The lines x = ay - 1 = z - 2 and x = 3y - 2 = | vedclass

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(A) First,rewrite the equations of the lines in symmetric form:Line $1$: $x = ay - 1 = z - 2 \Rightarrow \frac{x}{1} = \frac{y - 1/a}{1/a} = \frac{z -

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

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(A) First,rewrite the equations of the lines in symmetric form:Line $1$: $x = ay - 1 = z - 2 \Rightarrow \frac{x}{1} = \frac{y - 1/a}{1/a} = \frac{z - 2}{1}$. This can be written as $\frac{x}{a} = \frac{y - 1/a}{1} = \frac{z - 2}{a}$.Wait,let's rewrite correctly: $x = ay - 1 \Rightarrow y = \frac{x+1}{a}$,and $x = z - 2 \Rightarrow z = x + 2$. So,$\frac{x+1}{a} = y = \frac{z-2}{1}$.Line $2$: $x = 3y - 2 = bz - 2 \Rightarrow y = \frac{x+2}{3}$,and $z = \frac{x+2}{b}$. So,$\frac{x+2}{3} = y = \frac{z-2/b}{1/b} \Rightarrow \frac{x+2}{3} = y = \frac{z-2/b}{1/b}$.Actually,the standard form is $\frac{x-x_1}{l_1} = \frac{y-y_1}{m_1} = \frac{z-z_1}{n_1}$.Line $1$: $x = ay - 1 = z - 2 \Rightarrow \frac{x+1}{a} = y = \frac{z-1}{a}$ is incorrect. Let's use the points and vectors from the image.Line $1$ passes through $P_1(-1, 0, 1)$ with direction vector $\vec{v_1} = a\hat{i} + \hat{j} + a\hat{k}$.Line $2$ passes through $P_2(-2, 0, 0)$ with direction vector $\vec{v_2} = 3\hat{i} + \hat{j} + \frac{3}{b}\hat{k}$.For lines to be coplanar,the scalar triple product of the vector connecting the points and the two direction vectors must be zero: $(\vec{P_2P_1}) \cdot (\vec{v_1} 	imes \vec{v_2}) = 0$.$\vec{P_2P_1} = (-1 - (-2))\hat{i} + (0 - 0)\hat{j} + (1 - 0)\hat{k} = \hat{i} + \hat{k}$.The determinant is:$\left|\begin{array}{ccc} 1 & 0 & 1 \ a & 1 & a \ 3 & 1 & 3/b \end{array}ight| = 0$$1(\frac{3}{b} - a) - 0 + 1(a - 3) = 0$$\frac{3}{b} - a + a - 3 = 0 \Rightarrow \frac{3}{b} = 3 \Rightarrow b = 1$.Since $a$ is in the denominator of the original line equation $x = ay - 1$,$a 
eq 0$. Thus,$b = 1, a \in R - \{0\}$.

The lines $x = ay - 1 = z - 2$ and $x = 3y - 2 = bz - 2$ $(ab \neq 0)$ are coplanar,if:

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