Subjective Type Questions: Vector Algebra and Three Dimensional Geometry - 1 | JEE Advanced

# Subjective Type Questions: Vector Algebra and Three Dimensional Geometry - 1 | JEE Advanced | Mathematics for NDA PDF Download

Q. 1. From a point O inside a triangle ABC, perpendiculars OD, OE, OF are drawn to the sides BC, CA, AB respectively. Prove that the perpendiculars from A, B, C to the sides EF, FD, DE are concurrent.            (1978)

Solution. Let with respect to O, position vectors of points A, B, C, D, E, F be

Let perpendiculars from A to EF  and from B to DF meet each other at H. Let position vector of H be   we join CH.
In order to prove the statement given in question, it is sufficient to prove that CH is perpendicular to DE.

Adding (4) and (5), we get

(using (1), (2) and (3))

Q. 2. A1, A2,...................... An are the vertices of a regular plane polygon with n sides and O is its centre. Show that          (1982 - 2 Marks)

Solution.   all vectors are of same magnitude, say ‘a’  and angle between any two consecutive vector  is same  that is  be the unit vectors ⊥ to the plane of the polygon.

Q. 3. Find all values of λ such that x, y,z, ≠ (0, 0, 0) and   are unit vectors along the coordinate axes.             (1982 - 3 Marks)

Ans. λ = 0, - 1

Solution.

All the above three equations are satisfied for x, y, z not all zero if

Q. 4. A vector  has components A1, A2, A3 in a right -handed rectangular Cartesian coordinate system oxyz. The coordinate system is rotated about the x-axis through an angle  Find the components of A in the new coordinate system, in terms of A1, A2, A3.

Ans.

Solution. Since vector  has components A1 , A2 ,A3, in the coordinate system OXYZ,

When given system is rotated through   the new x-axis is along old y-axis and new y-axis is along the old negative x-axis z remains same as before.
Hence the components of A in the new system are

A2 , - A1,A3

Q. 5. The position vectors of the points A, B, C an d D are respectively. If the points A, B, C and D lie on a plane, find the value of λ .

Ans. 146/17

Solution.

We know that A, B, C, D lie in a plane if   are coplanar i.e.

Q. 6. If A, B, C , D are any four points in space, prove that –

(area of triangle ABC)

Solution. Let the position vectors of points A, B, C, D be a, b, c, and d respectively with respect to some origin O.

Then,

…(1)
Also Area of ΔABC is

… (2)

From (1) and (2), we ge

Q. 7. Let OA CB be a parallelogram with O at the origin and OC a diagonal. Let D be the midpoint of OA. Using vector methods prove that BD and CO intersect in the same ratio. Determine this ratio.

Solution. OACB is a parallelogram with O as  origin. Let with respect to O position vectors of A and B be   respectively..
Then p.v. of C is

Also D is mid pt. of OA, therefore position vector of D is

CO and BD intersect each other at P.

Let P divides CO in the ratio λ : 1 and BD in the ratio μ : 1 Then by section theorem, position vector of pt. P dividing CO in ratio

…(1)
And position vector of pt. P dividing BD in the ratio μ : 1 is

…(2)

As (1) and (2) represent the position vector of same point, we should have

Equating the coefficients of    we get

… (i)

…(ii)

From (ii) we get λ = μ ⇒ P divides CO and BD in the same ratio.

Putting λ = μ in eq. (i) we get μ = 2

Thus required ratio is 2 : 1.

Q. 8. If vectors  are coplanar, show that

Solution. Given that  are  three coplanar vectors.

∴ There exists scalars x, y, z, not all zero, such that

… (1)

Taking dot product of  we get

… (2)

Again taking dot product of  we get

… (3)

Now equations (1), (2), (3) form a homogeneous system of equations, where x, y, z are not all zero.

∴ system must have non trivial solution and for this, determinant of coefficient matrix should be zero

Hence Proved.

Q. 9. In a triangle OAB, E is the midpoint of BO and D is a point on AB such that AD : DB = 2 : 1. If OD and AE intersect at P, determine the ratio OP : PD using vector methods.

Solution. With O as origin let   be the position vectors of A and B respectively.

Then the position vector of E, the mid point of OB is

Again since AD : DB = 2 : 1, the  position vector of D is

∴ Equation of OD is

…(1)

and  Equation of AE is

…(2)

If OD and AE intersect at P, then we will have identical values of  Hence comparing the coefficients of   we get

Putting value of t in eq. (1) we get position vector of point of intersection P as

… (3)

Now if P divides OD in the ratio λ : 1, then p.v. of P is

… (4)

From (3) and (4) we get

Q. 10.  Determine a vector   Satisfying

Ans.

Solution. We are given that   and  to determine a vector   such that   and

⇒ y -z =-10 … (1)
z -x =-11 … (2)
x -y= 7 … (3)

⇒ 2x +z=0 … (4)

Substituting y =x- 7 and z = -2x from (3) and (4) respectively in eq. (1) we get

x - 7 + 2x = -10 ⇒ 3x =-3

⇒ x =-1 , y =-8 and z = 2

Q. 11. Determine the value of ‘c’ so that for all real x, the vector   make an obtuse angle with each other.

Ans.

Solution. We have,

Now we know that

As angle between   is obtuse, therefore

Q. 12. In a triangle ABC, D an d E are points on BC and AC respectively, such that BD = 2 DC and AE = 3EC. Let P be the point of intersection of AD and BE. Find BP/PE using vector methods.

Ans. 8 : 3

Solution. Let   be the position vectors of pt A, B and C respectively with respect to some origin.

ATQ, D divides BC in the ratio 2 : 1 and E divides AC in the ratio 3 : 1.

∴ position vector of D is   and position vector of E is

Let pt. of intersection P of AD and BE divides BE in the ratio k : 1 and AD in the ratio m : 1, then position vectors of P in these two cases are

Equating the position vectors of P in two cases we get

… (1)

… (2)

Dividing (3) by (2) we get

the req. ratio is 8 : 3.

Q. 13. If the vectors   are not coplanar, then prove that the vector   parallel to

Solution. Given that   are  not coplanar

Consider,

Here,

…(2)

…(3)

[NOTE :  Here we have tried  to write the given expression in such a way that we can get terms involving  other terms similar which can get cancelled.]

Adding (1), (2) and (3), we get given vector

⇒ given vector = some constant multiple of

⇒ given vector is parallel to

Q. 14. The position vectors of the vertices A, B an d C of a tetrahedron ABCD are  respectively. The altitude from vertex D to the opposite face ABC meets the median line through A of the triangle ABC at a point E. If the length of the side AD is 4 and the volume of the tetrahedron is  find the position vector of the point E for all its possible positions.

Ans. (-1, 3, 3) or (3, - 1, - 1)

Solution. We are given AD = 4

Volume of tetrahedron

We have to find the P.V. of point E.  Let it divides median AF in the ratio λ : 1

…(3)

Q. 15. If A, B and C are vectors such that | B | = | C |. Prove that [(A + B) × (A + C)] × (B × C) (B + C)  = 0 .

Solution. We have,

[∵ (a x b) x c = (a.c)b- (b.c)a]

[∵ [A B C] = 0 if any two of A, B, C are equal.]

Thus, LHS of the given expression

[∵| B | = C|]

The document Subjective Type Questions: Vector Algebra and Three Dimensional Geometry - 1 | JEE Advanced | Mathematics for NDA is a part of the NDA Course Mathematics for NDA.
All you need of NDA at this link: NDA

## Mathematics for NDA

237 videos|243 docs|240 tests

## FAQs on Subjective Type Questions: Vector Algebra and Three Dimensional Geometry - 1 - JEE Advanced - Mathematics for NDA

 1. What is a vector in mathematics?
Ans. A vector is a mathematical object that has both magnitude and direction. It is represented by an arrow with a specified length and direction. In three-dimensional space, a vector is typically represented as an ordered triple of numbers (x, y, z), where x, y, and z are the components of the vector in the x, y, and z directions, respectively.
 2. How are vectors added and subtracted?
Ans. Vectors can be added and subtracted using the parallelogram law of vector addition. To add two vectors, we place them head to tail and draw a parallelogram using the two vectors as adjacent sides. The diagonal of the parallelogram represents the sum of the two vectors. To subtract two vectors, we add the negative of the vector being subtracted.
 3. What is the dot product of two vectors?
Ans. The dot product of two vectors is a scalar quantity that measures the degree of similarity or orthogonality between the two vectors. It is defined as the product of the magnitudes of the two vectors and the cosine of the angle between them. Mathematically, the dot product of two vectors A = (a1, a2, a3) and B = (b1, b2, b3) is given by A · B = a1b1 + a2b2 + a3b3.
 4. What is the cross product of two vectors?
Ans. The cross product of two vectors is a vector that is perpendicular to both of the original vectors. It is defined as the product of the magnitudes of the two vectors and the sine of the angle between them, multiplied by a unit vector perpendicular to the plane containing the two vectors. Mathematically, the cross product of two vectors A = (a1, a2, a3) and B = (b1, b2, b3) is given by A × B = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1).
 5. How can vectors be used in three-dimensional geometry?
Ans. Vectors play a crucial role in three-dimensional geometry. They can be used to represent points, lines, and planes in space. By using vector equations, we can determine the position, direction, and distance between objects in three-dimensional space. Vectors also help in solving problems related to distance, displacement, velocity, acceleration, and forces in three-dimensional systems.

## Mathematics for NDA

237 videos|243 docs|240 tests

### Up next

 Explore Courses for NDA exam

### Top Courses for NDA

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Track your progress, build streaks, highlight & save important lessons and more!
Related Searches

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;