Question

The equation of the line passing through $\left( { - 4,3,1} \right)$ parallel to the plane $x + 2y - z - 5 = 0$ and intersecting the line $\dfrac{{x + 1}}{{ - 3}} = \dfrac{{y - 3}}{2} = \dfrac{{z - 2}}{{ - 1}}$:
A. $\dfrac{{x + 4}}{{ - 1}} = \dfrac{{y - 3}}{1} = \dfrac{{z - 1}}{1}$
B. $\dfrac{{x + 4}}{3} = \dfrac{{y - 3}}{{ - 1}} = \dfrac{{z - 1}}{1}$
C. $\dfrac{{x + 4}}{{ - 1}} = \dfrac{{y - 3}}{1} = \dfrac{{z - 1}}{3}$
D. $\dfrac{{x - 4}}{2} = \dfrac{{y + 3}}{1} = \dfrac{{z + 1}}{4}$

Answer 1

Hint: We will first write the equation of the line by letting the direction ratios as $\left( {a,b,c} \right)$ for the required line and then using the given conditions to form relations in $a,b$ and $c$. We will use the properties of intersecting lines and planes parallel to line to make equations.

Complete step-by-step answer:
We will first find the direction ratios of the required line.
Let the direction ratios of the required line be $\left( {a,b,c} \right)$.
We are given that the line is parallel to the plane $x + 2y - z - 5 = 0$
The direction ratios of $x + 2y - z - 5 = 0$ are $\left( {1,2, - 1} \right)$
Therefore, we have, $a + 2b - c = 0$
If a line is passing through points $\left( {{x_1},{y_1},{z_1}} \right)$ and having direction ratios, $\left( {p,q,r} \right)$, then the equation of line is given by,
$\dfrac{{x - {x_1}}}{p} = \dfrac{{y - {y_1}}}{q} = \dfrac{{z - {z_1}}}{r}$
Hence, for the line passing through $\left( { - 4,3,1} \right)$ with direction ratios $\left( {a,b,c} \right)$ is:
$\dfrac{{x + 4}}{a} = \dfrac{{y - 3}}{b} = \dfrac{{z - 1}}{c}$
Also, we have the line $\dfrac{{x + 1}}{{ - 3}} = \dfrac{{y - 3}}{2} = \dfrac{{z - 2}}{{ - 1}}$ intersects the line $\dfrac{{x + 4}}{a} = \dfrac{{y - 3}}{b} = \dfrac{{z - 1}}{c}$
Therefore, the determinant of the direction ratios of $\dfrac{{x + 1}}{{ - 3}} = \dfrac{{y - 3}}{2} = \dfrac{{z - 2}}{{ - 1}}$ and $\dfrac{{x + 4}}{a} = \dfrac{{y - 3}}{b} = \dfrac{{z - 1}}{c}$ is 0.
That is ,

{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}} \\\ {{a_1}}&{{b_1}}&{{c_1}} \\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = 0$$ We have, $$\left| {\begin{array}{*{20}{c}} 3&0&1 \\\ { - 3}&2&{ - 1} \\\ a&b;&c; \end{array}} \right| = 0$$ On solving it further, we have, $ 3\left( {2c + b} \right) - 0 + 1\left( { - 3b - 2a} \right) = 0 \\\ 6c + 3b - 3b - 2a = 0 \\\ 6c = 2a \\\ a = 3c \\\ $ On substituting the above condition in equation (1), we get, $ 3c + 2b - c = 0 \\\ 2c + 2b = 0 \\\ c + b = 0 \\\ c = - b \\\ $ Hence, the ratio of $a:b:c$ is $ 3c: - c:c \\\ 3: - 1:1 \\\ \ $ Therefore, the equation of line is, $ \dfrac{{x - \left( { - 4} \right)}}{3} = \dfrac{{y - 3}}{{ - 1}} = \dfrac{{z - 1}}{1} \\\ \dfrac{{x + 4}}{3} = \dfrac{{y - 3}}{{ - 1}} = \dfrac{{z - 1}}{1} \\\ $ Hence, option B is correct. Note: While solving the determinant, take the sign of the coordinates into consideration to avoid chances of errors. The direction ratio of parallel lines is the same and the direction ratios of intersecting lines have determinant equals to 0.