Question

The volume of a right triangular prism $ ABC{A_1}{B_1}{C_1} $ is equal to 3. If the position vectors of the vertices of the base ABC are A(1, 0, 1), B(2, 0, 0) and C(0, 1, 0), then position vectors of the vertex $ {A_1} $ can be
A. (2, 2, 2)
B. (0, 2, 0)
C. (0, − 2, 2)
D. (0, − 2, 0)

Answer 1

Hint: Here, find the base area of the prism using vector product, and then compare the volume given by assuming height as h. In vector form compare the height f the prism obtained with the length of $ {A_1} $ and A. Then by solving you will get the coordinates of $ {A_1} $ .

Complete step-by-step answer:
We have position vectors of the base are A (1, 0, 1), B (2, 0, 0), C (0, 1, 0)

In vector form A, B and C can be written as
 $ \vec A = \hat i + \hat k, $ $ \vec B = 2\hat i, $ $ \vec C = \hat j $
As we know, the volume of a triangular prism is equal to the area of the base of the prism multiplied by height of the prism.
Volume of prism = (Area of base) × Height
As ABC is the base of the prism, then height of the prism will be perpendicular to the lane containing ABC.
Let $ \vec n = \overrightarrow {AB} \times \overrightarrow {AC} $
Now, $ \overrightarrow {AB} = \vec B - \vec A = \left( {2\hat i} \right) - (\hat i + \hat k) = \hat i - \hat k $
and $ \overrightarrow {AC} = \vec C - \vec A = \left( {\hat j} \right) - (\hat i + \hat k) = - \hat i + \hat j - \hat k $
Then, \[\overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  1&0&{ - 1} \\
  { - 1}&1&{ - 1}
\end{array}} \right| = \hat i + 2\hat j + \hat k\]
 $ \Rightarrow \left| {\vec n} \right| = \left| {\hat i + 2\hat j + \hat k} \right| = \sqrt 6 $
Let coordinates of $ {A_1} $ is (x, y, z)
 $ \left| {{A_1} - A} \right| = \lambda \left| {\vec n} \right| $
 $ \Rightarrow \left| {{A_1} - A} \right| = \lambda \sqrt 6 $ …(i)
Let height be h
Area of base = $ \dfrac{1}{2} \times \overrightarrow {AB} \times \overrightarrow {AC} = \dfrac{1}{2}\left| {\vec n} \right| = \dfrac{{\sqrt 6 }}{2} $
Volume = Area × height
 $ 3 = \dfrac{{\sqrt 6 }}{2} \times h \Rightarrow h = \sqrt 6 $ …(ii)
From equations (i) and (ii),
Height of the prism = $ \left| {{A_1} - A} \right| = \lambda \sqrt 6 = \sqrt 6 \Rightarrow \lambda = 1 $
Now, \[\left| {{A_1} - A} \right| = \pm \left| {\vec n} \right| = \pm \left( {\hat i + 2\hat j + \hat k} \right)\]
 $ \Rightarrow {A_1} = A \pm \left( {\hat i + 2\hat j + \hat k} \right) $
 $ \Rightarrow {A_1} = \left( {\hat i + \hat k} \right) \pm \left( {\hat i + 2\hat j + \hat k} \right) $
 $ \Rightarrow {A_1} = 2\hat i + 2\hat j + 2\hat k $ or $ {A_1} = - 2\hat j $
In Cartesian form $ {A_1}(2,2,2) $ or $ {A_1}(0, - 2,0) $ .

So, the correct answer is “Option A AND D”.

Note: In these types of questions, change the vertices in vector form to make the calculations easy. You must be aware about vector notation and operations of vectors. Modulus of any vector means the magnitude of the vector, so we say height of a prism means magnitude of the vector which represents the height of the prism. Magnitude of cross product of two vectors means the area of the parallelogram, in which given vectors represents two adjacent sides of the parallelogram.