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 Concept:
Two lines are coplanar if they intersect or are parallel. For non-parallel lines, they are coplanar if the scalar triple product of $(\vec{a}_2 - \vec{a}_1)$, $\vec{b}_1$, and $\vec{b}_2$ is zero.

Step 2: Formula Application:
Line 1: $x/1 = (y-1/a)/(1/a) = (z-2)/1$. Point $P_1(0, 1/a, 2)$, Direction $\vec{v}_1(1, 1/a, 1)$. Line 2: $x/1 = (y-2/3)/(1/3) = (z-2/b)/(1/b)$. Point $P_2(0, 2/3, 2/b)$, Direction $\vec{v}_2(1, 1/3, 1/b)$.

Step 3: Explanation:
Both lines pass through the x-axis area at $x=0$. If $b=1$, the direction vectors become $(1, 1/a, 1)$ and $(1, 1/3, 1)$. Since both lines then pass through the point $(0, y, 2)$ contextually or share a common structure in the $x-z$ plane, they intersect. Specifically, if $b=1$, both lines lie in the plane $x - z + 2 = 0$.

Step 4: Final Answer:
The lines are coplanar if $b=1$.