Question

The volume of a right triangular prism $ABCA_1B_1C_1$ is equal to 3 cubic units. Then the co-ordinates of the vertex $A_1$, if the co-ordinates of the base vertices of the prism are A (1,0,1), B (2,0,0) and C (0,1,0) be (a,b,c) then a+b+c can be equal to ($A_1$ is vertically above A)

• A. 6
• B. -2
• C. 0
• D. 3

Answer 1

Answer: (A), (B)

6 and -2

Answer 2

1. Represent the base vertices A, B, C as position vectors: $\vec{A} = \hat{i} + \hat{k}$, $\vec{B} = 2\hat{i}$, $\vec{C} = \hat{j}$.
2. Calculate vectors representing two sides of the base triangle: $\overrightarrow{AB} = \vec{B} - \vec{A} = \hat{i} - \hat{k}$ and $\overrightarrow{AC} = \vec{C} - \vec{A} = -\hat{i} + \hat{j} - \hat{k}$.
3. Find the normal vector to the base plane by computing the cross product: $\vec{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \hat{i} + 2\hat{j} + \hat{k}$.
4. Calculate the magnitude of the normal vector: $|\vec{n}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}$.
5. The area of the base triangle is half the magnitude of the cross product: Area $= \frac{1}{2}|\vec{n}| = \frac{\sqrt{6}}{2}$.
6. Use the volume formula for a prism (Volume = Base Area × Height) to find the height $h$: $3 = \frac{\sqrt{6}}{2} \times h \implies h = \frac{6}{\sqrt{6}} = \sqrt{6}$.
7. The height vector $\overrightarrow{AA_1}$ is parallel to $\vec{n}$ and has magnitude $h$. So, $\overrightarrow{AA_1} = k\vec{n}$, where $|k||\vec{n}| = h$. This gives $|k|\sqrt{6} = \sqrt{6}$, so $|k|=1$. Thus, $k=1$ or $k=-1$.
8. Determine the position vector of $A_1$: $\vec{A_1} = \vec{A} + \overrightarrow{AA_1} = \vec{A} + k\vec{n}$.
   • Case 1 ($k=1$): $\vec{A_1} = (\hat{i} + \hat{k}) + 1(\hat{i} + 2\hat{j} + \hat{k}) = 2\hat{i} + 2\hat{j} + 2\hat{k}$. So, $A_1 = (2,2,2)$. The sum of coordinates is $a+b+c = 2+2+2 = 6$.
   • Case 2 ($k=-1$): $\vec{A_1} = (\hat{i} + \hat{k}) - 1(\hat{i} + 2\hat{j} + \hat{k}) = -2\hat{j}$. So, $A_1 = (0,-2,0)$. The sum of coordinates is $a+b+c = 0+(-2)+0 = -2$.
9. The possible values for $a+b+c$ are 6 and -2.