Key Concepts and Formulas

• Vector Representation of a Triangle Side: The vector representing a side of a triangle, say from point $P$ to point $Q$, is given by the difference in their position vectors: $\vec{PQ} = \text{Position vector of } Q - \text{Position vector of } P$.
• Condition for a Right-Angled Triangle: A triangle $ABC$ is right-angled at vertex $C$ if the vectors representing the sides meeting at $C$, i.e., $\vec{CA}$ and $\vec{CB}$, are perpendicular. This condition is mathematically expressed using the dot product: $\vec{CA} \cdot \vec{CB} = 0$.
• Dot Product of Vectors: For two vectors $\vec{u} = u_x\widehat i + u_y\widehat j + u_z\widehat k$ and $\vec{v} = v_x\widehat i + v_y\widehat j + v_z\widehat k$, their dot product is $\vec{u} \cdot \vec{v} = u_x v_x + u_y v_y + u_z v_z$.

Step-by-Step Solution

Step 1: Identify the Position Vectors of the Vertices We are given the position vectors of the vertices $A$, $B$, and $C$. Let these be denoted by $\vec{p}_A$, $\vec{p}_B$, and $\vec{p}_C$ respectively.

• Position vector of $A$, $\vec{p}_A = 2\widehat i - \widehat j + \widehat k$
• Position vector of $B$, $\vec{p}_B = \widehat i - 3\widehat j - 5\widehat k$
• Position vector of $C$, $\vec{p}_C = a\widehat i - 3\widehat j + \widehat k$

Step 2: Calculate the Vectors Representing the Sides Meeting at Vertex C Since the triangle is right-angled at $C$, we need to find the vectors $\vec{CA}$ and $\vec{CB}$. These vectors represent the sides of the triangle that form the right angle. Using the formula $\vec{PQ} = \vec{p}_Q - \vec{p}_P$:

First, calculate $\vec{CA}$:

$$\vec{CA} = \vec{p}_A - \vec{p}_C \\ \vec{CA} = (2\widehat i - \widehat j + \widehat k) - (a\widehat i - 3\widehat j + \widehat k) \\ \vec{CA} = (2 - a)\widehat i + (-1 - (-3))\widehat j + (1 - 1)\widehat k \\ \vec{CA} = (2 - a)\widehat i + (-1 + 3)\widehat j + 0\widehat k \\ \vec{CA} = (2 - a)\widehat i + 2\widehat j$$

Note that the $\widehat k$ component is $0$.

Next, calculate $\vec{CB}$:

$$\vec{CB} = \vec{p}_B - \vec{p}_C \\ \vec{CB} = (\widehat i - 3\widehat j - 5\widehat k) - (a\widehat i - 3\widehat j + \widehat k) \\ \vec{CB} = (1 - a)\widehat i + (-3 - (-3))\widehat j + (-5 - 1)\widehat k \\ \vec{CB} = (1 - a)\widehat i + (-3 + 3)\widehat j - 6\widehat k \\ \vec{CB} = (1 - a)\widehat i + 0\widehat j - 6\widehat k \\ \vec{CB} = (1 - a)\widehat i - 6\widehat k$$

Note that the $\widehat j$ component is $0$.

Step 3: Apply the Condition for a Right Angle at C For the triangle $ABC$ to be right-angled at $C$, the dot product of the vectors $\vec{CA}$ and $\vec{CB}$ must be zero.

$$\vec{CA} \cdot \vec{CB} = 0$$

Step 4: Compute the Dot Product and Form an Equation Substitute the expressions for $\vec{CA}$ and $\vec{CB}$ into the dot product equation. Remember to include the zero components for clarity in the dot product calculation.

$$\vec{CA} = (2 - a)\widehat i + 2\widehat j + 0\widehat k$$ $$\vec{CB} = (1 - a)\widehat i + 0\widehat j - 6\widehat k$$

Now, calculate the dot product:

$$\vec{CA} \cdot \vec{CB} = ((2 - a)\widehat i + 2\widehat j + 0\widehat k) \cdot ((1 - a)\widehat i + 0\widehat j - 6\widehat k) \\ \vec{CA} \cdot \vec{CB} = (2 - a)(1 - a) + (2)(0) + (0)(-6) \\ \vec{CA} \cdot \vec{CB} = (2 - a)(1 - a) + 0 + 0 \\ \vec{CA} \cdot \vec{CB} = (2 - a)(1 - a)$$

Setting the dot product to zero gives us the equation:

$$(2 - a)(1 - a) = 0$$

Step 5: Solve the Algebraic Equation for 'a' The equation $(2 - a)(1 - a) = 0$ is a product of two factors that equals zero. This implies that at least one of the factors must be zero. We have two possible cases:

Case 1: $2 - a = 0$ Solving for $a$, we get $a = 2$.

Case 2: $1 - a = 0$ Solving for $a$, we get $a = 1$.

Therefore, the possible values of $a$ for which the triangle $ABC$ is right-angled at $C$ are $2$ and $1$.

Step 6: Match the Results with the Given Options The calculated values for $a$ are $2$ and $1$. We compare this set of values with the given options: (A) $2$ and $1$ (B) $-2$ and $-1$ (C) $-2$ and $1$ (D) $2$ and $-1$

Our result $\{2, 1\}$ matches option (A).

Common Mistakes & Tips

• Vector Direction: Ensure that the vectors for the sides are correctly oriented. For a right angle at $C$, use vectors originating from $C$, i.e., $\vec{CA}$ and $\vec{CB}$. Using vectors like $\vec{AC}$ and $\vec{BC}$ is also valid but requires careful handling of signs if mixed.
• Dot Product Calculation Errors: Pay close attention to signs when subtracting position vectors and when multiplying components in the dot product. Remember that the dot product of vectors with missing components (e.g., a vector without an $\widehat k$ term) involves multiplying by zero for that component.
• Algebraic Solution: When solving equations like $(x-p)(x-q)=0$, remember that both $x=p$ and $x=q$ are valid solutions.

Summary

To determine the values of $a$ for which triangle $ABC$ is right-angled at $C$, we utilized the property that the dot product of the vectors forming the right angle at $C$, namely $\vec{CA}$ and $\vec{CB}$, must be zero. We first computed these vectors by subtracting the position vector of $C$ from the position vectors of $A$ and $B$, respectively. Subsequently, we applied the dot product condition, $\vec{CA} \cdot \vec{CB} = 0$, which led to the algebraic equation $(2-a)(1-a) = 0$. Solving this equation yielded the values $a=2$ and $a=1$.

The final answer is $\boxed{2 \text{ and } 1}$ which corresponds to option (A).