
If the plane \[x - 3y + 5z = d\] passes through the point \[\left( {1,2,4} \right)\], then find the lengths of intercepts cut by it on the axes of \[x,y,z\].
A. \[15, - 5,3\]
B. \[1, - 5,3\]
C. \[ - 15,5, - 3\]
D. \[1, - 6,20\]
Answer
162k+ views
Hint: Here, an equation of the plane passing through the point \[\left( {1,2,4} \right)\] is given. First, substitute the co-ordinates in the equation of the plane and calculate the value of \[d\]. Then, convert the equation of the plane in the intercept form by dividing it by the value of \[d\] and get the required answer.
Formula Used:The equation of the plane whose intercepts are \[a,b,c\] is: \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]
Complete step by step solution:Given:
The plane \[x - 3y + 5z = d\] passes through the point \[\left( {1,2,4} \right)\].
Since the plane passes through the point \[\left( {1,2,4} \right)\]. So, the point satisfies the equation of plane.
Substitute the co-ordinates of the point in the equation \[x - 3y + 5z = d\].
We get,
\[1 - 3\left( 2 \right) + 5\left( 4 \right) = d\]
\[ \Rightarrow 1 - 6 + 20 = d\]
\[ \Rightarrow d = 15\]
Now substitute the value \[d = 15\] in the equation of the plane.
\[x - 3y + 5z = 15\]
We know that the intercept form of the plane is \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\], where \[a,b,c\] are the intercepts.
We have,
\[x - 3y + 5z = 15\]
To convert it into the intercept form, divide both sides by \[15\].
\[\dfrac{x}{{15}} - \dfrac{{3y}}{{15}} + \dfrac{{5z}}{{15}} = \dfrac{{15}}{{15}}\]
\[ \Rightarrow \dfrac{x}{{15}} - \dfrac{y}{5} + \dfrac{z}{3} = 1\]
\[ \Rightarrow \dfrac{x}{{15}} + \dfrac{y}{{ - 5}} + \dfrac{z}{3} = 1\]
Thus, the lengths of intercepts cut by it on the axes of \[x,y,z\] are \[15, - 5,3\].
Option ‘A’ is correct
Note: Remember the following forms of the equation of a plane:
General equation of a plane: \[ax + by + cz + d = 0\] , where \[d \ne 0\]
Equation of plane whose intercepts are \[a,b,c\] is: \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]
Equation of a plane passing through the point \[\left( {{x_1},{y_1},{z_1}} \right)\] with direction ratios \[\left( {a,b,c} \right)\]: \[a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\]
Formula Used:The equation of the plane whose intercepts are \[a,b,c\] is: \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]
Complete step by step solution:Given:
The plane \[x - 3y + 5z = d\] passes through the point \[\left( {1,2,4} \right)\].
Since the plane passes through the point \[\left( {1,2,4} \right)\]. So, the point satisfies the equation of plane.
Substitute the co-ordinates of the point in the equation \[x - 3y + 5z = d\].
We get,
\[1 - 3\left( 2 \right) + 5\left( 4 \right) = d\]
\[ \Rightarrow 1 - 6 + 20 = d\]
\[ \Rightarrow d = 15\]
Now substitute the value \[d = 15\] in the equation of the plane.
\[x - 3y + 5z = 15\]
We know that the intercept form of the plane is \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\], where \[a,b,c\] are the intercepts.
We have,
\[x - 3y + 5z = 15\]
To convert it into the intercept form, divide both sides by \[15\].
\[\dfrac{x}{{15}} - \dfrac{{3y}}{{15}} + \dfrac{{5z}}{{15}} = \dfrac{{15}}{{15}}\]
\[ \Rightarrow \dfrac{x}{{15}} - \dfrac{y}{5} + \dfrac{z}{3} = 1\]
\[ \Rightarrow \dfrac{x}{{15}} + \dfrac{y}{{ - 5}} + \dfrac{z}{3} = 1\]
Thus, the lengths of intercepts cut by it on the axes of \[x,y,z\] are \[15, - 5,3\].
Option ‘A’ is correct
Note: Remember the following forms of the equation of a plane:
General equation of a plane: \[ax + by + cz + d = 0\] , where \[d \ne 0\]
Equation of plane whose intercepts are \[a,b,c\] is: \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]
Equation of a plane passing through the point \[\left( {{x_1},{y_1},{z_1}} \right)\] with direction ratios \[\left( {a,b,c} \right)\]: \[a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\]
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