Question

The plane containing the line $\dfrac{{x - 1}}{1} = \dfrac{{y - 2}}{2} = \dfrac{{z - 3}}{3}$ and parallel to the line $\dfrac{x}{1} = \dfrac{y}{1} = \dfrac{z}{4}$ passes through the point :
(A) $( - 1, - 2,0)$
(B) $(1,0,5)$
(C) $(1, - 2,5)$
(D) $(0,3, - 5)$

Answer 1

Since the required plane is parallel to both the given lines, the normal to the plane will be perpendicular to both the lines. Also note that the given plane passes through the point $(1,2,3)$.

Formula used: We get two equations from two points $(1,2,3)$ and $(1,1,4)$. We can solve these equations by cross multiplication method. Let ${A_1}x + {B_1}y + {C_1} = 0$ and ${A_2}x + {B_2}y + {C_2} = 0$ are two equations. These can solve by cross-multiplication method as
$\dfrac{x}{{{B_1}{C_2} - {B_2}{C_1}}} = \dfrac{y}{{{C_1}{A_2} - {C_2}{A_1}}} = \dfrac{1}{{{B_2}{A_1} - {B_1}{A_2}}}$
For calculating values of x and y, the denominator should not be zero.
After solving for x and y, we calculate coordinates of the point through which the plane is passing.

Complete step-by-step answer:
Let the direction ratios of the normal to the plane be $a,b,c$.
Since this normal is perpendicular to both the given lines, we have :
$\Rightarrow a + 2b + 3c = 0 \\\ \Rightarrow a + b + 4c = 0$
Applying cross – multiplication we get :
$\Rightarrow \dfrac{a}{{8 - 3}} = \dfrac{b}{{3 - 4}} = \dfrac{c}{{1 - 2}} \\\ \Rightarrow \dfrac{a}{5} = \dfrac{b}{{ - 1}} = \dfrac{c}{{ - 1}} = k(let) \\\ \Rightarrow a = 5k;b = - k;c = - k \to (1)$
Since the plane passes through the point $(1,2,3)$ let the equation of the plane be :
$\Rightarrow$ $a(x - 1) + b(y - 2) + c(z - 3) = 0$
Substituting the values of a , b , c from (1) , the equation of the plane will be :
$\Rightarrow$ $5(x - 1) - (y - 2) - (z - 3) = 0$
Now substituting the values of x , y , z of each point given in the options.
For $( - 1, - 2,0)$ we have $5( - 1 - 1) - ( - 2 - 2) - (0 - 3) = - 10 + 4 + 3 = - 3 \ne 0$
For $(1,0,5)$ we have $5(1-1)-(0-2)-(5-3)=2-2=0$
Hence the point $(1,0,5)$ lies on the plane.

Thus (B) is the correct option.

Note: The equation of a plane passing through the point $(p,q,r)$ is given by :
$\dfrac{{x - p}}{a} = \dfrac{{y - q}}{b} = \dfrac{{z - r}}{c}$ where , a , b , c are the direction ratios of the normal to the plane. Students make sure about using accurate formulas for solving equations and find the coordinates of points.