Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 3 - Three Dimensional Coordinate System>Three Dimensional Coordinate System Exercise 8>Q 224

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JEE Advanced

IMPORTANT

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Consider the lines $L_{1} : \frac{x - 1}{2} = \frac{y}{- 1} = \frac{z + 3}{1}, L_{2} : \frac{x - 4}{1} = \frac{y + 3}{1} = \frac{z + 3}{2}$ and the planes $P_{1} : 7 x + y + 2 z = 3, P_{2} : 3 x + 5 y - 6 z = 4$ . Let $a x + b y + c z = d$ be the equation of the plane passing through the point of intersection of lines $L_{2}$ and $L_{1}$ perpendicular to planes $P_{1}$ and $P_{2}$ .

Match list $I$ with list $II$ and select the correct answer using the code given below the lists.

List $I$ List $II$

$P .$ $a =$ $1 .$ $13$

$Q .$ $b =$ $2 .$ $- 3$

$R .$ $c =$ $3 .$ $1$

$S .$ $d =$ $4 .$ $- 2$

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Important Questions on Three Dimensional Coordinate System

MEDIUM

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 3 - Three Dimensional Coordinate System>Three Dimensional Coordinate System Exercise 8>Q 225

If the straight lines

\frac{x - 1}{2} = \frac{y + 1}{k} = \frac{z}{2}

and

\frac{x + 1}{5} = \frac{y + 1}{2} = \frac{z}{k}

are coplanar, then plane

(s)

containing these two lines is/are

HARD

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 3 - Three Dimensional Coordinate System>Three Dimensional Coordinate System Exercise 8>Q 226

If the distance between the plane

A x - 2 y + z = d

and the plane containing the lines

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}

and

\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}

is

\sqrt{6}

, then

|d|

is equal to

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JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 3 - Three Dimensional Coordinate System>Three Dimensional Coordinate System Exercise 8>Q 233

If the image of the point

P (1, - 2, 3)

in the plane

2 x + 3 y - 4 z + 22 = 0

measured parallel to the line

\frac{x}{1} = \frac{y}{4} = \frac{z}{5}

is

Q

, then

P Q

is equal to

HARD

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 3 - Three Dimensional Coordinate System>Three Dimensional Coordinate System Exercise 8>Q 234

The distance of the point

(1, 3, - 7)

from the plane passing through the point

(1, - 1, - 1)

having normal perpendicular to both the lines

\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}

and

\frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}

is

MEDIUM

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 3 - Three Dimensional Coordinate System>Three Dimensional Coordinate System Exercise 8>Q 235

The distance (in units) of the point

(1, - 5, 9)

from plane

x - y + z = 5

measured along the line

x = y = z

is

HARD

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 3 - Three Dimensional Coordinate System>Three Dimensional Coordinate System Exercise 8>Q 236

If the line,

\frac{x - 3}{2} = \frac{y + 2}{- 1} = \frac{z + 4}{3}

lies in the plane

l x + m y - z = 9,

then

l^{2} + m^{2}

is equal to

MEDIUM

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 3 - Three Dimensional Coordinate System>Three Dimensional Coordinate System Exercise 8>Q 237

The distance of the point

(1, 0, 2)

from the point of intersection of the line

\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}

and the plane

x - y + z = 16

, is

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JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 3 - Three Dimensional Coordinate System>Three Dimensional Coordinate System Exercise 8>Q 238

The equation of the plane containing the line

2 x - 5 y + z = 3, x + y + 4 z = 5

and parallel to the plane

x + 3 y + 6 z = 1

is