambda(1-2) -2 + 5 + (-\lambda)3-\lambda$$ so, Given volume $$= \pm 1$$ taking (+) sign $$3-\lambda = 1$$ $$\lambda = 2$$ Angle between $$\vec{u} $$ and $$\vec{w}$$ $$\cos \theta = \dfrac{\vec{u}.\vec{w}}{|\vec{u}|. |\vec{w}|} = \dfrac{(i + j + 2k).(2i + j + k)}{\sqrt{1+1+4} . \sqrt{4+1+1}}$$ $$= \dfrac{2+1+2}{\sqrt{5}. \sqrt{5}}$$ $$= \dfrac{5}{5} =1$$ Taking $$(-)$$ sign $$3-\lambda = -1$$ $$\lambda=4$$ $$\vec{u} = \hat{i} + \hat{j} + 4\hat{k}$$ $$\vec{w} = 2i+ j + \hat{k}$$ $$\cos \theta = \dfrac{\vec{u}.\vec{w}}{|\vec{u}| | \vec{w}|} = \dfrac{(\hat{i}+\hat{j}+4\hat{k}).(2i + j + k)}{\sqrt{1+1+16}\sqrt{4+1+1}}$$ $$= \dfrac{2+1+4}{\sqrt{18} \sqrt{6}} = \dfrac{7}{6\sqrt{3}}$$ $$\therefore$$ $$\boxed{\cos \theta= \dfrac{7}{6\sqrt{3}}}....Answer$$ Hence option $$D$$ is the answer
Let $$\vec{PR}=3\hat{i}+\hat{j}-2\hat{k}$$ and $$\vec{SQ}=\hat{i}-3\hat{j}-4\hat{k}$$ determine diagonals of a parallelogram $$PQRS$$ and $$ \vec{PT}=\hat{i}+2\hat{j}+3\hat{k}$$ be another vector. Then the volume of the parallelepiped determined by the vectors $$\vec{PT},\vec{PQ}$$ and $$\vec{PS}$$ is
If the volume of parallelopiped whose coterminous edges are $$\displaystyle \overline { a } =3\overline { i } -\overline { j } +4\overline { k } ,\overline { b } =2\overline { i } +3\overline { j } -\overline { k } \\ $$ and $$\displaystyle \overline { c } =-5\overline { i } +2\overline { j } +3\overline { k } $$ is three times the volume of parallelopiped whose coterminous edges are $$\displaystyle \overline { p } =\overline { i } +\overline { j } +3\overline { k } ,\overline { q } =\overline { i } -2\overline { j } +\lambda \overline { k } $$ and $$\displaystyle \overline { r } =2\overline { i } +3\overline { j } $$ then the value of $$\displaystyle \lambda $$ is
The value of $$a$$ so that the volume of parallelopiped formed by $$\hat {i}+a\hat {j}+\hat {k},\hat {j}+a\hat {k}$$ and $$a\hat {i}+\hat {k}$$ becomes minimum is
Three vectors $$\widehat{i} + \widehat{j}$$, $$\widehat{j} + \widehat{k}$$ and $$\widehat{k} + \widehat{i}$$ taken two at a time form three planes. The three unit vector drawn perpendicular to these three planes form a parallelepiped of volume
If the two diagonals of one of its faces are $$6 \widehat{i} + 6 \widehat{k}$$ and $$4\widehat{j} + 2 \widehat{k} $$ and of the edges not containing the given diagonals is $$\vec{c} = 4 \widehat{i} - 8 \widehat{k}$$, then the volume of a parallelepiped is
The volume of a tetrahedron formed by the coterminus edges $$\vec{a}, \vec{b} $$ and $$\vec{c}$$ is $$3$$, then the volume of the parallelepiped formed by the coterminus edges $$\vec{a} + \vec{b}, \vec{b} + \vec{c}$$ and $$\vec{c} + \vec{a}$$ is
Find the volume of a parallelopiped having three coterminus vectors of equal magnitude $$|\vec{a}|$$ and equal inclination $$\theta$$ with each other.
The volume of the parallelopiped whose sides are given by $$ \overrightarrow{O A}=2 i-2 j, \overrightarrow{O B}=i+j-k $$ and $$ \overrightarrow{O C} $$ $$ =3 i-k $$ is
Find the volume of the parallelepiped whose coterminous edges are represented by the vectors. $$\vec{a}=-3\hat{i}+7\hat{j}+5\hat{k}, \vec{b}=-5\hat{i}+7\hat{j}-3\hat{k}, \vec{c}=7\hat{i}-5\hat{j}-3\hat{k}$$.
Find the volume of the parallelepiped whose coterminous edges are represented by the vectors. $$\vec{a}=6\hat{i}, \vec{b}=2\hat{j}, \vec{c}=5\hat{k}$$.