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NCERT Solutions class 12 Maths Three Dimensional Geometry

1.Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points

Ans. We know that direction ratios of the line joining the origin (0, 0, 0) to the point are

= 2 – 0, 1 – 0, 1 – 0 = 2, 1, 1 =

Similarly, direction ratios of the line joining the points  and  are

=  =  =

For these two lines,

=  = 2 – 2 + 0 = 0

Therefore, the two given lines are perpendicular to each other.

2.If  and  are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are

Ans.  and  are direction cosines of two mutually perpendicular of two given lines L1 and L2. (say)

Let  and  be the unit vectors along these lines L1 and L2.

and

Let L be the line perpendicular to both the lines L1 and L2 and let  be a unit vector along line L perpendicular both lines L1 and L2.

Cross-product of two vectors =

[ L1  L2 (given,  angle between them is ]

Since, is a unit vector, therefore its components are its direction cosines.

Thus, direction cosines of  are

direction cosines of line L are

3.Find the angle between the lines whose direction ratios are  and

Ans. Direction ratios of one line are

A vector along this line is

Direction ratios of second line are

A vector along second line is

Let  be the angle between the two lines, then

=

=  = 0 =

4.Find the equation of the line parallel to axis and passing through the origin.

Ans. We know that a unit vector along axis is

Direction cosines of axis are coefficients of  in the unit vector

i.e., 1, 0, 0 =

Equation of the required line passing through the origin (0, 0, 0) and parallel to axis is

Vector equation of the required line is

[ and  ]

5.If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7),  and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Ans. Given: Points A B C and D

Direction ratios of line AB are

A vector along the line AB is

Similarly, direction ratios of line CD are

A vector along the line AB is

Let  be the angle between the two lines, then

= =  =  =  = 1

=

Therefore, lines AB and CD are parallel.

6.If the lines  and  are perpendicular, find the value of

Ans. Given: Equation of one line is

Direction ratios of this line are its denominators, i.e.,  =

A vector along this line is

Again, equation of second line is

Direction ratios of this line are its denominators, i.e.,  =

A vector along this line is

Since these given lines are perpendicular.

7.Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane

Ans. The required line passes through the point P (1, 2, 3).

Position vector  (say) of point P is (1, 2, 3)

Equation of the given plane is

Comparing with

Since, the required line is perpendicular to the given plane, therefore, vector  along the required line is

Equation of the required line is

8.Find the equation of the plane passing through  and parallel to the plane

Ans. Equation of any plane parallel to the plane  is …..(i)

Plane (i) passes through

Putting  in eq. (i), we get

Putting the value of  in eq. (i), to get the required plane is

9.Find the shortest distance between lines  and

Ans. Given: Vector equation of one line is

Comparing with  we get

and

Again given: Vector equation of another line is

Comparing with  we get

and

We know that length of shortest distance between two (skew) lines is  ..(i)

Now =  =

Again

Expanding along first row,

=

=

And =

Putting these values in eq. (i), length of shortest distance =

10.Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.

Ans. Given: A line through the points A (5, 1, 6) and B (3, 4, 1)

Direction ratios of this line AB are

3 – 5, 4 – 1, 1 – 6

Equation of the line AB is

……….(i)

Now we have to find the coordinates of the point where this line AB crosses the YZ-plane

i.e.,  ……….(ii)

Putting  in eq. (i), we get

and

and   and

and

Thus, required point is P

11.Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.

Ans. Given: A line through the points A (5, 1, 6) and B (3, 4, 1)

Direction ratios of this line AB are

3 – 5, 4 – 1, 1 – 6

Equation of the line AB is

……….(i)

Now we have to find the coordinates of the point where this line AB crosses the ZX-plane

i.e.,  ……….(ii)

Putting  in eq. (i), we get

and

and   and

and

Thus, required point is P

12.Find the coordinates of the point where the line through  and  crosses the plane

Ans. Direction ratios of the line joining the points A and B are

Equation of the line AB are  ………(i)

Equation of the plane is  ………(ii)

Now to find the point where line (i) crosses plane (ii),

From eq. (i) (say)

……….(iii)

Putting the values of  in eq. (ii), we get

Putting  in eq. (iii), point of intersection of line (i) and plane (ii) is

Thus, required point of intersection is

13.Find the equation of the plane passing through the point  and perpendicular to each of the planes  and

Ans. Since equation of any plane through the point  is

……….(i)

This required plane is perpendicular to the plane

Product of coefficients  ……….(ii)

Again the required plane is perpendicular to the plane

Product of coefficients  ……….(iii)

Solving eq. (ii) and (iii), we get

Putting these values of  in eq. (i), we get

14. If the points  and  be equidistant from the plane  then find the value of

Ans. Equation of the given plane is

[  = Position vector of any point  on the plane ]

……….(i)

Also, the point  and  are equidistant from plane (i)

(Perpendicular) distance of point  from plane (i)

= Distance of point  from plane (i)

[ If  then]

Taking positive sign,

Taking negative sign,

Hence, the values of  are 1 or

15.Find the equation of the plane passing through the line of intersection of the planes  and  and parallel to axis.

Ans. Equation of one plane is

……….(i)

Equation of the second plane is ……….(ii)

Since, equation of any plane passing through the line intersection of these two planes is

L.H.S. of I +  (L.H.S. of II) = 0

……….(i)

Comparing  we have

Now required plane (i) is parallel to axis ( a vector  along axis is  = )

Putting  in eq. (i), the equation of required plane,

16.If O be the origin and the coordinates of P be  then find the equation of the plane passing through P and perpendicular to OP.

Ans. Given: Origin O (0, 0, 0) and point P

To find: Equation of the plane passing through P =

Direction ratios of normal OP to the plane are

Equation of the required plane is

17.Find the equation of the plane which contains the line of intersection of the planes  and which is perpendicular to the plane

Ans. Equation of any plane passing through (or containing) the line of intersection of the planes  and  is L.H.S. of I +  (L.H.S. of II) = 0

……….(i)

Comparing with  we have,

Now plane (i) is perpendicular to the given plane

Comparing with  we have,

For perpendicular planes

Putting  in eq. (i), equation of required plane is

18.Find the distance of the point  from the point of intersection of the line  and the plane

Ans. Given: A point P (say)

and equation of the line  ……….(i)

equation of the plane is

Putting the value of  from eq. (i) in eq. (ii),

Putting  in eq. (i),

Therefore, Point of intersection is  =

Distance of the given point P from the point of intersection is

=

=

19.Find the vector equation of the line passing through  and parallel to the plane  and

Ans. The required line passes through the point A (1, 2, 3) =

= Position vector of point A =

Let  be any vector along the required line.

Vector equation of required line is

……….(i)

Since required line is parallel to the plane

and

Comparing with  we have,

And Comparing with  we have,

Since  is perpendicular to both  and

=

Expanding along first row,

=

Putting this value of  in eq. (i), vector equation of required line,

20.Find the vector equation of the line passing through the point  and perpendicular to the two lines:  and

Ans. Given: A point on the required line is A

Position vector of point A is

Also given equations of two lines

and

Direction ratios of given two lines are  and

Now =

Expanding along first row,

=  =

Equation of the required line is

Again replacing  by

21.Prove that if a plane has the intercepts  and is at a distance of  units from the origin, then

Ans. We know that equation of plane making intercepts  (on the axes) is

Given: Perpendicular distance of the origin (0, 0, 0) from plane =

=

Squaring both sides,

Choose the correct answer in Exercise Q. 22 and 23.

22.Distance between the two planes:  and  is

(A) 2 units(B) 4 units(C) 8 units(D)  units

Ans. Equation of one plane is

Equation of second plane is

Here

Since,  therefore, the given two lines are parallel.

We know that the distance of the parallel lines =

=

=

Therefore, option (D) is correct.

23. The planes:  and  are

(A) Perpendicular(B) Parallel

(C) intersect axis(D) passes through

Ans. Equations of the given planes are

and

For perpendicular =  =

Planes are not perpendicular.

For parallel

given planes are parallel.

Therefore, option (B) is correct.

NCERT Solutions class 12 Maths Miscellaneous

NCERT Solutions Class 12 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2. NCERT Solutions for CBSE Class 12 Maths have total 13 chapters. 12 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 12 solutions PDF and Maths ncert class 12 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide

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