Question

Let the vectors $\vec P\vec R = 3\hat i + \hat j - 2\hat k$ and $\vec S\vec Q = \hat i - 3\hat j - 4\hat k$ determines diagonal of a parallelogram PQRS and $\vec P\vec T = \hat i + 2\hat j + 3\hat k$ be another vector. Then the volume of the parallelepiped determined by the vectors $\vec P\vec T,\vec P\vec Q,{\text{ and }}\vec P\vec S$ is
$\left( a \right)5$
$\left( b \right)20$
$\left( c \right)10$
$\left( d \right)30$

Answer 1

Hint: In this particular question first draw the pictorial representation of the above problem it will give us a clear picture of what we have to find out, then use the concept that the volume of the parallelepiped is the product of the area of the base and the height of the parallelepiped, so use these concepts to reach the solution of the question.

Complete step-by-step solution:

Given data:
$\vec P\vec R = 3\hat i + \hat j - 2\hat k$and,$\vec S\vec Q = \hat i - 3\hat j - 4\hat k$ are the two diagonals of the parallelogram PQRS as shown in the above figure.
Now as we know that the area of the parallelogram when the diagonals of the parallelogram is given is given as,
$ \Rightarrow A = \dfrac{1}{2}\left| {\vec P\vec R \times \vec S\vec Q} \right|$
Ow substitute the values we have,
$ \Rightarrow A = \dfrac{1}{2}\left| {\left( {3\hat i + \hat j - 2\hat k} \right) \times \left( {\hat i - 3\hat j - 4\hat k} \right)} \right|$
Now simplify it we have,
$ \Rightarrow A = \dfrac{1}{2}\left| {\left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  3&1&{ - 2} \\
  1&{ - 3}&{ - 4}
\end{array}} \right|} \right|$
$ \Rightarrow A = \dfrac{1}{2}\left| {\left( { - 4 - 6} \right)\hat i - \left( { - 12 + 2} \right)\hat j + \left( { - 9 - 1} \right)\hat k} \right|$
$ \Rightarrow A = \dfrac{1}{2}\left| { - 10\hat i + 10\hat j - 10\hat k} \right|$
$ \Rightarrow A = 5\left| { - \hat i + \hat j - \hat k} \right|$
Now as we know that the modulus of $\left| {a\hat i + b\hat j + c\hat k} \right| = \sqrt {{a^2} + {b^2} + {c^2}} $ so the area becomes,
$ \Rightarrow A = 5\sqrt {{{\left( { - 1} \right)}^2} + {1^2} + {{\left( { - 1} \right)}^2}} = 5\sqrt 3 $ Sq. units.
Now we have to find the height of the parallelepiped as shown in the above figure, so for this we have to take the projection of vector PT on the cross product of the diagonals of the parallelogram.
Now as we know that the projection of $a\hat i + b\hat j + c\hat k$ on $p\hat i + q\hat j + r\hat k$ is given as
\[ \Rightarrow \left( {a\hat i + b\hat j + c\hat k} \right).\dfrac{{\left( {p\hat i + q\hat j + r\hat k} \right)}}{{\sqrt {{p^2} + {q^2} + {r^2}} }}\].
So the projection of PT i.e. $\vec P\vec T = \hat i + 2\hat j + 3\hat k$ on $ - \hat i + \hat j - \hat k$ is,
$ \Rightarrow h = \left| {\left( {\hat i + 2\hat j + 3\hat k} \right).\dfrac{{\left( { - \hat i + \hat j - \hat k} \right)}}{{\sqrt {{{\left( { - 1} \right)}^2} + {1^2} + {{\left( { - 1} \right)}^2}} }}} \right|$
Now simplify we have,
$ \Rightarrow h = \left| {\dfrac{{ - 1 + 2 - 3}}{{\sqrt 3 }}} \right| = \dfrac{{\left| { - 2} \right|}}{{\sqrt 3 }} = \dfrac{2}{{\sqrt 3 }}$
So the height of the parallelepiped is $\dfrac{2}{{\sqrt 3 }}$ units.
Now as we know that the volume of the parallelepiped is the product of area of the base and the height of the parallelepiped, so we have,
$ \Rightarrow V = A\left( h \right) = 5\sqrt 3 \left( {\dfrac{2}{{\sqrt 3 }}} \right) = 10$ Cubic units.
So this is the required answer.
Hence option (c) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always, recall the formula of the area of the parallelogram when the diagonals of the parallelogram are given and a projection of a vector on another vector which is all stated above so first find out the area of the parallelogram and the projection of the vector PT on the cross product of the diagonals of the parallelogram as above, then multiply these two values we will get the required volume of the parallelepiped.