Question

Find the equation of the plane which is parallel to y-axis and cuts off intercepts of length 2 and 3 from x-axis and z-axis.
A. \[3x + 2z = 1\]
B. \[3x + 2z = 6\]
C. \[2x + 3z = 6\]
D. \[3x + 2z = 0\]

Answer 1

Hint: First, consider the equation of plane as \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]. Then, apply the given condition (parallel to y-axis) and simplify the equation of the plane. After that, substitute the values of the intercepts in the equation of plane and solve it to get the required answer.



Formula Used:The equation of plane: \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]



Complete step by step solution:Given:
The plane is parallel to the y-axis.
The x-intercept: 2 units
The z-intercept: 3 units

Let consider,
The general equation of a plane is \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\].
It is given that the plane is parallel to the y-axis.
So, substitute \[y = 0\] in the equation of plane.
We get,
\[\dfrac{x}{a} + \dfrac{0}{b} + \dfrac{z}{c} = 1\]
\[ \Rightarrow \dfrac{x}{a} + \dfrac{z}{c} = 1\] \[.....\left( 1 \right)\]
We have, x-intercept: \[a = 2\] and z-intercept: \[c = 3\].
Substitute these values in the equation \[\left( 1 \right)\].
\[\dfrac{x}{2} + \dfrac{z}{3} = 1\]
\[ \Rightarrow \dfrac{{2x + 3z}}{{2 \times 3}} = 1\]
\[ \Rightarrow \dfrac{{2x + 3z}}{6} = 1\]
\[ \Rightarrow 2x + 3z = 6\]
Thus, the equation of the plane is \[2x + 3z = 6\].



Option ‘B’ is correct



Note: Remember the following equations of the plane:
If the plane is parallel to the x-axis: \[by + cz = d\], because \[x = 0\]
If the plane is parallel to the y-axis: \[ax + cz = d\], because \[y = 0\]
If the plane is parallel to the z-axis: \[ax + by = d\], because \[z = 0\]