
The distance of the point $\left( {3,8,2} \right)$ from the line $\dfrac{{x - 1}}{2} = \dfrac{{y - 3}}{4} = \dfrac{{z - 2}}{3}$ measured parallel to the plane $3x + 2y - 2z + 15 = 0$ is
$
\left( a \right)2 \\
\left( b \right)3 \\
\left( c \right)6 \\
\left( d \right)7 \\
$
Answer
513.3k+ views
Hint- Use a family of parallel planes . Equation of all planes parallel to $ax + by + cz + d = 0$ lies on this family $ax + by + cz + e = 0$.
Now, the equation of all planes parallel to $3x + 2y - 2z + 15 = 0$ lies on the family of planes $3x + 2y - 2z + d = 0$.
But we require a unique plane pass through the point $\left( {3,8,2} \right)$. So, we put the point in the family of planes.
$
3 \times 3 + 2 \times 8 - 2 \times 2 + d = 0 \\
\Rightarrow 9 + 16 - 4 + d = 0 \\
\Rightarrow d = - 21 \\
$
So, the required plane is $3x + 2y - 2z - 21 = 0..........\left( 2 \right)$
The required plane also passes through the given line. So, we find an intersection point.
Equation of line $\dfrac{{x - 1}}{2} = \dfrac{{y - 3}}{4} = \dfrac{{z - 2}}{3} = p$
$ \Rightarrow x = 2p + 1,y = 4p + 3,z = 3p + 2$
These points also satisfy the equation of the plane. So,
$
\Rightarrow 3\left( {2p + 1} \right) + 2(4p + 3) - 2(3p + 2) - 21 = 0 \\
\Rightarrow 6p + 3 + 8p + 6 - 6p - 4 - 21 = 0 \\
\Rightarrow 8p = 16 \\
\Rightarrow p = 2 \\
$
Intersection points of the plane and line are $x = 2 \times 2 + 1 = 5$ ,$y = 4 \times 2 + 3 = 11$ , $z = 3 \times 2 + 2 = 8$ .
Coordinate of intersection $\left( {5,11,8} \right)$ .
Distance of point $\left( {3,8,2} \right)$ from the line $\dfrac{{x - 1}}{2} = \dfrac{{y - 3}}{4} = \dfrac{{z - 2}}{3}$ is equal to distance between $\left( {3,8,2} \right)$ and $\left( {5,11,8} \right)$.
Distance $
= \sqrt {{{\left( {5 - 3} \right)}^2} + {{\left( {11 - 8} \right)}^2} + {{\left( {8 - 2} \right)}^2}} \\
\\
$
$
\Rightarrow \sqrt {4 + 9 + 36} \\
\Rightarrow \sqrt {49} = 7 \\
$
So, the correct option is (d).
Note-Whenever we face such types of problems we use some important points. First find the equation of plane passing through the point and parallel to another plane with the help of the family of planes then find the point of intersection of plane and line then using distance formula we get the required answer.
Now, the equation of all planes parallel to $3x + 2y - 2z + 15 = 0$ lies on the family of planes $3x + 2y - 2z + d = 0$.
But we require a unique plane pass through the point $\left( {3,8,2} \right)$. So, we put the point in the family of planes.
$
3 \times 3 + 2 \times 8 - 2 \times 2 + d = 0 \\
\Rightarrow 9 + 16 - 4 + d = 0 \\
\Rightarrow d = - 21 \\
$
So, the required plane is $3x + 2y - 2z - 21 = 0..........\left( 2 \right)$
The required plane also passes through the given line. So, we find an intersection point.
Equation of line $\dfrac{{x - 1}}{2} = \dfrac{{y - 3}}{4} = \dfrac{{z - 2}}{3} = p$
$ \Rightarrow x = 2p + 1,y = 4p + 3,z = 3p + 2$
These points also satisfy the equation of the plane. So,
$
\Rightarrow 3\left( {2p + 1} \right) + 2(4p + 3) - 2(3p + 2) - 21 = 0 \\
\Rightarrow 6p + 3 + 8p + 6 - 6p - 4 - 21 = 0 \\
\Rightarrow 8p = 16 \\
\Rightarrow p = 2 \\
$
Intersection points of the plane and line are $x = 2 \times 2 + 1 = 5$ ,$y = 4 \times 2 + 3 = 11$ , $z = 3 \times 2 + 2 = 8$ .
Coordinate of intersection $\left( {5,11,8} \right)$ .
Distance of point $\left( {3,8,2} \right)$ from the line $\dfrac{{x - 1}}{2} = \dfrac{{y - 3}}{4} = \dfrac{{z - 2}}{3}$ is equal to distance between $\left( {3,8,2} \right)$ and $\left( {5,11,8} \right)$.
Distance $
= \sqrt {{{\left( {5 - 3} \right)}^2} + {{\left( {11 - 8} \right)}^2} + {{\left( {8 - 2} \right)}^2}} \\
\\
$
$
\Rightarrow \sqrt {4 + 9 + 36} \\
\Rightarrow \sqrt {49} = 7 \\
$
So, the correct option is (d).
Note-Whenever we face such types of problems we use some important points. First find the equation of plane passing through the point and parallel to another plane with the help of the family of planes then find the point of intersection of plane and line then using distance formula we get the required answer.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

What is a transformer Explain the principle construction class 12 physics CBSE

What are the major means of transport Explain each class 12 social science CBSE
