
If for a plane, the intercepts on the coordinate axes are 8, 4, 4. Then find the length of the perpendicular from the origin on to the plane.
A. \[\dfrac{8}{3}\]
B. \[\dfrac{3}{8}\]
C. 3
D. \[\dfrac{4}{3}\]
E. \[\dfrac{4}{5}\]
Answer
162.6k+ views
Hint: Here, the intercepts of the plane on the coordinate axes are given. First, by using the intercepts calculate the equation of the plane. Then, calculate the distance between the origin and the plane to get the required answer.
Formula used: The equation of the plane whose intercepts are \[a,b,c\] is: \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]
The smallest distance between a point \[\left( {{x_1},{y_1},{z_1}} \right)\] and a plane \[ax + by + cz = d\] is: \[D = \left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\]
Complete step by step solution: Given:
The intercepts of the plane on the coordinate axes are: 8, 4, 4
Let’s calculate the equation of the plane in the intercept form.
8, 4, 4 are the intercepts of the plane.
So, the equation of the plane is:
\[\dfrac{x}{8} + \dfrac{y}{4} + \dfrac{z}{4} = 1\]
\[ \Rightarrow \dfrac{x}{8} + \dfrac{{y \times 2}}{{4 \times 2}} + \dfrac{{z \times 2}}{{4 \times 2}} = 1\]
\[ \Rightarrow \dfrac{x}{8} + \dfrac{{2y}}{8} + \dfrac{{2z}}{8} = 1\]
\[ \Rightarrow \dfrac{{x + 2y + 2z}}{8} = 1\]
\[ \Rightarrow x + 2y + 2z = 8\]
Now calculate the distance between the plane \[x + 2y + 2z = 8\] and the origin \[\left( {0,0,0} \right)\].
So, use the formula of the smallest distance between a plane and a point \[D = \left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\].
Here, \[\left( {{x_1},{y_1},{z_1}} \right) = \left( {0,0,0} \right)\] and the equation of plane is \[x + 2y + 2z = 8\].
Substitute the values in the formula. We get
\[D = \left| {\dfrac{{\left( 1 \right)\left( 0 \right) + \left( 2 \right)\left( 0 \right) + \left( 2 \right)\left( 0 \right) - 8}}{{\sqrt {{1^2} + {2^2} + {2^2}} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{ - 8}}{{\sqrt {1 + 4 + 4} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{ - 8}}{{\sqrt 9 }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{ - 8}}{3}} \right|\]
\[ \Rightarrow D = \dfrac{8}{3}\]
Thus, the distance between the origin and the plane is \[\dfrac{8}{3}\].
Thus, Option (A) is correct.
Note: Sometimes students use the formula of the smallest distance between a plane and a point as \[D = \dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]. In this type of question, they will get only one value or the wrong value. So, always remember to apply the modulus function.
Formula used: The equation of the plane whose intercepts are \[a,b,c\] is: \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]
The smallest distance between a point \[\left( {{x_1},{y_1},{z_1}} \right)\] and a plane \[ax + by + cz = d\] is: \[D = \left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\]
Complete step by step solution: Given:
The intercepts of the plane on the coordinate axes are: 8, 4, 4
Let’s calculate the equation of the plane in the intercept form.
8, 4, 4 are the intercepts of the plane.
So, the equation of the plane is:
\[\dfrac{x}{8} + \dfrac{y}{4} + \dfrac{z}{4} = 1\]
\[ \Rightarrow \dfrac{x}{8} + \dfrac{{y \times 2}}{{4 \times 2}} + \dfrac{{z \times 2}}{{4 \times 2}} = 1\]
\[ \Rightarrow \dfrac{x}{8} + \dfrac{{2y}}{8} + \dfrac{{2z}}{8} = 1\]
\[ \Rightarrow \dfrac{{x + 2y + 2z}}{8} = 1\]
\[ \Rightarrow x + 2y + 2z = 8\]
Now calculate the distance between the plane \[x + 2y + 2z = 8\] and the origin \[\left( {0,0,0} \right)\].
So, use the formula of the smallest distance between a plane and a point \[D = \left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\].
Here, \[\left( {{x_1},{y_1},{z_1}} \right) = \left( {0,0,0} \right)\] and the equation of plane is \[x + 2y + 2z = 8\].
Substitute the values in the formula. We get
\[D = \left| {\dfrac{{\left( 1 \right)\left( 0 \right) + \left( 2 \right)\left( 0 \right) + \left( 2 \right)\left( 0 \right) - 8}}{{\sqrt {{1^2} + {2^2} + {2^2}} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{ - 8}}{{\sqrt {1 + 4 + 4} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{ - 8}}{{\sqrt 9 }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{ - 8}}{3}} \right|\]
\[ \Rightarrow D = \dfrac{8}{3}\]
Thus, the distance between the origin and the plane is \[\dfrac{8}{3}\].
Thus, Option (A) is correct.
Note: Sometimes students use the formula of the smallest distance between a plane and a point as \[D = \dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]. In this type of question, they will get only one value or the wrong value. So, always remember to apply the modulus function.
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