Question

Volume of parallelepiped with sides in the form of position vector which is represented by$( - 12\hat i + \alpha \hat k)$,$(3\hat j - \hat k)$and $(2\hat i +\hat j - 15\hat k)$ is $(546)$.Then $(\alpha = ?)$ [IIT Screening $(1989)$; MNR $(1987)$].
A) 3
B) 2
C) - 3
D) - 2

Answer 1

Hint: Parallelepiped is a $(3)$-D structure consists of six symmetrical parallelograms. In this question we have to use the concept of the volume of parallelepiped which is found by using the scalar triple product of edges. In this question we take the scalar triple product of vectors formed by three concurrent edges.

Formula UsedScalar triple product of vectors $( = \vec a{\rm{.}}\left( {\vec b \times \vec c} \right))$
Scalar triple product can also be represented as $(\left[ {a\;b\;c\;} \right])$
Where $(\vec a,\vec b)$and $(\vec c)$are three concurrent edges of parallelepiped.
$(\vec a{\rm{.}}\left( {\vec b \times \vec c} \right) = \left| a \right|\left| {\left( {\vec b \times \vec c} \right)} \right|\cos \left( \theta \right))$
Where $(\theta )$is the angle between $(\vec a)$and $(\left( {\vec b \times \vec c} \right))$
$(\vec a.\left( {\vec b \times \vec c} \right) = ( {a_1}\hat i + {a_2}\hat j + {a_3}\hat k) .\begin{bmatrix}{\hat i} & {\hat j} & {\hat k} \\{{b_1}} & {{b_2}} & {{b_3}} \\{{c_1}} & {{c_2}} & {{c_3}} \end{bmatrix}$
$( {a_1}\hat i + {a_2}\hat j + {a_3}\hat k) \lbrace(b_2c_3-b_3c_2)\hat i-(b_1c_3-b_3c_1)\hat i+(b_1c_2-b_2c_1)\hat i\rbrace$
$( = {a_1}\left( {{b_2}{c_3} - {b_3}{c_2}} \right) - {a_2}\left( {{b_1}{c_3} - {b_3}{c_1}} \right) + {a_3}\left( {{b_1}{c_2} - {b_2}{c_1}} \right))$


Complete step by step solution:Given: vectors a, b and c are the three concurrent edges of parallelepiped.
Volume of parallelepiped is 546
Edges of parallelepiped is :
$(\vec a = - 12\hat i + \alpha \hat k)$
$(\vec b = 3\hat j - \hat k)$
$(\vec c = 2\hat i + \hat j - 15\hat k)$
Where,
$({a_1} = - 12,{a_2} = 0,{a_3} = \alpha )$
$({b_1} = 0,\;{b_2} = 3,{b_3} = - 1)$
$({c_1} = 2,{c_2} = 1,{c_3} = - 15)$
Volume of parallelepiped is given by :
$(\vec a.\left( {\vec b \times \vec c} \right) = ( {a_1}\hat i + {a_2}\hat j + {a_3}\hat k) .\begin{bmatrix}{\hat i} & {\hat j} & {\hat k} \\{{b_1}} & {{b_2}} & {{b_3}} \\{{c_1}} & {{c_2}} & {{c_3}} \end{bmatrix}$
$( {a_1}\hat i + {a_2}\hat j + {a_3}\hat k) \lbrace(b_2c_3-b_3c_2)\hat i-(b_1c_3-b_3c_1)\hat i+(b_1c_2-b_2c_1)\hat i\rbrace$
$( = {a_1}\left( {{b_2}{c_3} - {b_3}{c_2}} \right) - {a_2}\left( {{b_1}{c_3} - {b_3}{c_1}} \right) + {a_3}\left( {{b_1}{c_2} - {b_2}{c_1}} \right))$
$ = - 12(3 \times (- 15 ) - ( - 1) \times 1) + \alpha ( 0 \times 1 - 3 \times 2$
$- 12 ( - 45 + 1) + \alpha ( - 6 ) = 546$
$(528 - 6\alpha = 546)$
$(6\alpha = 528 - 546)$
$(6\alpha = - 18)$
$(\alpha = - 3)$



Option ‘C’ is correct



Note: Here question asked to find the value of unknown variable $\alpha$. IIn order to find the volume of parallelepiped must be known. Scalar triple product formula is used to find the volume of parallelepiped. Scalar triple product means product of three vectors i.e. dot product of one of the vectors with cross product of other two vectors.
Scalar triple products are represented as $(\left[ {a\;b\;c\;} \right])$.
The resultant scalar triple product is always scalar. Whenever we get the value of a scalar triple product as zero we can say that three vectors are coplanar.
Scalar triple product may be zero, negative and positive.