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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 midpoints 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 midpoints of parallel sides AB and DC then
Equation of RS by r = a + s(ba) 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} = a^{2}
L.H.S. = (1 + a^{2}) (1 + b^{2})
R.H.S. = (1  ab cos qθ)^{2} + (u + v)^{2} (u × v)^{2}
= 1 + a^{2}b^{2} cos^{2} θ  2ab cos θ +a^{2} +b^{2} + 2abcos θ + a^{2}b^{2} sin^{2} 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 3dimensional 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 xy plane.
is along the positive direction of xaxis
then
makes an angle 135° with lines in xy plane,
Also keeping in mind
are some possible answers.
Q. 21. where f_{1}, f_{2}, g_{1}, g_{2} 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 = (a_{1}b_{2}c_{3} + a_{2}b_{3}c_{1}+ a_{3}b_{1}c_{2})  (a_{1}b_{3}c_{2} + a_{2}b_{1}c_{3}+ a_{3}b_{2}c_{1}) ....(1)
Now we know that AM > GM
[same reason]
L^{3} > 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)
Eq^{n} 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 noncoplanar 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 def^{n} 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 eq^{n} (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 L_{1} and L_{2} with direction ratios as (1, 0, –1) and (1, –1, 0)
∴ A vector perpendicular to L_{1} and L_{2} 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. P_{1} and P_{2} are planes passing through origin. L_{1} and L_{2} are two line on P_{1} and P_{2} 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 L_{1}, B on P_{1} but not on L_{1} and C not on P_{1} (ii) A' is on L_{2}, B' on P_{2} but not on L_{2} and C' not on P_{2} (2004  4 Marks)
Solution. Following fig. shows the possible situation for planes P_{1} and P_{2} and the lines L_{1} and L_{2}
Now if we choose pts A, B, C as follows.
A on L_{1} , B on the line of intersection of P_{1} and P_{2} but other than origin and C on L_{2} 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+z3 = 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
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