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

Answer: (A), (D)

The correct options are
A $(2,2,2)$
D $(0,-2,0)$

Multiple answers are available.

$A(1,0,1),B(2,0,0),C(0,1,0)$

$\overrightarrow{n}=\overrightarrow{AB}\times \overrightarrow{AC}$

$=(\hat{i}-\hat{k})\times (-\hat{i}+\hat{j}-\hat{k})$

$\overrightarrow{n}=\left|\begin{array}{ccc}i& j& k\\ 1& 0& -1\\ -1& 1& -1\end{array}\right|=\hat{i}+2\hat{j}+\hat{k}$

$A^{\prime }(x,y,z),A(1,0,1)$

$|A^{\prime }-A|=\lambda |\overrightarrow{n}|$

$=\lambda \sqrt{1+4+1}=\sqrt{6}\lambda$

$\therefore |\overrightarrow{n}|=\sqrt{6}\lambda$

Volume=Area(height)

$3=\frac{1}{2}|\overrightarrow{n}|\times h$

$3=\frac{1}{2}\sqrt{6}\times h$

$h=\sqrt{6}$

$|A^{\prime }-A|=\pm |\overrightarrow{n}|=\pm (\hat{i}+2\hat{j}+\hat{k})$

$A^{\prime }=(\hat{i}+\hat{k})\pm (\hat{i}+2\hat{j}+\hat{k})$

$\therefore A^{\prime }(2,2,2),A^{\prime }(0,-2,0)$