Question 3: In a pentagon ABCDE Show that + + + + = 2

Step-by-step solution provided.

Maharashtra State Board Class 12 (HSC Board) Mathematics & Statistics Solutions Chapter 5 Vectors Miscellaneous Exercise 5 - Digest Answers | Maharashtra Board Solutions

Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 5 Vectors Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Miscellaneous Exercise 5

I) Select the correct option from the given alternatives :
Question 1.
If || = 2, || = 3 || = 4 then [ + + – ] is equal to
(A) 24
(B) -24
(C) 0
(D) 48
Solution:
(C) 0

Question 2 Maharashtra Board Solution

If || = 3, || = 4, then the value of λ for which + λ is perpendicular to – λ, is

Solution & Step-by-Step Answer:

(b)

Question 3 Maharashtra Board Solution

If sum of two unit vectors is itself a unit vector, then the magnitude of their difference is (A) (B) (C) 1 (D) 2

Solution & Step-by-Step Answer:

(B)

Question 4 Maharashtra Board Solution

If || = 3, || = 5, || = 7 and + + = 0, then the angle between and is

Solution & Step-by-Step Answer:

(b)

Question 5 Maharashtra Board Solution

The volume of tetrahedron whose vertices are (1, -6, 10), (-1, -3, 7), (5, -1, λ) and (7, -4, 7) is 11 cu. units then the value of λ is (A) 7 (B) (C) 1 (D) 5

Solution & Step-by-Step Answer:

(A) 7

Question 6 Maharashtra Board Solution

If α, β, γ are direction angles of a line and α = 60º, β = 45º, the γ = (A) 30º or 90º (B) 45º or 60º (C) 90º or 30º (D) 60º or 120º

Solution & Step-by-Step Answer:

(D) 60º or 120º

Question 7 Maharashtra Board Solution

The distance of the point (3, 4, 5) from Y- axis is (A) 3 (B) 5 (C) (D)

Solution & Step-by-Step Answer:

(C)

Question 8 Maharashtra Board Solution

The line joining the points (-2, 1, -8) and (a, b, c) is parallel to the line whose direction ratios are 6, 2, 3. The value of a, b, c are (A) 4, 3, -5 (B) 1, 2, (C) 10, 5, -2 (D) 3, 5, 11

Solution & Step-by-Step Answer:

(A) 4, 3, -5

Question 9 Maharashtra Board Solution

If cos α, cos β, cos γ are the direction cosines of a line then the value of sin2 α + sin2β + sin2γ is (A) 1 (B) 2 (C) 3 (D) 4

Solution & Step-by-Step Answer:

(B) 2

Question 10 Maharashtra Board Solution

If l, m, n are direction cosines of a line then is (A) null vector (B) the unit vector along the line (C) any vector along the line (D) a vector perpendicular to the line

Solution & Step-by-Step Answer:

(B) the unit vector along the line

Question 11 Maharashtra Board Solution

If || = 3 and –1 ≤ k ≤ 2, then |k| lies in the interval (A) [0, 6] (B) [-3, 6] (C) [3, 6] (D) [1, 2]

Solution & Step-by-Step Answer:

(A) [0, 6]

Question 12 Maharashtra Board Solution

Let α, β, γ be distinct real numbers. The points with position vectors , , (A) are collinear (B) form an equilateral triangle (C) form a scalene triangle (D) form a right angled triangle

Solution & Step-by-Step Answer:

(B) form an equilateral triangle

Question 13 Maharashtra Board Solution

Let and be the position vectors of P and Q respectively, with respect to O and || = p, || = q. The points R and S divide PQ internally and externally in the ratio 2 : 3 respectively. If OR and OS are perpendicular then. (A) 9p2 = 4q2 (B) 4p2 = 9q2 (C) 9p = 4q (D) 4p = 9q

Solution & Step-by-Step Answer:

(A) 9p2 = 4q2

Question 14 Maharashtra Board Solution

The 2 vectors and represents the two sides AB and AC, respectively of a ∆ABC. The length of the median through A is (A) (B) (C) (D) None of these

Solution & Step-by-Step Answer:

(A)

Question 15 Maharashtra Board Solution

If and are unit vectors, then what is the angle between and for – to be a unit vector ? (A) 30º (B) 45º (C) 60º (D) 90º

Solution & Step-by-Step Answer:

(A) 30º

Question 16 Maharashtra Board Solution

If θ be the angle between any two vectors and , then = , when θ is equal to (A) 0 (B) (C) (D) π

Solution & Step-by-Step Answer:

(B)

Question 17 Maharashtra Board Solution

The value of (A) 0 (B) -1 (C) 1 (D) 3

Solution & Step-by-Step Answer:

Question 18 Maharashtra Board Solution

Let a, b, c be distinct non-negative numbers. If the vectors , and lie in a plane, then c is (A) The arithmetic mean of a and b (B) The geometric mean of a and b (C) The harmonic man of a and b (D) 0

Solution & Step-by-Step Answer:

(B) The geometric mean of a and b

Question 19 Maharashtra Board Solution

Let = , = , = . If is a unit vector such that , then equals.

Solution & Step-by-Step Answer:

(a)

Question 20 Maharashtra Board Solution

If are non coplanar unit vectors such that = then the angle between and is (A) (B) (C) (D) π

Solution & Step-by-Step Answer:

(A)

II Answer the following :
1) ABCD is a trapezium with AB parallel to DC and DC = 3AB. M is the mid-point of DC,
= and = . Find in terms of and .
(i)
Solution:

(ii)
Solution:

(iii)
Solution:

(iv)
Solution:

Question 2 Maharashtra Board Solution

The points A, B and C have position vectors , and respectively. The point P is midpoint of AB. Find in terms of , and the vector

Solution & Step-by-Step Answer:

P is the mid-point of AB. ∴ = , where is the position vector of P.

Question 3 Maharashtra Board Solution

In a pentagon ABCDE Show that + + + + = 2

Solution & Step-by-Step Answer:

Question 4 Maharashtra Board Solution

If in parallelogram ABCD, diagonal vectors are = and = , then find the adjacent side vectors and

Solution & Step-by-Step Answer:

ABCD is a parallelogram

Question 5 Maharashtra Board Solution

If two sides of a triangle are and , then find the length of the third side.

Solution & Step-by-Step Answer:

Let ABC be a triangle with = , = . By triangle law of vectors Hence, the length of third side is 3 units.

Question 6 Maharashtra Board Solution

If || = | | = 1 . = 0 and + + = 0 then find ||

Solution & Step-by-Step Answer:

+ + = 0 ∴ – = + Taking dot product of both sides with itself, we get

Question 7 Maharashtra Board Solution

Find the lengths of the sides of the triangle and also determine the type of a triangle. (i) A(2, -1, 0), B(4, 1, 1,), C(4, -5, 4)

Solution & Step-by-Step Answer:

The position vectors , , of the points A, B, C are ∴ ∆ ABC is right angled at A.

(ii) L(3, -2, -3), M(7, 0, 1), N (1, 2, 1)
Solution:
The position vectors bar , , of the points L M, N are

l(LM) = 6, l(MN) = 2, l(NL) = 6
∆LMN is sosceles

Question 8 Maharashtra Board Solution

Find the component form of if a if (i) It lies in YZ plane and makes 60º with positive Y-axis and || = 4

Solution & Step-by-Step Answer:

Let α, β, γ be the direction angles of Since lies in YZ-plane, it is perpendicular to X-axis ∴ α = 90° It is given that β= 60° ∵ cos2α + cos2β + cos2γ = 1 ∴ cos290° + cos260° + cos2γ = 1 ∴ 0 + + cos2γ = 1 ∴ cos2γ = 1 – ∴ cos γ = Unit vector along a is given by

(ii) It lies in XZ plane and makes 45º with positive Z-axis and || = 10
Solution:

Question 9 Maharashtra Board Solution

Two sides of a parallelogram are and . Find the unit vectors parallel to the diagonals.

Solution & Step-by-Step Answer:

Question 10 Maharashtra Board Solution

If D, E, F are the mid-points of the sides BC, CA, AB of a triangle ABC, prove that + + = 0

Solution & Step-by-Step Answer:

Let , , , , , be the position vectors of the points A, B, C, D, E, F respectively. Since D, E, F are the midpoints of BC, CA, AB respec-tively, by the midpoint formula

Question 11 Maharashtra Board Solution

Find the unit vectors that are parallel to the tangent line to the parabola y = x2 at the point (2, 4)

Solution & Step-by-Step Answer:

Differentiating y = x2 w.r.t. x, we get = 2x Slope of tangent at P(2, 4) = = 2 × 2 = 4 ∴ the equation of tangent at P is y – 4 = 4(-2) ∴ y = 4x – 4 ∴ y = 4x is equation of line parallel to the tangent at P and passing through the origin O. 4x = y, z = 0 ∴ , z = 0 ∴ the direction ratios of this line are 1, 4, 0 ∴ its direction cosines are

Question 12 Maharashtra Board Solution

Express the vector as a linear combination of the vectors , and

Solution & Step-by-Step Answer:

By equality of vectors, 2x + 2y – z = 1 -x – 2y + 3z = 4 3x + 4y – 5z = -4 We have to solve these equations by using Cramer’s Rule. D = = 2(10 – 12) – 2(5 – 9) – 1(-4 + 6) = -4 + 8 – 2 = 2 ≠ 0 Dx = = 1(10 – 12) – 2(-20 + 12) – 1 (16 – 8) = -2 + 16 – 8 = 6 Dy = = 2(-20 + 12) – 1(5 – 9) – 1(4 – 12) = -16 – 4 – 8 = -28 Dz = = 2(8 – 16) – 2(4 – 12) + 1(-4 + 6) = -16 – 16 + 2 = -30

Question 13 Maharashtra Board Solution

If = and = then show that the vector along the angle bisector of angle AOB is given by = λ Question is modified If = and = then show that the vector along the angle bisector of ∠AOB is given by = λ

Solution & Step-by-Step Answer:

Choose any point P on the angle bisector of ∠AOB. Draw PM parallel to OB. ∴ ∠OPM = ∠POM = ∠POB Hence, OM = MP ∴ OM and MP is the same scalar multiple of unit vectors and along these directions,

Question 14 Maharashtra Board Solution

The position vectors f three consecutive vertices of a parallelogram are , and Find the position vector of the fourth vertex.

Solution & Step-by-Step Answer:

Let ABCD be a parallelogram. Let , , , be the position vectors of the vertices A, B, C, D of the parallelogram, Hence, the position vector of the fourth vertex is 7( + + ).

Question 15 Maharashtra Board Solution

A point P with position vector divides the line joining A(-1, 6, 5) and B in the ratio 3 : 2 then find the point B.

Solution & Step-by-Step Answer:

Let A, B and P have position vectors , and respectively. ∴ coordinates of B are (-4, 9, 6).

Question 16 Maharashtra Board Solution

Prove that the sum of the three vectors determined by the medians of a triangle directed from the vertices is zero.

Solution & Step-by-Step Answer:

Let , and are the position vectors of the vertices A, B and C respectively. Then we know that the position vector of the centroid O of the triangle is Therefore sum of the three vectors , and , is Hence, Sum os the three vectors determined by the medians of a triangle directed from the vertices is zero.

Question 17 Maharashtra Board Solution

ABCD is a parallelogram E, F are the mid points of BC and CD respectively. AE, AF meet the diagonal BD at Q and P respectively. Show that P and Q trisect DB.

Solution & Step-by-Step Answer:

LHS is the position vector of the point on AE and RHS is the position vector of the point on DB. But AE and DB meet at Q. LHS is the position vector of the point on AF and RHS is the position vector of the point on DB. But AF and DB meet at P. ∴ ∴ P divides DB in the ratio 1 : 2 … (5) From (4) and (5), if follows that P and Q trisect DB.

Question 18 Maharashtra Board Solution

If aBC is a triangle whose orthocenter is P and the circumcenter is Q, then prove that + + = 2

Solution & Step-by-Step Answer:

Let G be the centroid of the ∆ ABC. Let A, B, C, G, Q have position vectors , , , , w.r.t. P. We know that Q, G, P are collinear and G divides segment QP internally in the ratio 1 : 2.

Question 19 Maharashtra Board Solution

If P is orthocenter, Q is circumcenter and G is centroid of a triangle ABC, then prove that = 3

Solution & Step-by-Step Answer:

Let and be the position vectors of P and G w.r.t. the circumcentre Q. i.e. = p and = g. We know that Q, G, P are collinear and G divides segment QP internally in the ratio 1 : 2 ∴ by section formula for internal division,

Question 20 Maharashtra Board Solution

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 & Step-by-Step Answer:

Let A, B, D, E, P have position vectors , , , , respectively w.r.t. O. ∵ AD : DB = 2 : 1. ∴ D divides AB internally in the ratio 2 : 1. Using section formula for internal division, we get LHS is the position vector of the point which divides OD internally in the ratio 3 : 2. RHS is the position vector of the point which divides AE internally in the ratio 4 : 1. But OD and AE intersect at P ∴ P divides OD internally in the ratio 3 : 2. Hence, OP : PD = 3 : 2.

Question 21 Maharashtra Board Solution

Dot-product of a vector with vectors and are respectively -1, 6 and 5. Find the vector.

Solution & Step-by-Step Answer:

∴ 3x – 5z= -1 … (1) ∴ 2x + 7y = 6 … (2) ∴ x + y + z = 5 … (3) From (3), z = 5 – x – y Substituting this value of z in (1), we get ∴ 3x – 5(5 – x – y)= -1 ∴ 8x + 5y = 24 … (4) Multiplying (2) by 4 and subtracting from (4), we get 8x + 5y – 4(2x + 7y) = 24 – 6 × 4 ∴ -23y = 0 ∴ y = 0 Substituting y = 0 in (2), we get ∴ 2x = 6 ∴ x = 3 Substituting x = 3 in (1), we get ∴ 3(3) – 5z = -1 ∴ 5z = -10 ∴ z = 2 ∴ Hence, the required vector is

Question 22 Maharashtra Board Solution

If , , are unit vectors such that + + = 0, then find the value of . + . + .

Solution & Step-by-Step Answer:

, , are unit vectors Adding (2), (3), (4) and using the fact that scalar product commutative, we get

Question 23 Maharashtra Board Solution

If a parallelogram is constructed on the vectors , and and angle between and is show that the ratio of the lengths of the sides is :

Solution & Step-by-Step Answer:

Hence, the ratio of the lengths of the sides is : .

Question 24 Maharashtra Board Solution

Express the vector as a sum of two vectors such that one is parallel to the vector and other is perpendicular to .

Solution & Step-by-Step Answer:

By equality of vectors 3m + x = 5 … (1) y = -2 and m – 3x = 5 From (1) and (2) 3m + x = m – 3x ∴ 2m = -4x m ∴ m = -2x Substituting m = -2x in (1), we get ∴ -6x + x = 5 ∴ -5x = 5 ∴ x = -1 ∴ m = -2x = 2

Question 25 Maharashtra Board Solution

Find two unit vectors each of which makes equal angles with , and . , and

Solution & Step-by-Step Answer:

Question 26 Maharashtra Board Solution

Find the acute angles between the curves at their points of intersection. y = x2, y = x3

Solution & Step-by-Step Answer:

The angle between the curves is same as the angle between their tangents at the points of intersection. We find the points of intersection of y = x2 … (1) and y = x3 … (2) From (1) and (2) x3 = x2 ∴ x3 – x2 = 0 ∴ x2(x – 1) = 0 ∴ x = 0 or x = 1 When x = 0, y = 0. When x = 1, y = 1. ∴ equation of tangent to y = x3 at P is y = 0. ∴ the tangents to both curves at (0, 0) are y = 0 ∴ angle between them is 0. Angle at P = (1, 1) Slope of tangent to y = x2 at P ∴ equation of tangent to y = x3 at P is y – 1 = 3(x – 1) y = 3x – 2 We have to find angle between y = 2x – 1 and y = 3x – 2 Lines through origin parallel to these tagents are y = 2x and y = 3x ∴ and These lines lie in XY-plane. ∴ the direction ratios of these lines are 1, 2, 0 and 1, 3, 0. The angle θ between them is given by

Question 27 Maharashtra Board Solution

Find the direction cosines and direction angles of the vector. (i)

Solution & Step-by-Step Answer:

Let =

(ii)
Solution:

Question 28 Maharashtra Board Solution

Let = and be two vectors perpendicular to each other in the XY-plane. Find vectors in the same plane having projection 1 and 2 along and , respectively, are given y.

Solution & Step-by-Step Answer:

Question 29 Maharashtra Board Solution

Show that no line in space can make angle and with X- axis and Y-axis.

Solution & Step-by-Step Answer:

Let, if possible, a line in space make angles and with X-axis and Y-axis. ∴ cos2γ = 1 – This is not possible, because cos γ is real ∴ cos2γ cannot be negative. Hence, there is no line in space which makes angles and with X-axis and Y-axis.

Question 30 Maharashtra Board Solution

Find the angle between the lines whose direction cosines are given by the equation 6mn – 2nl + 5lm = 0, 3l + m + 5n = 0

Solution & Step-by-Step Answer:

Given 6mn – 2nl + 5lm = o 3l + m +5n = 0. From (2), m = 3l – 5n Putting the value of m in equation (1), we get, ⇒ 6n(-3l – 5n) – 2nl + 5l(-3l – 5n) = 0 ⇒ -18nl- 30n – 2nl- 15l2 – 25nl = 0 ⇒ – 30n2 – 45nl – 15l2 = 0 ⇒ 2n2 + 3nl + l2 = 0 ⇒ 2n2 + 2nl + nl + l2 = 0 ⇒ (2n + l) (n + l) = 0 ∴ 2n + l = 0 OR n + l = 0 ∴ l = -2n OR l = -n ∴ l = -2n From (2), 3l + m + 5n = 0 ∴ -6n + m + 5n = 0 ∴ m = n i.e. (-2n, n, n) = (-2, 1, 1) ∴ l = -n ∴ -3n + m + 5n = 0 ∴ m = -2n i.e. (-n, -2n, n) = (1, 2, -1) (a1, b1, c1) = (-2, 1, 1) and (a2, b3, c3) = (1, 2, -1)

Question 31 Maharashtra Board Solution

If Q is the foot of the perpendicular from P(2, 4, 3) on the line joining the points A(1, 2, 4) and B(3, 4, 5), find coordinates of Q.

Solution & Step-by-Step Answer:

Let PQ be the perpendicular drawn from point P(2, 4, 3) to the line joining the points A(1, 2, 4) and B (3, 4, 5). Let Q divides AB internally in the ratio λ : 1 Now, direction ratios of AB are, 3 – 1, 4 – 2, 5 – 4 i.e., 2, 2, 1. Coordinates of Q are,

Question 32 Maharashtra Board Solution

Show that the area of a triangle ABC, the position vectors of whose vertices are a, b and c is Question is modified. Show that the area of a triangle ABC, the position vectors of whose vertices are , and is .

Solution & Step-by-Step Answer:

Consider the triangle ABC. Complete the parallelogram ABDC. Vector area of ∆ABC

Question 33 Maharashtra Board Solution

Find a unit vector perpendicular to the plane containing the point (a, 0, 0), (0, b, 0), and (0, 0, c). What is the area of the triangle with these vertices?

Solution & Step-by-Step Answer:

Question 34 Maharashtra Board Solution

State whether each expression is meaningful. If not, explain why ? If so, state whether it is a vector or a scalar. (a)

Solution & Step-by-Step Answer:

This is the scalar product of two vectors. Therefore, this expression is meaningful and it is a scalar.

(b)
Solution:
This expression is meaningless because is a vector, is a scalar and vector product of vector and scalar is not defined.

(d)
Solution:
This is meaningless because is a vector, is a scalar and scalar product of vector and scalar is not defined.

(e)
Solution:
This is meaningless because are scalars and cross product of two scalars is not defined.

(f)
Solution:
This is scalar product of two vectors. Therefore, this expression is meaningful and it is a scalar.

(g)
Solution:
This is meaningless because is a vector, scalar and scalar product of vector and scalar is not defined.

(h)
Solution:
This is a scalar multiplication of a vector. Therefore, this expression is meaningful and it is a vector.

(i)
Solution:
This is the product of two scalars. Therefore, this expression is meaningful and it is a scalar.

(j)
Solution:
This is the scalar product of two vectors. Therefore, this expression is meaningful and it is a scalar.

(k)
Solution:
This is the sum of scalar and vector which is not defined. Therefore, this expression is meaningless.

(l)
Solution:
This is meaningless because is a vector, is a scalar and the scalar product of vector and scalar is not defined.

Question 35 Maharashtra Board Solution

Show that, for any vectors Question is modified. For any vectors show that {aligned} &({a}+{b}+{c}) × {c}+({a}+{b}+{c}) × {b}+({b}-{c}) × {a} \\ &=2 {a} × {c}. {aligned}

Solution & Step-by-Step Answer:

Question 36 Maharashtra Board Solution

Suppose that = 0. (a) If then is ?

Solution & Step-by-Step Answer:

(b) If then is ?
Solution:

Question 37 Maharashtra Board Solution

If A(3, 2, -1), B(-2, 2, -3), C(3, 5, -2), D(-2, 5, -4) then (i) verify that the points are the vertices of a parallelogram and

Solution & Step-by-Step Answer:

∴ opposite sides AB and DC of ABCD are parallel and equal. ∴ ABCD is a parallelogram.

(ii) find its area.
Solution:

Question 38 Maharashtra Board Solution

Let A, B, C, D be any four points in space. Prove that = 4 (area of ∆ABC)

Solution & Step-by-Step Answer:

Let A, B, C, D have position vectors , , , respectively. Consider

Question 39 Maharashtra Board Solution

Let be unit vectors such that and the angle between and be. Prove that

Solution & Step-by-Step Answer:

∴ is perpendicular to and both ∴ is parallel to × ∴ = m( × ), m is a scalar.

Question 40 Maharashtra Board Solution

Find the value of ‘a’ so that the volume of parallelopiped a formed by aand becomes minimum. Question is modified. Find the value of ‘a’ so that the volume of parallelopiped formed by and becomes minimum.

Solution & Step-by-Step Answer:

Let = , = , = Let V be the volume of the parallelopiped formed by .

Question 41 Maharashtra Board Solution

Find the volume of the parallelepiped spanned by the diagonals of the three faces of a cube of side a that meet at one vertex of the cube.

Solution & Step-by-Step Answer:

Take origin O as one vertex of the cube and OA, OB and OC as the positive directions of the X-axis, the Y-axis and the Z-axis respectively. Here, the sides of the cube are OA = OB = OC = a ∴ the coordinates of all the vertices of the cube will be O = (0, 0, 0) A = (a, 0, 0) B = (0, a, 0) C = (0, 0, a) N = (a, a, 0) L = (0, a, a) M = (a, 0, a) P = (a, a, a) ON, OL, OM are the three diagonals which meet at the vertex O

Question 42 Maharashtra Board Solution

If are three non-coplanar vectors, then show that = 0

Solution & Step-by-Step Answer:

Question 43 Maharashtra Board Solution

Prove that

Solution & Step-by-Step Answer:

Question 44 Maharashtra Board Solution

Find the volume of a parallelopiped whose coterminus edges are represented by the vector and . Also find volume of tetrahedron having these coterminous edges.

Solution & Step-by-Step Answer:

Let = , = and = be the co-terminus edges of a parallelopiped. Then volume of the parallelopiped = = = 0(0 – 1) – 1(0 – 1) + 1(1 – 0) = 0 + 1 + 1 = 2cu units. Also, volume of tetrahedron = = cubic units.

Question 45 Maharashtra Board Solution

Using properties of scalar triple product, prove that .

Solution & Step-by-Step Answer:

Question 46 Maharashtra Board Solution

If four points A(), B(), C() and D() are coplanar then show that

Solution & Step-by-Step Answer:

Question 47 Maharashtra Board Solution

If and are three non coplanar vectors, then .

Solution & Step-by-Step Answer:

Question 48 Maharashtra Board Solution

If in a tetrahedron, edges in each of the two pairs of opposite edges are perpendicular, then show that the edges in the third pair are also perpendicular.

Solution & Step-by-Step Answer:

Let O-ABC be a tetrahedron. Then o (OA, BC), (OB, CA) and (OC, AB) are the pair of opposite edges. Take O as the origin of reference and let and ∴ the third pair (OC, AB) is perpendicular.

Chapter 5 Vectors Miscellaneous Exercise 5 Solutions