# 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 \\

$

Last updated date: 24th Mar 2023

•

Total views: 309.3k

•

Views today: 4.86k

Answer

Verified

309.3k+ 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

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?