Find the point on the line x + 2 3 = y + 1 2 = z - 3 2 at a distance of 3 2 from the point 1 , 2 , 3

The coordinates of any point on the line x + 2 3 = y + 1 2 = z - 3 2 = λ are given by 3 λ - 2 , 2 λ - 1 , 2 λ + 3 The distance between the points x 1 , y 1 , z 1 and x 2 , y 2 , z 2 is x 1 - x 2 2 + y 1 - y 2 2 + z 1 - z 2 2 Given , the distance between the above point and 1 , 2 , 3 is 3 2 ∴ ( 3 λ - 2 - 1 ) 2 + ( 2 λ - 1 - 2 ) 2 + ( 2 λ + 3 - 3 ) 2 = 3 2 ⇒ 3 λ - 3 2 + 2 λ - 3 2 + 2 λ 2 = 9 × 2 ⇒ 9 λ 2 - 18 λ + 9 + 4 λ 2 - 12 λ + 9 + 4 λ 2 = 18 ⇒ 17 λ 2 - 30 λ = 0 ⇒ λ = 30 17 , 0 ∴ Required points are 0 - 2 , 0 - 1 , 0 + 3 = - 2 , - 1 , 3 or 3 × 30 17 - 2 , 2 × 30 17 - 1 , 2 × 30 17 + 3 = 56 17 , 43 17 , 111 17 .

A line passes through 2 , - 1 , 3 and is perpendicular to the lines r → = i ^ + j ^ - k ^ + λ 2 i ^ - 2 j ^ + k ^ and r → = 2 i ^ - j ^ - 3 k ^ + μ i ^ + 2 j ^ + 2 k ^ obtain its equation.

The directions of the given lines are b → 1 = 2 i ^ - 2 i ^ + k ^ and b → 2 = i ^ + 2 j ^ + 2 k ^ . The required line is perpendicular to the lines which are parallel to vectors b → 1 = 2 i ^ - 2 i ^ + k ^ and b → 2 = i ^ + 2 j ^ + 2 k ^ , respectively. So, b → = b → 1 × b → 2 . Now, b → = b → 1 × b → 2 = i ^ j ^ k ^ 2 - 2 1 1 2 2 ⇒ b → = i ^ - 4 - 2 - j ^ 4 - 1 + k ^ 4 + 2 ⇒ b → = - 6 i ^ - 3 j ^ + 6 k ^ Thus, the required line passes through the point 2 , - 1 , 3 or a → = 2 i ^ - j ^ + 3 k ^ and is parallel to the vector b → = - 6 i ^ - 3 j ^ + 6 k ^ . So, its vector equation is r → = 2 i ^ - j ^ + 3 k ^ + λ - 6 i ^ - 3 j ^ + 6 k ^ or r → = 2 i ^ - j ^ + 3 k ^ + μ 2 i ^ + j ^ - 2 k ^ , where μ = - 3 λ .

Prove that the distance of the point of intersection of the line x – 2 3 = y + 1 4 = z – 2 12 and the plane x - y + z = 5 from the point - 1 , - 5 , - 10 is 13 .

Let x – 2 3 = y + 1 4 = z – 2 12 = r Any point on the line is 3 r + 2 , 4 r - 1 , 12 r + 2 . If it lies on the plane x - y + z = 5 , then 3 r + 2 - 4 r - 1 + 12 r + 2 = 5 . ⇒ r = 0 Now, putting the value of r , we are getting. Point of intersection as 2 , - 1 , 2 . Its distance from - 1 , - 5 , - 10 is 2 + 1 2 + – 1 + 5 2 + 2 + 10 2 . = 9 + 16 + 144 = 169 = 13

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A variable plane which remains at a constant distance $p$ from the origin $(0)$ cuts the coordinate axes in $A, B, C$

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Important Questions on Three Dimensional Coordinate System

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 7>Q 208

Find the angle between the lines whose direction cosines has the relation

l + m + n = 0

and

2 l^{2} + 2 m^{2} - n^{2} = 0

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 7>Q 209

Prove that the two lines whose direction cosines are connected by the two relations

a l + b m + c n = 0

and

u l^{2} + v m^{2} + w n^{2} = 0

are perpendicular if

a^{2} (v + w) + b^{2} (w + u) + c^{2} (u + v) = 0

and parallel if

\frac{a^{2}}{u} + \frac{b^{2}}{v} + \frac{c^{2}}{w} = 0

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 7>Q 210

Find the point on the line

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

at a distance of

3 \sqrt{2}

from the point

(1, 2, 3)

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 7>Q 211

A line passes through

(2, - 1, 3)

and is perpendicular to the lines

\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + λ (2 \hat{i} - 2 \hat{j} + \hat{k})

and

\vec{r} = (2 \hat{i} - \hat{j} - 3 \hat{k}) + μ (\hat{i} + 2 \hat{j} + 2 \hat{k})

obtain its equation.

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 7>Q 212

Find the equations of the two lines through the origin which intersect the line

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

at angle of

\frac{π}{3}

each

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 7>Q 213

Vertices

B

and

C

Δ A B C

lie along the line

\frac{x + 2}{2} = \frac{y - 1}{1} = \frac{z - 0}{4} .

Find the area of the triangle given that

A

has coordinates

(1, - 1, 2)

and line segment

B C

has length

5

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 7>Q 214

A point

P

moves on the plane

\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1

which is fixed. The plane through

P

perpendicular to

O P

meets the axes in

A, B & C

. The planes through

A, B, C

parallel to

y z, z x, x y

planes intersect in

Q

. Prove that if the axis is rectangular, then the locus of

Q

\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} = \frac{1}{a x} + \frac{1}{b y} + \frac{1}{c z}

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 7>Q 215

Prove that the distance of 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 = 5

from the point

(- 1, - 5, - 10)

13

A variable plane which remains at a constant distance p from the origin 0 cuts the coordinate axes in A,B,C

Important Questions on Three Dimensional Coordinate System

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A variable plane which remains at a constant distance $p$ from the origin $(0)$ cuts the coordinate axes in $A, B, C$