If $A\left( {6, - 3} \right),B\left( { - 3,5} \right),C\left( {4, - 2} \right),P\left( {{\rm{\alpha }},{\rm{\beta }}} \right),$ then the ratio of the areas of the triangles $PBC,ABC$ is

$\left| {{\rm{\alpha }} + {\rm{\beta }} - 2} \right|$

$\left| {{\rm{\alpha }} + {\rm{\beta }}} \right|$

$\left| {{\rm{\alpha }} - {\rm{\beta }}} \right|$

$\left| {{\rm{\alpha }} + {\rm{\beta }} + 2} \right|$

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JEE • MATHEMATICS • 3D GEOMETRY SYSTEM

3d geometry system practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 140 questions

1 Easy

1 of 140

Application +4 | -1

The mid point of the line joining the points

\left( { - 10,\;8} \right)

and

\left( { - 6,\;12} \right)

divides the line joining the points

\left( {4,\; - 2} \right)

and

\left( { - 2,\;4} \right)

in the ratio

View answer and explanation

Correct option: D

2:1

externally

Option D (

2:1

externally) is correct because the midpoint divides the line externally in the ratio 2:1, as calculated using the section formula.

2 Medium

2 of 140

Application +4 | -1

The equation of the plane perpendicular to the

y z

-plane and passing through the points

(1,-2,4)

and

(3,-4

5)

View answer and explanation

Correct option: C

y+2 z=6

Option D (

2y+z=6

) is incorrect; the correct answer is

y + 2z = 6

because it satisfies the conditions of being perpendicular to the yz-plane and passing through the given points.

3 Easy

3 of 140

Application +4 | -1

The value of

p

for which the lines

2 x+y-3=0

3 x-y-2=0

and

x-p y+5=0

may intersect at a point is

View answer and explanation

Correct option: D

Option D (6) is correct because the lines intersect at a single point only when p=6, which is found by solving the system of equations.

4 Easy

4 of 140

Analyzing +4 | -1

If the vertices of a triangle at

O\left( {0,0} \right),A\left( {a,0} \right)

and

B\left( {0,a} \right).

Then, the distance between its circumcentre and orthocentre is

View answer and explanation

Correct option: B

\frac{a}{{\sqrt 2 }}

Option B (

\frac{a}{{\sqrt 2 }}

) is correct because the orthocenter is at the origin and the circumcenter is the midpoint of the hypotenuse, allowing us to calculate the distance between them.

5 Easy

5 of 140

Application +4 | -1

The coordinates of the foot of the perpendicular drawn from the point

A(1,0,3)

to the join of the points

B(4,7,1)

and

C(3,5,3)

are

View answer and explanation

Correct option: A

\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)

Option A (

\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)

) is correct because it represents the coordinates of the foot of the perpendicular calculated using section formula and dot product.

6 Easy

6 of 140

Analyzing +4 | -1

The area of the triangle formed by the three lines

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

and

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

View answer and explanation

Correct option: A

\frac{3 \sqrt{3}}{2}

Option A (

\frac{3 \sqrt{3}}{2}

) is correct because the area of the triangle formed by the given lines is calculated using the coordinates of the vertices.

7 Medium

7 of 140

Application +4 | -1

\mathrm{A}(1,-1,2), \mathrm{B}(5,7,-6), \mathrm{C}(3,4,-10)

and

\mathrm{D}(-1,-4,-2)

are the vertices of a quadrilateral

\mathrm{ABCD},

then its area is:

View answer and explanation

Correct option: A

12 \sqrt{29}

Option A (

12 \sqrt{29}

) is correct. The area of the quadrilateral is found by dividing it into two triangles and using the cross product to find the area of each triangle.

8 Medium

8 of 140

Application +4 | -1

The equation of the plane through the point

(-1,3,2)

and perpendicular to the planes

x+2 y+2 z=5

and

3 x+3 y+2 z=8

, is

View answer and explanation

Correct option: A

2 x-4 y+3 z+8=0

Option A (

2x-4y+3z+8=0

) is correct because it is the equation of the plane passing through the given point and perpendicular to the two given planes.

9 Easy

9 of 140

Application +4 | -1

The direction cosines of the line which is perpendicular to the lines whose direction cosines are proportional to

(1,-1,2)

and

(2,1,-1)

, are

View answer and explanation

Correct option: B

-\frac{1}{\sqrt{35}}, \frac{5}{\sqrt{35}}, \frac{3}{\sqrt{35}}

Option B (

-\frac{{1}}{{\sqrt{{35}}}}, \frac{{5}}{{\sqrt{{35}}}}, \frac{{3}}{{\sqrt{{35}}}}

) is correct because it represents the direction cosines of a line perpendicular to the two given lines.

10 Medium

10 of 140

Application +4 | -1

If a parallelopiped is formed by planes drawn through the points

(2,3,5)

and

(5,9,7)

parallel to the coordinate plane, then the length of edges and diagonal of parallelopiped are

View answer and explanation

Correct option: B

(3,6,2), 7

units

Option B ((3,6,2), 7 units) is correct because the lengths of the edges are the differences in coordinates, and the length of the diagonal is

\sqrt{{3^2 + 6^2 + 2^2}} = 7

11 Easy

11 of 140

Application +4 | -1

Equation of the line passing through

(1,1,1)

and parallel to the plane

2 x+3 y+z+5=0

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because the direction ratios of the line are perpendicular to the normal vector of the plane.

12 Easy

12 of 140

Application +4 | -1

If the origin is the centroid of the triangle with vertices

{A}(2 {a}, 2,6), {B}(-4,3 {b},-10)

and

{C}(8,14,2 {c})

, then the sum of value of

a

and

c

View answer and explanation

Correct option: A

Option A (0) is correct because the sum of a and c is 0, calculated using the centroid formula.

13 Medium

13 of 140

Analyzing +4 | -1

The equation

\sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 1} \right)}^2}} + \sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {y - 4} \right)}^2}} = 5

represents

View answer and explanation

Correct option: C

Line segment

The correct answer is C (Line segment) because the sum of distances from a point to two fixed points equals the distance between the fixed points, which is the definition of a line segment.

14 Easy

14 of 140

Application +4 | -1

The locus of a point which moves so that the difference of the squares of its distances from two given points is constant, is a

View answer and explanation

Correct option: B

plane

The correct answer is B (plane) because the locus results in a linear equation in terms of the position vector.

15 Easy

15 of 140

Application +4 | -1

The line passing through the points

(5,1 a)

, and

(3, b, 1)

crosses the

Y Z

-plane at the point

\left(0, \frac{17}{2}, \frac{-13}{2}\right)

. Then,

View answer and explanation

Correct option: D

a=6, b=4

Option D (

a=6, b=4

) is correct because these are the values of a and b for which the line passing through the given points crosses the YZ-plane at the specified point, found by setting up the equation of the line and solving for a and b.

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16 Easy

16 of 140

Application +4 | -1

Ratio in which the

z x

-plane divides the join of

(1,2,3)

and

(4,2,1)

View answer and explanation

Correct option: B

1: 1

externally

Option B is correct because the zx-plane divides the join externally in the ratio found by setting the y-coordinate of the division point to zero.

17 Easy

17 of 140

Application +4 | -1

The ratio in which the plane

{r} \cdot(\hat{i}-2 \hat{j}+2 \hat{k})=17

divides the line joining the points

-2 \hat{i}+4 \hat{j}+7 \hat{k}

and

3 \hat{i}-5 \hat{j}+8 \hat{k}

View answer and explanation

Correct option: C

3: 10

Option C (

3: 10

) is correct because it represents the ratio in which the plane divides the line segment, calculated using the section formula and the plane's equation.

18 Easy

18 of 140

Application +4 | -1

Point

\left( {\frac{1}{2},\; - \frac{{13}}{4}} \right)

divides the line joining the points

\left( {3,\; - 5} \right)

and

\left( { - 7,\;2} \right)

in the ratio of

View answer and explanation

Correct option: A

1:3

internally

Option A (

1:3

internally) is correct because it is the ratio in which the point divides the line segment, calculated using the section formula.

19 Medium

19 of 140

Application +4 | -1

The co-ordinates of the point

P

on the line

\mathbf{r}=(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})+\lambda(-\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})

which is nearest to the origin is

View answer and explanation

Correct option: A

\left(\frac{2}{3}, \frac{4}{3}, \frac{2}{3}\right)

Option A (

\left( \frac{{2}}{{3}}, \frac{{4}}{{3}}, \frac{{2}}{{3}} \right)

) is correct because it is the point on the line closest to the origin, found by minimizing the distance from the origin to a general point on the line.

20 Medium

20 of 140

Application +4 | -1

The equation of the plane through the line

x+y+z+3=0=2 x-y+3 z+1

and parallel to the line

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

View answer and explanation

Correct option: A

x-5 y+3 z=7

Option A (

x-5y+3z=7

) is correct because it is the equation of the plane through the line and parallel to the given line.

21 Easy

21 of 140

Understanding +4 | -1

The locus represented by

x y+y z-0

View answer and explanation

Correct option: D

a pair of perpendicular planes

Option D (a pair of perpendicular planes) is correct because the equation can be factored into y(x+z) = 0, representing two planes y=0 and x+z=0, which are perpendicular.

22 Easy

22 of 140

Application +4 | -1

Let

P

be the point (1, 0) and

{\cal Q}

be the point on

{y^2} = 8x

. The locus of mid point of

P{\cal Q}

View answer and explanation

Correct option: D

{y^2} - 4x + 2 = 0

Option D

{y^2} - 4x + 2 = 0

is correct because it is the locus of the midpoint derived from the given parabola equation.

23 Easy

23 of 140

Application +4 | -1

A\left( {6, - 3} \right),B\left( { - 3,5} \right),C\left( {4, - 2} \right),P\left( {\mathrm{\alpha },\mathrm{\beta }} \right),

then the ratio of the areas of the triangles

PBC,ABC

View answer and explanation

Correct option: D

\left| {\mathrm{\alpha } + \mathrm{\beta } - 2} \right|

Option D is correct because the ratio of areas of triangles PBC and ABC is found by calculating the determinants and simplifying to |α + β - 2|.

24 Medium

24 of 140

Application +4 | -1

Let the point, on the line passing through the points

{P}(1,-2,3)

and

Q(5,-4,7)

, farther from the origin and at a distance of 9 units from the point

P

, be

(\alpha, \beta, \gamma)

. Then

\alpha^2+\beta^2+\gamma^2

is equal to:

View answer and explanation

Correct option: A

155

Option D (

\frac{\pi}{32}

) is correct because, after solving the differential equation and applying the initial condition, y(2) equals

\frac{\pi}{32}

25 Medium

25 of 140

Application +4 | -1

The coordinates of the point where the line segment joining

A(5,1,6)

and

B(3,4,1)

crosses the

Y Z

-plane are

View answer and explanation

Correct option: C

\left(0, \frac{17}{2},-\frac{13}{2}\right)

Option A

\left(0, \frac{{17}}{{2}},-\frac{{13}}{{2}}\right)

is correct because it represents the point where the line segment AB intersects the YZ-plane, found by setting the x-coordinate to zero.

26 Easy

26 of 140

Application +4 | -1

The locus of a point

P

which moves such that

2PA = 3PB

, where coordinates of points

A

and

B

are (0,0) and (4,-

- 3

), is

View answer and explanation

Correct option: B

5{x^2} + 5{y^2} - 72x + 54y + 225 = 0

Option B (

5{x^2} + 5{y^2} - 72x + 54y + 225 = 0

) is correct because it represents the locus of point P satisfying the given distance condition.

27 Medium

27 of 140

Application +4 | -1

The equation of the plane passing through the straight line

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

and perpendicular to the plane

x+

2 y+z=12

View answer and explanation

Correct option: B

9 x-2 y-5 z+4=0

Option B (

9x-2y-5z+4=0

) is correct because it is the equation of the plane passing through the straight line and perpendicular to the given plane.

28 Medium

28 of 140

Application +4 | -1

The length of the perpendicular from the origin to the plane passing though three non-collinear points a, b, c is

View answer and explanation

Correct option: A

\frac{[\mathbf{a b} \mathbf{c}]}{|\mathbf{a} \times \mathbf{b}+\mathbf{c} \times \mathbf{a}+\mathbf{b} \times \mathbf{c}|}

Option A is correct because it represents the formula for the length of the perpendicular from the origin to the plane passing through three non-collinear points a, b, c.

29 Easy

29 of 140

Understanding +4 | -1

The direction cosines of any normal to the

x y

-plane are

View answer and explanation

Correct option: D

0,0,1

Option D (0,0,1) is correct because the normal vector to the xy-plane is along the z-axis, and its direction cosines are (0,0,1).

30 Hard

30 of 140

Application +4 | -1

Projection of the line

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

4 z-6

on the plane

z=0

View answer and explanation

Correct option: B

\frac{x}{1}=\frac{y-6}{-2}=\frac{z}{0}

Option B,

\frac{x}{1}=\frac{y-6}{-2}=\frac{z}{0}

, is the correct answer because it represents the projection of the given line onto the plane z=0, found by eliminating z and finding the direction ratios.

Revision notes and key takeaways

Review 3D GEOMETRY SYSTEM in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 140 single-correct-answer MCQs on 3D GEOMETRY SYSTEM.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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