Question

If $x=ay-1=z-2$ and $x=3y-2=bz-2$ lie in the same plane, then the values of $a,b$ are

• A. $a=2,b=3$
• B. $a=1,b=1$
• C. $b=1,a\in R-\left\{0\right\}$
• D. $a=3,b=2$

Answer 1

The correct option is C

$b=1,a\in R-\left\{0\right\}$

Explanation for the correct option.

Step 1. Modify the equations of the line.

The equation of the line $x=ay-1=z-2$ can also be written as: $\frac{x+1}{a}=y=\frac{z-1}{a}$. So the line passes through the point $\left(-1,0,1\right)$ and is parallel along the vector $ai+j+ak$.

Again, the equation of the line $x=3y-2=bz-2$ can also be written as: $\frac{x+2}{3}=y=\frac{z}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$b$}\right.}}$. So the line passes through the point $\left(-2,0,0\right)$ and is parallel along the vector $3i+j+\frac{3}{b}k$.

Step 2. Set the condition of coplanarity and find the values of $a$ and $b$.

Now as the two lines are coplanar, so

$\left|\begin{array}{ccc}a& 1& a\\ 3& 1& \frac{3}{b}\\ -1& 0& -1\end{array}\right|=0$

Now, expand the determinant using the third row.

$$\begin{gathered} -1\left(\frac{3}{b}-a\right)+0-\left(a-3\right)=0 \\ \Rightarrow -\frac{3}{b}+a-a+3=0 \\ \Rightarrow -\frac{3}{b}+3=0 \end{gathered}$$

This will be true when $b=1$ and $a$ can be any real number except $0$.

Thus, $b=1,a\in R-\left\{0\right\}$.

Hence, the correct option is C.