Question

If the lines $x = ay - 1 = z - 2$ and $x = 3y - 2 = bz - 2$ ($ab \neq 0$) are coplanar, then \dots

• A. $a = 1, b = 1/2$
• B. $a = 2, b = 2$
• C. $a = 1/2, b = 1/2$
• D. $b = 1, a \in \mathbb{R} - \{0\}$

Hint:

When converting continuous equality lines like $x = 3y-2 = \dots$ into standard symmetric form, always factor out the coefficient of $y$ and $z$ so that the numerator variables sit alone (e.g., $3(y - 2/3)$ gives a denominator of $1/3$).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the equations of two lines in 3D space. We are told they are coplanar (they lie in the exact same flat plane). This condition sets the scalar triple product of specific vectors related to the lines equal to zero.

Step 2: Key Formula or Approach:
1. Rewrite both lines into the standard symmetric form: $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$.
2. Extract the anchor points $\vec{p_1}, \vec{p_2}$ and direction vectors $\vec{d_1}, \vec{d_2}$ of the two lines.
3. The condition for coplanarity is: $(\vec{p_2} - \vec{p_1}) \cdot (\vec{d_1} \times \vec{d_2}) = 0$. This is equivalent to setting a $3 \times 3$ determinant to zero.

Step 3: Detailed Explanation:
Line 1: $x = a(y - 1/a) = z - 2$
Standard form: $\frac{x-0}{1} = \frac{y - 1/a}{1/a} = \frac{z-2}{1}$
Point $P_1(0, 1/a, 2)$, Direction $\vec{d_1}(1, 1/a, 1)$.
Line 2: $x = 3(y - 2/3) = b(z - 2/b)$
Standard form: $\frac{x-0}{1} = \frac{y - 2/3}{1/3} = \frac{z - 2/b}{1/b}$
Point $P_2(0, 2/3, 2/b)$, Direction $\vec{d_2}(1, 1/3, 1/b)$.
Vector connecting points: $\vec{P_1 P_2} = (0 - 0, \frac{2}{3} - \frac{1}{a}, \frac{2}{b} - 2)$.
Set the determinant of coplanarity to 0:
$$\begin{vmatrix} 0 & \frac{2}{3} - \frac{1}{a} & \frac{2}{b} - 2 \\ 1 & \frac{1}{a} & 1 \\ 1 & \frac{1}{3} & \frac{1}{b} \end{vmatrix} = 0$$
Expand along the first column:
$$-1 \left[ \left(\frac{2}{3} - \frac{1}{a}\right)\frac{1}{b} - (1)\left(\frac{2}{b} - 2\right) \right] + 1 \left[ \left(\frac{2}{3} - \frac{1}{a}\right)(1) - \frac{1}{a}\left(\frac{2}{b} - 2\right) \right] = 0$$
To simplify algebra, let $Y = \frac{1}{a}$ and $Z = \frac{1}{b}$:
$$- \left[ \left(\frac{2}{3} - Y\right)Z - \left(2Z - 2\right) \right] + \left[ \left(\frac{2}{3} - Y\right) - Y(2Z - 2) \right] = 0$$
$$-\frac{2}{3}Z + YZ + 2Z - 2 + \frac{2}{3} - Y - 2YZ + 2Y = 0$$
Group the terms:
$$\left(2Z - \frac{2}{3}Z\right) + (2Y - Y) - YZ - \frac{4}{3} = 0$$
$$\frac{4}{3}Z + Y - YZ - \frac{4}{3} = 0$$
Factor by grouping:
$$\frac{4}{3}(Z - 1) - Y(Z - 1) = 0$$
$$(Z - 1)\left(\frac{4}{3} - Y\right) = 0$$
This gives two possibilities: $Z = 1$ or $Y = 4/3$.
If $Z = 1$, then $1/b = 1 \implies b = 1$. In this case, $Y$ (and thus $a$) can be any real number except 0.
If $Y = 4/3$, then $a = 3/4$, and $b$ can be anything.
Checking the options, option (d) specifies $b=1$ and $a \in \mathbb{R} - \{0\}$.

Step 4: Final Answer:
The condition $b=1, a \in \mathbb{R} - \{0\}$ is correct, matching option (d).