Question

if you are genius Answer this

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Answer 1

Solution :

O ( 0,0,0 ) is the origin and S ( 1/2 . 1/2 , 1/2 ) is the centre of the cube . T is the vertex opposite to that of O .

From the question the coordinates of the points P , Q , R and T will be :

P = ( 1/2 , -1/2 , - 1/2 )

Q = ( -1/2 , 1/2 , - 1/2 )

R = ( - 1/2 , - 1/2 , 1/2 )

S = ( 1/2 , 1/2 , 1/2 )

Draw the figure so that you can understand the coordinates and yes , if you do not know 3-D Geometry , then you cannot understand this . First study 3-D Geometry and then only you can proceed .

The values of the vectors if written down from the question are :

$\vec{p}=\dfrac{1}{2}\hat{i}-\dfrac{1}{2}\hat{j}-\dfrac{1}{2}\hat{k}$

$\vec{q}=-\dfrac{1}{2}\hat{i}+\dfrac{1}{2}\hat{j}-\dfrac{1}{2}\hat{k}$

$\vec{r}=-\dfrac{1}{2}\hat{i}-\dfrac{1}{2}\hat{j}+\dfrac{1}{2}\hat{k}$

$\vec{t}=\dfrac{1}{2}\hat{i}+\dfrac{1}{2}\hat{j}+\dfrac{1}{2}\hat{k}$

Now we should know the cross product of vectors . If you don't know , please refer to Youtube or any other class 11 physics/maths book otherwise the problem cannot be solved and is useless to solve .

$|(\vec{p}\times \vec{q})\times (\vec{r}\times \vec{t})|$

$\vec{p}\times \vec{q}=\left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\dfrac{1}{2}&-\dfrac{1}{2}&\dfrac{1}{2}\\-\dfrac{1}{2}&\dfrac{1}{2}&-\dfrac{1}{2}\end{array}\right|\\\\\implies [(\dfrac{1}{4}+\dfrac{1}{4})\hat{i}-(-\dfrac{1}{4}-\dfrac{1}{4})\hat{j}+(\dfrac{1}{4}-\dfrac{1}{4})\hat{k}]\\\\\implies \hat{\dfrac{i}{2}}+\hat{\dfrac{j}{2}}+0\hat{k}$

$\vec{r}\times \vec{t}=\left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\-\dfrac{1}{2}&-\dfrac{1}{2}&\dfrac{1}{2}\\\dfrac{1}{2}&\dfrac{1}{2}&\dfrac{1}{2}\end{array}\right|\\\\\implies [(-\dfrac{1}{4}-\dfrac{1}{4})\hat{i}-(-\dfrac{1}{4}-\dfrac{1}{4})\hat{j}+(\dfrac{1}{4}-\dfrac{1}{4})\hat{k}\\\\\implies -\hat{\dfrac{i}{2}}+\hat{\dfrac{j}{2}}+0\hat{k}$

Now we will take the cross product of the resultant vectors .

$\left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\\dfrac{1}{2}&\dfrac{1}{2}&0\\\ -\dfrac{1}{2}&\dfrac{1}{2}&0\end{array}\right|\\\\\implies 0\hat{i}+0\hat{j}+(\dfrac{1}{4}+\dfrac{1}{4})\hat{k}\\\\\implies \dfrac{1}{2}units\\\\\implies 0.5 units$

The required value is 0.5 units .