Subjective Type Questions: Vector Algebra and Three Dimensional Geometry - 2 | JEE Advanced | 35 Years Chapter wise Previous Year Solved Papers for JEE PDF Download

Q. 16. Prove, by vector methods or otherwise, that the point of intersection of the diagonals of a trapezium lies on the line passing through the mid-points of the parallel sides. (You may assume that the trapezium is not a parallelogram.)            (1998 - 8 Marks)

Solution. The P.Vs. of the points A, B, C, D are

Equations of AC and BD are

For point of intersection say T compare the coefficients

                ...(1)

Let R and S be mid-points of parallel sides AB and DC then

Equation of RS by r = a + s(b-a) is

Comparing the coefficients, we get

which is true . Hence proved.

Q. 17. For any two vectors u and v, prove that            (1998 - 8 Marks)

Solution. We have,  Where θ is the angle between  is a unit vector perpendicular to both   and is such that  form a right handed system.

(b)  is perpendicular to both u and v , | a |² = a²

L.H.S. = (1 + a²) (1 + b²)
R.H.S. = (1 - ab cos qθ)² + (u + v)² (u × v)² 

= 1 + a²b² cos² θ - 2ab cos θ +a² +b² + 2abcos θ + a²b² sin² θ.1+ 0

Q. 18. Let u an d v be unit vector s. If w is a vector such th at w +(w × u) = v, then prove that |(u × v) · w| < 1/2 and  that the equality holds if and only if u is perpendicular to v.            (1999 - 10 Marks)

Solution. 

(θ is angle between )

Equality holds when 

Q. 19. Show, by vector methods, that the angular bisectors of a triangle are concurrent and find an expression for the position vector of the point of concurrency in terms of the position vectors of the vertices.              (2001 - 5 Marks)

Solution. Let   be the position vectors by A, B, and C respectively,

Let AB, BE and CF be the bisectors of ∠A, ∠B , and ∠C respectively.

a, b, c are the lengths of sides BC, CA and AB respectively.

Now we know by angle bisector thm that AD divides, BC in the ratio

BD : DC = AB : AC = c : b.

∴ The position vector of D is  

Let I be the point of intersection of BE and AD. Then in ΔABD, BI is bisector of ∠B.

∴ DI : IA = BD : BA

As p.v. of I is symm. in   and a,b,c.

∴ It must lie on CF as well.

[We can also see that p.v. of intersection of

Above prove that all the ∠ bisectors pass through I, i.e., these are concurrent.

Q. 20. Find 3-dimensional vectors   satisfying

              (2001 - 5 Marks)

Ans.  are some possible values.

Solution. Given data is insufficient to uniquely determine the three vectors as there are only 6 equations involving 9 variables.

∴ We can obtain infinitely many set of three vectors,

  satisfying these conditions. 

From the given data, we get

[where θ is the angle between ]

Now since any two vectors are always coplanar, let us suppose that   are in x-y plane.

 is along the positive direction of x-axis  

then     

 makes an angle 135° with   lines in x-y plane, 

Also keeping in mind 

are some possible answers.

Q. 21.    where f₁, f₂, g₁, g₂ are continuous functions. If   and   are nonzero vectors for all t and    Then show that    are parallel for some t.           (2001 - 5 Marks)

Solution. is parallel to   if and only if 

Since h is a continuous function, and h(0).h(1) < 0

⇒ There is some t ∈ [0,1] for which h(t) = 0 i.e.,   and   are parallel vectors for this t.

Q. 22. Let V be the volume of the parallelopiped formed by the vectors    where r = 1, 2, 3, are n on negative real numbers and   show that               (2002 - 5 Marks)

Solution. 

⇒ V = (a₁b₂c₃ + a₂b₃c₁+ a₃b₁c₂) - (a₁b₃c₂ + a₂b₁c₃+ a₃b₂c₁) ....(1)

Now we know that  AM > GM

[same reason]

L³ > V from (1) Hence Proved.

Q. 23. (i) Find the equation of the plane passing through the points (2, 1, 0), (5, 0, 1) and (4, 1, 1).          (2003 - 4 Marks)

(ii) If P is the point (2, 1, 6) then find the point Q such that PQ is perpendicular to the plane in (i) and the mid point of PQ lies on it.

Ans. (i) x + y - 2z = 3
(ii) Q(6, 5, - 2)

Solution. (i) Plane passing thr ough (2, 1, 0) , (5, 0, 1) and (4, 1, 1) is

⇒ ( x - 2)(-1 - 0) - (y - 1)(3- 2) + z(0 - (-2)) = 0
⇒ -x + 2 - y +1+ 2z = 0 ⇒ x + y - 2z = 3

(ii) As per question we have to find a pt. Q such that PQ is ⊥ to the plane x + y - 2z = 3      ...(1)

And mid pt. of PQ lies on the plane, (Clearly we have to find image of pt. P with respect to plane).

Let Q be (α, β,γ)

Eqⁿ of PM passing through P(2, 1, 6) and ⊥ to plane x + y - 2z = 3 , is given by

For some value of   lies on PM

But M lies on plane (1)

Q. 24. If  are three non-coplanar unit vectors and are the angles between  and respectively and are unit vectors along the bisectors of the angles   respectively. Prove that                 (2003 - 4 Marks)

Solution. Given that   are three non coplanar unit vectors. Angle   and between   represent 

Let P be a pt. on angle bisector of ∠AOB such that OAPB is a parallelogram

Also ∠POA =∠BOP = α /2

∴ ∠APO = ∠BOP = α /2 (alt. int. ∠ 's)

∴ In ΔOAP, OA= AP a unit vector in the direction of 

∴A unit vector in the direction of

[Using defⁿ of vector triple product.]

          ...(1)

 when ever any two vectors are same)

From (1) and (2),

Q. 25. If  are distinct vectors such that                     

Prove that                 (2004 - 2 Marks)

Solution. Given that 

Such that          ...(1)
                    ...(2)
To prove 

Subtracting eqⁿ (2) from (1) we get

⇒ Angle between  is either 0 or 180° . 

as a, b, c, d all are different. Hence Proved.

Q. 26. Find the equation of plane passing through (1, 1, 1) & parallel to the lines L1, L2 having direction ratios (1,0, –1), (1, –1,0). Find the volume of tetrahedron formed by origin and the points where these planes intersect the coordinate axes.           (2004 - 2 Marks)

Ans. 

Solution. ∴ The plane is parallel to the lines L₁ and L₂ with direction ratios as (1, 0, –1) and (1, –1, 0)

∴ A vector perpendicular to L₁ and L₂ will be parallel to the normal   to the plane.

∴ Eqn. of plane through (1, 1, 1) and having normal vector 

Now the pts where this plane meets the axes are A(3, 0, 0), B(0, 3, 0),C (0, 0, 3)
∴  Vol.of  tetrahedron OABC

(Note that ΔABC is an equilateral Δ here.)

Q. 27. A parallelopiped ‘S’ h as base points A, B, C an d D and upper face points A', B', C' and D'. This parallelopiped is compressed by upper face A'B'C'D' to form a new parallelopiped ‘T’ having upper face points A'', B'', C'' and D''. Volume of parallelopiped T is 90 percent of the volume of parallelopiped S. Prove that the locus of ‘A''’, is a plane.           (2004 - 2 Marks)

Solution. ATQ ‘S’ is the parallelopiped with base points A, B, C and D and upper face points A', B',C' and D ' . Let its vol. be V_s .
By compr essi n g it by upper face A', B', C ',D', a n ew parallelopiped ‘T’ is formed whose upper face pts are now A'',B'',C'' and D ''. Let its vol. be V_T .

Let h be the height of original parallelopiped S.
Then V_S = (ar ABCD)x h     ...(1)

Let equation of plane ABCD be  ax + by +cz +d= and A''(α,β,γ)

Then height of new parallelopipe T is the length of perpendicular from A'' to ABCD

But given that,

             ...(3)

From (1), (2) and (3) we get,

which is a plane parallel to ABCD .  Hence proved.

Q. 28. P₁ and P₂ are planes passing through origin. L₁ and L₂ are two line on P₁ and P₂ respectively such that their intersection is origin. Show that there exists points A, B, C, whose permutation A', B', C' can be chosen such that (i) A is on L₁, B on P₁ but not on L₁ and C not on P₁ (ii) A' is on L₂, B' on P₂ but not on L₂ and C' not on P₂        (2004 - 4 Marks)

Solution. Following fig. shows the possible situation for planes P₁ and P₂ and the lines L₁ and L₂

Now if we choose pts A, B, C as follows.
A on L₁ , B on the line of intersection of P₁ and P₂ but other than origin and C on L₂ again other than origin then we can consider

A corresponds to one of A',B',C' and
B corresponds to one of the remaining of A', B',C' and
C corresponds to third of A',B',C' e.g.

Hence one permutation of [ABC] is [CBA]. Hence Proved.
Q. 29. Find the equation of the plane containing the line 2x – y + z – 3 = 0, 3x + y + z = 5 and at a distance of from the point(2, 1, –1)           (2005 - 2 Marks)

Ans. 62 x + 29 y + 19z - 105 = 0

Solution. The given line is 2x - y + z - 3 = 0 = 3x + y + z - 5

Which is intersection line of two planes

2 x - y + z - 3 = 0 ...(i)
and 3x + y + z - 5 = 0 ...(ii)

Any plane containing this line will be the plane passing through the intersection of two planes (i) and (ii).

Thus the plane containing given line can be written as

As its distance from the pt. 

Squaring both sides, we get

∴ The required equations of planes are 2x- y+z-3 = 0

or 62x + 29 y + 19z - 105 = 0

Q. 30. If the incident ray on a surface is along the unit vectorthe reflected ray is along the unit vector   the normal is along unit vector   outwards. Express    in terms of and              (2005 - 4 Marks)

Ans.  

Solution. Given that incident ray is along reflected ray is along  normal is along , outwards. The given figure can be redrawn as shown.

We know  that incident ray, reflected ray and normal lie in a plane, and angle of incidence = angle of reflection.

Therefore , will be along the angle bisector of 

[∵ Angle  bisector will along a vector dividing in same ratio as the ratio of the sides forming that angle.]

But  is a unit vector.

Where 

Substituting this value in equation (1) we get