Let a vector V1¯=OP¯, Where O is origin and P is (2,0,2). Vector V1¯ is rotated about origin by an angle of π3 in a plane which is perpendicular to XOZ plane to get vector V2¯ , where V2¯.j∧>0. Again V1¯ is rotated about origin by right angle in XOZ plane to get vector V3¯. If V is volume of tetrahedron whose coterminous edges are V1¯,V2¯ and V3¯ then value of 3V6 is

Equation of plane perpendicular to XOZ plane is P:x−z=0 V2→ is coplanar V1→ and j ⇒V→2=xV→1+yj∧⇒V→2=x(2i∧+2k∧)+yj∧∵|V→2|=|V→1|=22∴4x2+y2+4x2=8 ..................................(1)Now, V→2.V→1=|V→2||V→1|cosπ3=44=(x(2i∧+2k∧)+yj∧).(2i∧+2k∧)⇒V→2=i∧+6j∧+k∧(∵V→2.j∧>0) Now, V→3=V→1+xi∧∵V→3.V→1=0⇒8+2x=0⇒x=−4 ⇒V→3=2i∧+2k∧−4i∧ V→3=−2i∧+2k∧ ⇒ Volume of tetrahedron ⇒V=1686⇒3V6=4 .

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Let a vector $\bar{V_{1}} = \bar{O P},$ Where O is origin and P is (2,0,2). Vector $\bar{V_{1}}$ is rotated about origin by an angle of $\frac{π}{3}$ in a plane which is perpendicular to XOZ plane to get vector $\bar{V_{2}}$ , where $\bar{V_{2}} . \overset{\land}{j} > 0$ . Again $\bar{V_{1}}$ is rotated about origin by right angle in XOZ plane to get vector $\bar{V_{3}}$ . If V is volume of tetrahedron whose coterminous edges are $\bar{V_{1}}, \bar{V_{2}}$ and $\bar{V_{3}}$ then value of $\frac{3 V}{\sqrt{6}}$ is

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Detailed Solution

Equation of plane perpendicular to XOZ plane is $P : x - z = 0$
$\vec{V_{2}}$ is coplanar $\vec{V_{1}}$ and j

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$\begin{array}{l} \Rightarrow {\vec{V}}_{2} = x {\vec{V}}_{1} + y \overset{\land}{j} \\ \Rightarrow {\vec{V}}_{2} = x (2 \overset{\land}{i} + 2 \overset{\land}{k}) + y \overset{\land}{j} \\ ∵ | {\vec{V}}_{2} | = | {\vec{V}}_{1} | = 2 \sqrt{2} \\ ∴ 4 x^{2} + y^{2} + 4 x^{2} = 8 \end{array}$
..................................(1)
Now, ${\vec{V}}_{2} . {\vec{V}}_{1} = | {\vec{V}}_{2} | | {\vec{V}}_{1} | \cos \frac{π}{3} = 4$ $4 = (x (2 \overset{\land}{i} + 2 \overset{\land}{k}) + y \overset{\land}{j}) . (2 \overset{\land}{i} + 2 \overset{\land}{k})$

$\Rightarrow {\vec{V}}_{2} = \overset{\land}{i} + \sqrt{6} \overset{\land}{j} + \overset{\land}{k} (∵ {\vec{V}}_{2} . \overset{\land}{j} > 0)$

Now, ${\vec{V}}_{3} = {\vec{V}}_{1} + x \overset{\land}{i}$

$∵ {\vec{V}}_{3} . {\vec{V}}_{1} = 0 \Rightarrow 8 + 2 x = 0 \Rightarrow x = - 4$
$\Rightarrow {\vec{V}}_{3} = 2 \overset{\land}{i} + 2 \overset{\land}{k} - 4 \overset{\land}{i}$ ${\vec{V}}_{3} = - 2 \overset{\land}{i} + 2 \overset{\land}{k}$

$\Rightarrow$ Volume of tetrahedron
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