Question

The distance of the point $(1,3, - 7)$ from the plane passing through the point $(1, - 1, - 1)$ having normal perpendicular to both the lines
$\dfrac{{x - 1}}{1} = \dfrac{{y + 2}}{{ - 2}} = \dfrac{{z - 4}}{3}{\text{ and }}\dfrac{{x - 2}}{2} = \dfrac{{y + 1}}{{ - 1}} = \dfrac{{z + 7}}{{ - 1}}{\text{ }}$
A. $\dfrac{{20}}{{\sqrt {74} }}$
B. $\dfrac{{10}}{{\sqrt {83} }}$
C. $\dfrac{5}{{\sqrt {83} }}$
D. $\dfrac{{10}}{{\sqrt {74} }}$

Answer 1

Here we need to know the concept of the plane in the three dimensions. We must know that if we have one point from where the plane is passing for example $({x_1},{y_1},{z_1})$ and we also know direction ratios of the normal which are $\left( {a,b,c} \right)$ then we can write the equation of the plane as the equation:
$a(x - {x_1}) + b(y - {y_1}) + c(z - {z_1}) = 0$
And also we must know that the distance of the point $(p,q,r)$ from the plane $ax + by + cz + d = 0$ is given by:
$\dfrac{{ap + bq + cz + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$
So by using these formulas we can get our required answer.

Complete step-by-step answer:
Here we are given that we need to find the distance of the point $(1,3, - 7)$ from the plane passing through the point $(1, - 1, - 1)$ having normal perpendicular to both the lines
$\dfrac{{x - 1}}{1} = \dfrac{{y + 2}}{{ - 2}} = \dfrac{{z - 4}}{3}{\text{ and }}\dfrac{{x - 2}}{2} = \dfrac{{y + 1}}{{ - 1}} = \dfrac{{z + 7}}{{ - 1}}{\text{ }}$
So firstly we need to find the equation of the plane which is passing through the point $(1, - 1, - 1)$ having normal perpendicular to both the lines
$\dfrac{{x - 1}}{1} = \dfrac{{y + 2}}{{ - 2}} = \dfrac{{z - 4}}{3}{\text{ and }}\dfrac{{x - 2}}{2} = \dfrac{{y + 1}}{{ - 1}} = \dfrac{{z + 7}}{{ - 1}}{\text{ }}$
So basically we need to find the two things which are the direction ratio of the normal of the plane and the point through which it is passing. Passing point is known to us which is $(1, - 1, - 1)$
Now we need to find the direction ratios of the normal of the plane. As we are given that plane normal is perpendicular to the lines $\dfrac{{x - 1}}{1} = \dfrac{{y + 2}}{{ - 2}} = \dfrac{{z - 4}}{3}{\text{ and }}\dfrac{{x - 2}}{2} = \dfrac{{y + 1}}{{ - 1}} = \dfrac{{z + 7}}{{ - 1}}{\text{ }}$
Let the normal of the plane have the direction ratios $(a,b,c)$ so as we are given that this normal is perpendicular to both the lines $\dfrac{{x - 1}}{1} = \dfrac{{y + 2}}{{ - 2}} = \dfrac{{z - 4}}{3}{\text{ and }}\dfrac{{x - 2}}{2} = \dfrac{{y + 1}}{{ - 1}} = \dfrac{{z + 7}}{{ - 1}}{\text{ }}$
Here we must know that the direction ratios of the lines are represented by the denominators which are $(1, - 2,3),(2, - 1, - 1)$ so we can write that
$a - 2b + 3c = 0$
$2a - b - c = 0$
Solving these we get that
$\Rightarrow \dfrac{a}{{( - 2)( - 1) - (3)( - 1)}} = \dfrac{{ - b}}{{1( - 1) - 3(2)}} = \dfrac{c}{{1( - 1) - 2( - 2)}} \\\ \Rightarrow \dfrac{a}{{2 + 3}} = \dfrac{b}{{1 + 6}} = \dfrac{c}{{ - 1 + 4}} \\\ \Rightarrow \dfrac{a}{5} = \dfrac{b}{7} = \dfrac{c}{3} \\\ \Rightarrow a = 5,b = 7,c = 3 \\\$
Here if we have one point from where the plane is passing for example $({x_1},{y_1},{z_1})$ and we also know direction ratios of the normal which are $\left( {a,b,c} \right)$ then we can write the equation of the plane as the equation:
$a(x - {x_1}) + b(y - {y_1}) + c(z - {z_1}) = 0$
So we have the point $(1, - 1, - 1)$
We get the equation of the plane as
$5(x - 1) + 7(y + 1) + 3(z + 1) = 0$
$5x + 7y + 3z + 5 = 0$
And also we must know that the distance of the point $(p,q,r)$ from the plane $ax + by + cz + d = 0$ is given by:
$\dfrac{{ap + bq + cz + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$
So we have the point as $(1,3, - 7)$ and we need to find the distance from the plane $5x + 7y + 3z + 5 = 0$
We get $\dfrac{{5(1) + 7(3) + 3( - 7) + 5}}{{\sqrt {{5^2} + {7^2} + {3^2}} }}$
$= \dfrac{{5 + 21 - 21 + 5}}{{\sqrt {25 + 49 + 9} }} \\\ = \dfrac{{10}}{{\sqrt {83} }} \\\$
Hence we get that option B is correct.

Note: Here we must be aware of the point that the direction ratio which we represent in the equation of the plane are not the direction ratios of the plane but actually they are the direction ratios of the normal of the plane. And also we must know that the distance of the point $(p,q,r)$ from the plane $ax + by + cz + d = 0$ is given by:
$\dfrac{{ap + bq + cz + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$