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Question: The volume of a right triangular prism \( ABC{A_1}{B_1}{C_1} \) is equal to 3. If the position vecto...

The volume of a right triangular prism ABCA1B1C1ABC{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 A1{A_1} can be
A. (2, 2, 2)
B. (0, 2, 0)
C. (0, − 2, 2)
D. (0, − 2, 0)

Explanation

Solution

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 A1{A_1} and A. Then by solving you will get the coordinates of A1{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
A⃗=i^+k^,\vec A = \hat i + \hat k, B⃗=2i^,\vec B = 2\hat i, C⃗=j^\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 n⃗=AB→×AC→\vec n = \overrightarrow {AB} \times \overrightarrow {AC}
Now, AB→=B⃗−A⃗=(2i^)−(i^+k^)=i^−k^\overrightarrow {AB} = \vec B - \vec A = \left( {2\hat i} \right) - (\hat i + \hat k) = \hat i - \hat k
and AC→=C⃗−A⃗=(j^)−(i^+k^)=−i^+j^−k^\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
⇒∣n⃗∣=∣i^+2j^+k^∣=6\Rightarrow \left| {\vec n} \right| = \left| {\hat i + 2\hat j + \hat k} \right| = \sqrt 6
Let coordinates of A1{A_1} is (x, y, z)
∣A1−A∣=λ∣n⃗∣\left| {{A_1} - A} \right| = \lambda \left| {\vec n} \right|
⇒∣A1−A∣=λ6\Rightarrow \left| {{A_1} - A} \right| = \lambda \sqrt 6 …(i)
Let height be h
Area of base = 12×AB→×AC→=12∣n⃗∣=62\dfrac{1}{2} \times \overrightarrow {AB} \times \overrightarrow {AC} = \dfrac{1}{2}\left| {\vec n} \right| = \dfrac{{\sqrt 6 }}{2}
Volume = Area × height
3=62×h⇒h=63 = \dfrac{{\sqrt 6 }}{2} \times h \Rightarrow h = \sqrt 6 …(ii)
From equations (i) and (ii),
Height of the prism = ∣A1−A∣=λ6=6⇒λ=1\left| {{A_1} - A} \right| = \lambda \sqrt 6 = \sqrt 6 \Rightarrow \lambda = 1
Now, ∣A1−A∣=±∣n⃗∣=±(i^+2j^+k^)\left| {{A_1} - A} \right| = \pm \left| {\vec n} \right| = \pm \left( {\hat i + 2\hat j + \hat k} \right)
⇒A1=A±(i^+2j^+k^)\Rightarrow {A_1} = A \pm \left( {\hat i + 2\hat j + \hat k} \right)
⇒A1=(i^+k^)±(i^+2j^+k^)\Rightarrow {A_1} = \left( {\hat i + \hat k} \right) \pm \left( {\hat i + 2\hat j + \hat k} \right)
⇒A1=2i^+2j^+2k^\Rightarrow {A_1} = 2\hat i + 2\hat j + 2\hat k or A1=−2j^{A_1} = - 2\hat j
In Cartesian form A1(2,2,2){A_1}(2,2,2) or A1(0,−2,0){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.