Question

The sum of the intercepts on the coordinate axes of the plane passing through the point $\left( { - 2, - 2,2} \right)$ and containing the line joining the points $\left( {1, - 1,2} \right)$ and $\left( {1,1,1} \right)$ is?
A) $12$
B) $ - 8$
C) $ - 4$
D) $4$

Answer 1

Hint: In these types of question, firstly we have to convert the equation of plane into standard equation of a plane in intercept form, i.e., \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\], where $a,b,c$ are the intercepts along x, y and z-axis and then find the sum of intercepts, i.e., $a + b + c$. The Cartesian equation of a plane passing through the point $\left( {{x_0},{y_0},{z_0}} \right)$ is given as, $a\left( {x - {x_0}} \right) + b\left( {y - {y_0}} \right) + c\left( {z - {z_0}} \right)$. Using this formula we will find the values of a,b and c. Later we will substitute the values of a, b and c in the equation to find the intercepts. Sum of all the intercepts will give us the final answer.

Complete step-by-step answer:
We know that the Cartesian equation of a plane passing through the point $\left( {{x_0},{y_0},{z_0}} \right)$ is given as, $a\left( {x - {x_0}} \right) + b\left( {y - {y_0}} \right) + c\left( {z - {z_0}} \right)$.
So, the equation of a plane passing through given points $\left( { - 2, - 2,2} \right)$will be $a\left( {x + 2} \right) + b\left( {y + 2} \right) + c\left( {z - 2} \right) = 0$ ….. (1)
This equation containing the line joining the points $\left( {1, - 1,2} \right)$ and $\left( {1,1,1} \right)$. So, these points will satisfies the equation.
Now put $x = 1,y = - 1,z = 2$ in the equation of plane (1),
\[a\left( {1 + 2} \right) + b\left( { - 1 + 2} \right) + c\left( {2 - 2} \right) = 0\]
$ \Rightarrow 3a + b = 0$
$ \Rightarrow 3a = - b$
$ \Rightarrow \dfrac{a}{{ - 1}} = \dfrac{b}{3}$ ….. (2)
Similarly, put $x = 1,y = 1,z = 1$ in the equation of plane (2),
$a\left( {1 + 2} \right) + b\left( {1 + 2} \right) + c\left( {1 - 2} \right) = 0$
$ \Rightarrow 3a + 3b - c = 0$
$ \Rightarrow - b + 3b - c = 0$ $\left[ {\because 3a = - b} \right]$
$ \Rightarrow 2b = c$
$ \Rightarrow \dfrac{b}{1} = \dfrac{c}{2}$
Divide both sides by $3$ to find the value of $\dfrac{b}{3}$ , so we can equate (1) and (2).
$ \Rightarrow \dfrac{b}{3} = \dfrac{c}{6}$ …… (3)
From (2) and (3), we get-
$\dfrac{a}{{ - 1}} = \dfrac{b}{3} = \dfrac{c}{6}$
Let $\dfrac{a}{{ - 1}} = \dfrac{b}{3} = \dfrac{c}{6} = k$
$ \Rightarrow a = - k,b = 3k,c = 6k$
Substitute the values of $a,b,c$ in (1), Equation of plane becomes
$a\left( {x + 2} \right) + b\left( {y + 2} \right) + c\left( {z - 2} \right) = 0$
 $ \Rightarrow $$ - k\left( {x + 2} \right) + 3k\left( {y + 2} \right) + 6k\left( {z - 2} \right) = 0$
$ \Rightarrow $$k\left[ { - \left( {x + 2} \right) + 3\left( {y + 2} \right) + 6\left( {z - 2} \right)} \right] = 0$
$ \Rightarrow $$ - \left( {x + 2} \right) + 3\left( {y + 2} \right) + 6\left( {z - 2} \right) = 0$
$ \Rightarrow $$ - x - 2 + 3y + 6 + 6z - 12 = 0$
$ \Rightarrow $$ - x + 3y + 6z = 8$
Divide both sides by $8$ to convert the equation in intercept form, i.e., \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\], where $a,b,c$ are the intercepts along x, y and z-axis.
$ \Rightarrow $$\dfrac{{ - x}}{8} + \dfrac{{3y}}{8} + \dfrac{{6z}}{8} = 1$
$ \Rightarrow $$\dfrac{x}{{ - 8}} + \dfrac{y}{{\left( {\dfrac{8}{3}} \right)}} + \dfrac{z}{{\left( {\dfrac{4}{3}} \right)}} = 1$
Therefore, intercepts are $ - 8,\dfrac{8}{3},\dfrac{4}{3}$.
Sum of intercepts = $ - 8 + \dfrac{8}{3} + \dfrac{4}{3}$
$ = \dfrac{{ - 8 \times 3 + 8 + 4}}{3}$
$ = \dfrac{{ - 24 + 12}}{3}$
$ = \dfrac{{ - 12}}{3}$
$ = - 4$

Hence, option (C) is the correct answer.

Note: The coordinate plane has two axes: the horizontal and vertical axes. These two axes intersect one another at a point called the origin. A coordinate system is a system that uses one or more numbers, or coordinates, to uniquely determine the position of the points or other geometric elements.