Question

The area of triangle with vertices $(1,2,0)$, $(1,0,a)$ and $(0,3,1)$ is $\sqrt{6}$ sq. units, then the values of '$a$' are

• A. $-8, 1$
• B. $2, -4$
• C. $-2, 4$
• D. $8, -1$

Hint:

Instead of solving the full quadratic equations during time-pressured exams, you can quickly test the options. Substituting $a=2$ into the simplified magnitude expression gives $\sqrt{2(2)^2+4(2)+8} = \sqrt{8+8+8} = \sqrt{24} = 2\sqrt{6}$. Dividing by 2 yields $\sqrt{6}$, verifying option (B) instantly!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given three vertices of a triangle in 3D space: $A(1,2,0)$, $B(1,0,a)$, and $C(0,3,1)$. The area of this triangle is given as $\sqrt{6}$ square units. We need to determine the possible real values of the parameter $a$.

Step 2: Key Formula or Approach:
The area of a triangle with vertices defined by position vectors can be computed using the vector cross product of two adjacent sides: $$\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}|$$ We will construct vectors $\vec{AB}$ and $\vec{AC}$, compute their cross product using a determinant, find the magnitude, and equate it to $\sqrt{6}$.

Step 3: Detailed Explanation:
Let the vertices be $A(1,2,0)$, $B(1,0,a)$, and $C(0,3,1)$. Find the components of vectors $\vec{AB}$ and $\vec{AC}$: $$\vec{AB} = (1-1)\hat{i} + (0-2)\hat{j} + (a-0)\hat{k} = 0\hat{i} - 2\hat{j} + a\hat{k}$$ $$\vec{AC} = (0-1)\hat{i} + (3-2)\hat{j} + (1-0)\hat{k} = -1\hat{i} + 1\hat{j} + 1\hat{k}$$ Now, find the cross product $\vec{AB} \times \vec{AC}$ using a matrix determinant environment: $$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -2 & a \\ -1 & 1 & 1 \end{vmatrix}$$ Expanding along the first row: $$\vec{AB} \times \vec{AC} = \hat{i}((-2)(1) - (a)(1)) - \hat{j}((0)(1) - (a)(-1)) + \hat{k}((0)(1) - (-2)(-1))$$ $$\vec{AB} \times \vec{AC} = \hat{i}(-2 - a) - \hat{j}(a) + \hat{k}(-2) = -(a+2)\hat{i} - a\hat{j} - 2\hat{k}$$ Next, compute the magnitude of this cross product vector: $$|\vec{AB} \times \vec{AC}| = \sqrt{(-(a+2))^2 + (-a)^2 + (-2)^2}$$ $$|\vec{AB} \times \vec{AC}| = \sqrt{(a^2 + 4a + 4) + a^2 + 4} = \sqrt{2a^2 + 4a + 8}$$ Since the area of the triangle is $\frac{1}{2}|\vec{AB} \times \vec{AC}| = \sqrt{6}$: $$\frac{1}{2} \sqrt{2a^2 + 4a + 8} = \sqrt{6}$$ Square both sides to eliminate the radicals: $$\frac{1}{4} (2a^2 + 4a + 8) = 6$$ $$2a^2 + 4a + 8 = 24$$ $$2a^2 + 4a - 16 = 0$$ Divide the entire equation by 2: $$a^2 + 2a - 8 = 0$$ Factor the quadratic equation: $$(a + 4)(a - 2) = 0$$ This yields two possible values for $a$: $$a = 2 \quad \text{or} \quad a = -4$$

Step 4: Final Answer:
The values of $a$ are $2, -4$, which corresponds to option (B).