Vectors Class 12 Maths 1 Miscellaneous Exercise 5 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Miscellaneous Exercise 5 Questions and Answers.

12th Maths Part 1 Vectors Miscellaneous Exercise 5 Questions And Answers Maharashtra Board

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

Question 2.
If |\(\bar{a}\)| = 3, |\(\bar{b}\)| = 4, then the value of λ for which \(\bar{a}\) + λ\(\bar{b}\) is perpendicular to \(\bar{a}\) – λ\(\bar{b}\), is
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-1
Solution:
(b) \(\frac{3}{4}\)

Question 3.
If sum of two unit vectors is itself a unit vector, then the magnitude of their difference is
(A) \(\sqrt {2}\)
(B) \(\sqrt {3}\)
(C) 1
(D) 2
Solution:
(B) \(\sqrt {3}\)

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Question 4.
If |\(\bar{a}\)| = 3, |\(\bar{b}\)| = 5, |\(\bar{c}\)| = 7 and \(\bar{a}\) + \(\bar{b}\) + \(\bar{c}\) = 0, then the angle between \(\bar{a}\) and \(\bar{b}\) is
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-2
Solution:
(b) \(\frac{\pi}{3}\)

Question 5.
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) \(\frac{\pi}{3}\)
(C) 1
(D) 5
Solution:
(A) 7

Question 6.
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:
(D) 60º or 120º

Question 7.
The distance of the point (3, 4, 5) from Y- axis is
(A) 3
(B) 5
(C) \(\sqrt {34}\)
(D) \(\sqrt {41}\)
Solution:
(C) \(\sqrt {34}\)

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Question 8.
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, \(\frac{-13}{2}\)
(C) 10, 5, -2
(D) 3, 5, 11
Solution:
(A) 4, 3, -5

Question 9.
If cos α, cos β, cos γ are the direction cosines of a line then the value of sin 2 α + sin 2 β + sin 2 γ is
(A) 1
(B) 2
(C) 3
(D) 4
Solution:
(B) 2

Question 10.
If l, m, n are direction cosines of a line then \(\hat{l}+m \hat{j}+n \hat{k}\) 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:
(B) the unit vector along the line

Question 11.
If |\(\bar{a}\)| = 3 and –1 ≤ k ≤ 2, then |k\(\bar{a}\)| lies in the interval
(A) [0, 6]
(B) [-3, 6]
(C) [3, 6]
(D) [1, 2]
Solution:
(A) [0, 6]

Question 12.
Let α, β, γ be distinct real numbers. The points with position vectors \(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}\), \(\beta \hat{i}+\gamma \hat{j}+\alpha \hat{k}\), \(\gamma \hat{i}+\alpha \hat{j}+\beta \hat{k}\)
(A) are collinear
(B) form an equilateral triangle
(C) form a scalene triangle
(D) form a right angled triangle
Solution:
(B) form an equilateral triangle

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Question 13.
Let \(\bar{p}\) and \(\bar{q}\) be the position vectors of P and Q respectively, with respect to O and |\(\bar{p}\)| = p, |\(\bar{q}\)| = 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) 9p 2 = 4q 2
(B) 4p 2 = 9q 2
(C) 9p = 4q
(D) 4p = 9q
Solution:
(A) 9p 2 = 4q 2

Question 14.
The 2 vectors \(\hat{j}+\hat{k}\) and \(3 \hat{i}-\hat{j}+4 \hat{k}\) represents the two sides AB and AC, respectively of a ∆ABC. The length of the median through A is
(A) \(\frac{\sqrt{34}}{2}\)
(B) \(\frac{\sqrt{48}}{2}\)
(C) \(\sqrt {18}\)
(D) None of these
Solution:
(A) \(\frac{\sqrt{34}}{2}\)

Question 15.
If \(\bar{a}\) and \(\bar{b}\) are unit vectors, then what is the angle between \(\bar{a}\) and \(\bar{b}\) for \(\sqrt{3} \bar{a}\) – \(\bar{b}\) to be a unit vector ?
(A) 30º
(B) 45º
(C) 60º
(D) 90º
Solution:
(A) 30º

Question 16.
If θ be the angle between any two vectors \(\bar{a}\) and \(\bar{b}\), then \(|\vec{a} \cdot \vec{b}|\) = \(|\vec{a} \times \vec{b}|\), when θ is equal to
(A) 0
(B) \(\frac{\pi}{4}\)
(C) \(\frac{\pi}{2}\)
(D) π
Solution:
(B) \(\frac{\pi}{4}\)

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Question 17.
The value of \(\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})\)
(A) 0
(B) -1
(C) 1
(D) 3
Solution:
(C) 1

Question 18.
Let a, b, c be distinct non-negative numbers. If the vectors \(\mathrm{a} \hat{i}+\mathrm{a} \hat{j}+\mathrm{c} \hat{k}\), \(\hat{i}+\hat{k}\) and \(\mathbf{c} \hat{i}+\mathrm{c} \hat{j}+\mathrm{b} \hat{k}\) 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:
(B) The geometric mean of a and b

Question 19.
Let \(\bar{a}\) = \(\hat{i} \hat{j}\), \(\bar{b}\) = \(\hat{j} \hat{k}\), \(\bar{c}\) = \(\hat{k} \hat{i}\). If \(\bar{d}\) is a unit vector such that \(\bar{a} . \bar{d}=0=\left[\begin{array}{lll}
\bar{b} & \bar{c} & \bar{d}
\end{array}\right]\), then \(\bar{d}\) equals.
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-3
Solution:
(a) \(\pm \frac{\hat{i}+\hat{j}-2 \hat{k}}{\sqrt{6}}\)

Question 20.
If \(\bar{a}\) \(\bar{b}\) \(\bar{c}\) are non coplanar unit vectors such that \(\bar{a} \times(\bar{b} \times \bar{c})\) =\(\frac{(\bar{b}+\bar{c})}{\sqrt{2}}\) then the angle between \(\bar{a}\) and \(\bar{b}\) is
(A) \(\frac{3 \pi}{4}\)
(B) \(\frac{\pi}{4}\)
(C) \(\frac{\pi}{2}\)
(D) π
Solution:
(A) \(\frac{3 \pi}{4}\)

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II Answer the following :
1) ABCD is a trapezium with AB parallel to DC and DC = 3AB. M is the mid-point of DC,
\(\overline{A B}\) = \(\bar{p}\) and \(\overline{B C}\) = \(\bar{q}\). Find in terms of \(\bar{p}\) and \(\bar{q}\).
(i) \(\overline{A M}\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-4

(ii) \(\overline{B D}\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-5

(iii) \(\overline{M B}\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-6

(iv) \(\overline{D A}\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-7

Question 2.
The points A, B and C have position vectors \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) respectively. The point P is midpoint of AB. Find in terms of \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) the vector \(\overline{P C}\)
Solution:
P is the mid-point of AB.
∴ \(\bar{p}\) = \(=\frac{\bar{a}+\bar{b}}{2}\), where \(\bar{p}\) is the position vector of P.
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-8

Question 3.
In a pentagon ABCDE
Show that \(\overline{A B}\) + \(\overline{A E}\) + \(\overline{B C}\) + \(\overline{D C}\) + \(\overline{E D}\) = 2\(\overline{A C}\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-9

Question 4.
If in parallelogram ABCD, diagonal vectors are \(\overline{A C}\) = \(2 \hat{i}+3 \hat{j}+4 \hat{k}\) and \(\overline{B D}\) = \(-6 \hat{i}+7 \hat{j}-2 \hat{k}\), then find the adjacent side vectors \(\overline{A B}\) and \(\overline{A D}\)
Solution:
ABCD is a parallelogram
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-10
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-11

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Question 5.
If two sides of a triangle are \(\hat{i}+2 \hat{j}\) and \(\hat{i}+\hat{k}\), then find the length of the third side.
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-12
Let ABC be a triangle with \(\overline{A B}\) = \(\hat{i}+2 \hat{j}\), \(\overline{B C}\) = \(\hat{i}+\hat{k}\).
By triangle law of vectors
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-13
Hence, the length of third side is 3 units.

Question 6.
If |\(\bar{a}\)| = |\(\bar{b}\) | = 1 \(\bar{a}\).\(\bar{b}\) = 0 and \(\bar{a}\) + \(\bar{b}\) + \(\bar{c}\) = 0 then find |\(\bar{c}\)|
Solution:
\(\bar{a}\) + \(\bar{b}\) + \(\bar{c}\) = 0
∴ –\(\bar{c}\) = \(\bar{a}\) + \(\bar{b}\)
Taking dot product of both sides with itself, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-14

Question 7.
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:
The position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) of the points A, B, C are
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-15
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-16
∴ ∆ ABC is right angled at A.

(ii) L(3, -2, -3), M(7, 0, 1), N (1, 2, 1)
Solution:
The position vectors bar \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) of the points L M, N are
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-17
l(LM) = 6, l(MN) = 2\(\sqrt {10}\) , l(NL) = 6
∆LMN is sosceles

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Question 8.
Find the component form of if a if
(i) It lies in YZ plane and makes 60º with positive Y-axis and |\(\bar{a}\)| = 4
Solution:
Let α, β, γ be the direction angles of \(\bar{a}\)
Since \(\bar{a}\) lies in YZ-plane, it is perpendicular to X-axis
∴ α = 90°
It is given that β= 60°
∵ cos 2 α + cos 2 β + cos 2 γ = 1
∴ cos 2 90° + cos 2 60° + cos 2 γ = 1
∴ 0 + \(\left(\frac{1}{2}\right)^{2}\) + cos 2 γ = 1
∴ cos 2 γ = 1 – \(\frac{1}{4}=\frac{3}{4}\)
∴ cos γ = \(\pm \frac{\sqrt{3}}{2}\)
Unit vector along a is given by
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-18
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-19

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

Question 9.
Two sides of a parallelogram are \(3 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(-2 \hat{j}+7 \hat{k}\). Find the unit vectors parallel to the diagonals.
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-20
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-21

Question 10.
If D, E, F are the mid-points of the sides BC, CA, AB of a triangle ABC , prove that \(\overline{A D}\) + \(\overline{B E}\) + \(\overline{C F}\) = 0
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-22
Solution:
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{d}\), \(\bar{e}\), \(\bar{f}\) 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
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-23

Question 11.
Find the unit vectors that are parallel to the tangent line to the parabola y = x 2 at the point (2, 4)
Solution:
Differentiating y = x 2 w.r.t. x, we get \(\) = 2x
Slope of tangent at P(2, 4) = \(\left(\frac{d y}{d x}\right)_{\text {at } \mathrm{P}(2, 4)}\) = 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 ∴ \(\frac{x}{1}=\frac{y}{4}\), z = 0
∴ the direction ratios of this line are 1, 4, 0
∴ its direction cosines are
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-24

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Question 12.
Express the vector \(\hat{i}+4 \hat{j}-4 \hat{k}\) as a linear combination of the vectors \(2 \hat{i}-\hat{j}+3 \hat{k}\), \(\hat{i}-2 \hat{j}+4 \hat{k}\) and \(-\hat{i}+3 \hat{j}-5 \hat{k}\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-25
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 = \(\left|\begin{array}{ccc}
2 & 2 & -1 \\
-1 & -2 & 3 \\
3 & 4 & -5
\end{array}\right|\)
= 2(10 – 12) – 2(5 – 9) – 1(-4 + 6)
= -4 + 8 – 2
= 2 ≠ 0
D x = \(\left|\begin{array}{ccc}
1 & 2 & -1 \\
4 & -2 & 3 \\
-4 & 4 & -5
\end{array}\right|\)
= 1(10 – 12) – 2(-20 + 12) – 1 (16 – 8)
= -2 + 16 – 8
= 6
D y = \(\left|\begin{array}{ccc}
2 & 1 & -1 \\
-1 & 4 & 3 \\
3 & -4 & -5
\end{array}\right|\)
= 2(-20 + 12) – 1(5 – 9) – 1(4 – 12)
= -16 – 4 – 8
= -28
D z = \(\left|\begin{array}{ccc}
2 & 2 & 1 \\
-1 & -2 & 4 \\
3 & 4 & -4
\end{array}\right|\)
= 2(8 – 16) – 2(4 – 12) + 1(-4 + 6)
= -16 – 16 + 2
= -30
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-26

Question 13.
If \(\overline{O A}\) = \(\bar{a}\) and \(\overline{O B}\) = \(\bar{b}\) then show that the vector along the angle bisector of angle AOB is
given by \(\bar{d}\) = λ\(\left(\frac{\bar{a}}{|\bar{b}|}+\frac{\bar{b}}{|\bar{b}|}\right)\)
Question is modified
If \(\overline{O A}\) = \(\bar{a}\) and \(\overline{O B}\) = \(\bar{b}\) then show that the vector along the angle bisector of ∠AOB is
given by \(\bar{d}\) = λ\(\left(\frac{\bar{a}}{|\bar{a}|}+\frac{\bar{b}}{|\bar{b}|}\right)\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-27
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 \(\hat{a}\) and \(\hat{b}\) along these directions,
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-28

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Question 14.
The position vectors f three consecutive vertices of a parallelogram are \(\hat{i}+\hat{j}+\hat{k}\), \(\hat{i}+3 \hat{j}+5 \hat{k}\) and \(7 \hat{i}+9 \hat{j}+11 \hat{k}\) Find the position vector of the fourth vertex.
Solution:
Let ABCD be a parallelogram.
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{d}\) be the position vectors of the vertices
A, B, C, D of the parallelogram,
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-29
Hence, the position vector of the fourth vertex is 7(\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)).

Question 15.
A point P with position vector \(\frac{-14 \hat{i}+39 \hat{j}+28 \hat{k}}{5}\) divides the line joining A(-1, 6, 5) and B in the ratio 3 : 2 then find the point B.
Solution:
Let A, B and P have position vectors \(\bar{a}\), \(\bar{b}\) and \(\bar{p}\) respectively.
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-30
∴ coordinates of B are (-4, 9, 6).

Question 16.
Prove that the sum of the three vectors determined by the medians of a triangle directed from the vertices is zero.
Solution:
Let \(\overrightarrow{\mathrm{a}}\), \(\overrightarrow{\mathrm{b}}\) and \(\overrightarrow{\mathrm{c}}\) 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 \(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\)
Therefore sum of the three vectors \(\overrightarrow{\mathrm{OA}}\), \(\overrightarrow{\mathrm{OB}}\) and \(\overrightarrow{\mathrm{OC}}\), is
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-31
Hence, Sum os the three vectors determined by the medians of a triangle directed from the vertices is zero.

Question 17.
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:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-32
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-33
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.
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-34
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.
∴ \(\bar{p}=\frac{\bar{b}+2 \bar{d}}{1+2}\)
∴ P divides DB in the ratio 1 : 2 … (5)
From (4) and (5), if follows that P and Q trisect DB.

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Question 18.
If aBC is a triangle whose orthocenter is P and the circumcenter is Q, then prove that \(\overline{P A}\) + \(\overline{P C}\) + \(\overline{P B}\) = 2 \(\overline{P Q}\)
Solution:
Let G be the centroid of the ∆ ABC.
Let A, B, C, G, Q have position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{g}\), \(\bar{q}\) w.r.t. P. We know that Q, G, P are collinear and G divides segment QP internally in the ratio 1 : 2.
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-35

Question 19.
If P is orthocenter, Q is circumcenter and G is centroid of a triangle ABC, then prove that \(\overline{Q P}\) = 3\(\overline{Q G}\)
Solution:
Let \(\bar{p}\) and \(\bar{g}\) be the position vectors of P and G w.r.t. the circumcentre Q.
i.e. \(\overline{\mathrm{QP}}\) = p and \(\overline{\mathrm{QG}}\) = 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,
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-36

Question 20.
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:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-37
Let A, B, D, E, P have position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{d}\), \(\bar{e}\), \(\bar{p}\) 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
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-38
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.
Dot-product of a vector with vectors \(3 \hat{i}-5 \hat{k}, 2 \hat{i}+7 \hat{j}\) and \(\hat{i}+\hat{j}+\hat{k}\) are respectively -1, 6 and 5. Find the vector.
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-39
∴ 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
∴ \(\bar{r}=3 \hat{i}+0 \cdot \hat{j}+2 \hat{k}=3 \hat{i}+2 \hat{k}\)
Hence, the required vector is \(3 \hat{i}+2 \hat{k}\)

Maharashtra-Board-Solutions

Question 22.
If \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are unit vectors such that \(\bar{a}\) + \(\bar{b}\) + \(\bar{c}\) = 0, then find the value of \(\bar{a}\).\(\bar{b}\) + \(\bar{b}\).\(\bar{c}\) + \(\bar{c}\).\(\bar{a}\)
Solution:
\(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are unit vectors
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-40
Adding (2), (3), (4) and using the fact that scalar product commutative, we get
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-41

Question 23.
If a parallelogram is constructed on the vectors \(\bar{a}=3 \bar{p}-\bar{q}\), \(\bar{b}=\bar{p}+3 \bar{q}\) and \(|\bar{p}|=|\bar{q}|=2\) and angle between \(\bar{p}\) and \(\bar{q}\) is\(\frac{\pi}{3}\) show that the ratio of the lengths of the sides is \(\sqrt {7}\) : \(\sqrt {13}\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-42
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-43
Hence, the ratio of the lengths of the sides is \(\sqrt {7}\) : \(\sqrt {13}\).

Question 24.
Express the vector \(\bar{a}=5 \hat{i}-2 \hat{j}+5 \hat{k}\) as a sum of two vectors such that one is parallel to the vector \(\bar{b}=3 \hat{i}+\hat{k}\) and other is perpendicular to \(\bar{b}\).
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-44
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
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-45

Maharashtra-Board-Solutions

Question 25.
Find two unit vectors each of which makes equal angles with \(\bar{u}\), \(\bar{v}\) and \(\bar{w}\). \(\bar{u}=2 \hat{i}+\hat{j}-2 \hat{k}\), \(\bar{v}=\hat{i}+2 \hat{j}-2 \hat{k}\) and \(\bar{W}=2 \hat{i}-2 \hat{j}+\hat{k}\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-46
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-47
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-48
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-49

Question 26.
Find the acute angles between the curves at their points of intersection. y = x 2 , y = x 3
Solution:
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 = x 2 … (1)
and y = x 3 … (2)
From (1) and (2)
x 3 = x 2
∴ x 3 – x 2 = 0
∴ x 2 (x – 1) = 0
∴ x = 0 or x = 1
When x = 0, y = 0.
When x = 1, y = 1.
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-50
∴ equation of tangent to y = x 3 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 = x 2 at P
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-51
∴ equation of tangent to y = x 3 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
∴ \(\frac{x}{1}=\frac{y}{2}\) and \(\frac{x}{1}=\frac{y}{3}\)
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
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-52

Question 27.
Find the direction cosines and direction angles of the vector.
(i) \(2 \hat{i}+\hat{j}+2 \hat{k}\)
Solution:
Let \(\bar{a}\) = \(2 \hat{i}+\hat{j}+2 \hat{k}\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-53

(ii) \((1 / 2) \hat{i}+\hat{j}+\hat{k}\)
Solution:

Maharashtra-Board-Solutions

Question 28.
Let \(\bar{b}\) = \(4 \hat{i}+3 \hat{j}\) and \(\bar{c}\) be two vectors perpendicular to each other in the XY-plane. Find vectors in the same plane having projection 1 and 2 along \(\bar{b}\) and \(\bar{c}\), respectively, are given y.
Solution:
\(\bar{b}\) = \(4 \hat{i}+3 \hat{j}\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-54
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-55
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-56

Question 29.
Show that no line in space can make angle \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\) with X- axis and Y-axis.
Solution:
Let, if possible, a line in space make angles \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\) with X-axis and Y-axis.
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-57
∴ cos 2 γ = 1 – \(\frac{3}{4}-\frac{1}{2}=-\frac{1}{4}\)
This is not possible, because cos γ is real
∴ cos 2 γ cannot be negative.
Hence, there is no line in space which makes angles \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\) with X-axis and Y-axis.

Question 30.
Find the angle between the lines whose direction cosines are given by the equation 6mn – 2nl + 5lm = 0, 3l + m + 5n = 0
Solution:
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
⇒ – 30n 2 – 45nl – 15l 2 = 0
⇒ 2n 2 + 3nl + l 2 = 0
⇒ 2n 2 + 2nl + nl + l 2 = 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)
(a 1 , b 1 , c 1 ) = (-2, 1, 1) and (a 2 , b 3 , c 3 ) = (1, 2, -1)
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-89

Maharashtra-Board-Solutions

Question 31.
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:
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
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-59
Now, direction ratios of AB are, 3 – 1, 4 – 2, 5 – 4 i.e., 2, 2, 1.
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-60
Coordinates of Q are,
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-61

Question 32.
Show that the area of a triangle ABC, the position vectors of whose vertices are a, b and c is \(\frac{1}{2}[\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}]\)
Question is modified.
Show that the area of a triangle ABC, the position vectors of whose vertices are \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) is \(\frac{1}{2}[\bar{a} \times \bar{b}+\bar{b} \times \bar{c}+\bar{c} \times \bar{a}]\).
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-62
Consider the triangle ABC.
Complete the parallelogram ABDC.
Vector area of ∆ABC
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-63
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-64

Question 33.
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:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-65
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-66

Question 34.
State whether each expression is meaningful. If not, explain why ? If so, state whether it is a vector or a scalar.
(a) \(\bar{a} \cdot(\bar{b} \times \bar{c})\)
Solution:
This is the scalar product of two vectors. Therefore, this expression is meaningful and it is a scalar.

(b) \(\bar{a} \times(\bar{b} \cdot \bar{c})\)
Solution:
This expression is meaningless because \(\bar{a}\) is a vector, \(\bar{b} \cdot \bar{c}\) is a scalar and vector product of vector and scalar is not defined.

(c) \(\bar{a} \times(\bar{b} \times \bar{c})\)
Solution:
This is vector product of two vectors. Therefore, this expression is meaningful and it is a vector.

(d) \(\bar{a} \cdot(\bar{b} \cdot \bar{c})\)
Solution:
This is meaningless because \(\bar{a}\) is a vector, \(\bar{b} \cdot \bar{c}\) is a scalar and scalar product of vector and scalar is not defined.

Maharashtra-Board-Solutions

(e) \((\bar{a} \cdot \bar{b}) \times(\bar{c} \cdot \bar{d})\)
Solution:
This is meaningless because \(\bar{a} \cdot \bar{b}, \bar{c} \cdot \bar{d}\) are scalars and cross product of two scalars is not defined.

(f) \((\bar{a} \times \bar{b}) \cdot(\bar{c} \times \bar{d})\)
Solution:
This is scalar product of two vectors. Therefore, this expression is meaningful and it is a scalar.

(g) \((\bar{a} \cdot \bar{b}) \cdot \bar{c}\)
Solution:
This is meaningless because \(\bar{c}\) is a vector, \(\bar{a} \cdot \bar{b}\) scalar and scalar product of vector and scalar is not defined.

(h) \((\bar{a} \cdot \bar{b}) \bar{c}\)
Solution:
This is a scalar multiplication of a vector. Therefore, this expression is meaningful and it is a vector.

(i) \((|\bar{a}|)(\bar{b} \cdot \bar{c})\)
Solution:
This is the product of two scalars. Therefore, this expression is meaningful and it is a scalar.

(j) \(\bar{a} \cdot(\bar{b}+\bar{c})\)
Solution:
This is the scalar product of two vectors. Therefore, this expression is meaningful and it is a scalar.

(k) \(\bar{a} \cdot \bar{b}+\bar{c}\)
Solution:
This is the sum of scalar and vector which is not defined. Therefore, this expression is meaningless.

(l) \(|\bar{a}| \cdot(\bar{b}+\bar{c})\)
Solution:
This is meaningless because \(\bar{a}\) is a vector, \(\overline{\mathrm{b}}+\overline{\mathrm{c}}\) is a scalar and the scalar product of vector and scalar is not defined.

Maharashtra-Board-Solutions

Question 35.
Show that, for any vectors \(\bar{a}, \bar{b}, \bar{c}\)
\((\bar{a}+\bar{b}+\bar{c}) \times \bar{c}+(\bar{a}+\bar{b}+\bar{c}) \times \bar{b}+(\bar{b}+\bar{c}) \times \bar{a}=2 \bar{a} \times \bar{c}\)
Question is modified.
For any vectors \(\bar{a}, \bar{b}, \bar{c}\) show that
\(\begin{aligned}
&(\bar{a}+\bar{b}+\bar{c}) \times \bar{c}+(\bar{a}+\bar{b}+\bar{c}) \times \bar{b}+(\bar{b}-\bar{c}) \times \bar{a} \\
&=2 \bar{a} \times \bar{c} .
\end{aligned}\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-67

Question 36.
Suppose that \(\bar{a}\) = 0.
(a) If \(\bar{a} \cdot \bar{b}=\bar{a} \cdot \bar{c}\) then is \(\bar{b}=\bar{c}\)?
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-68

(b) If \(\bar{a} \times \bar{b}=\bar{a} \times \bar{c}\) then is \(\bar{b}=\bar{c}\)?
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-69

(c) If \(\bar{a} \cdot \bar{b}=\bar{a} \cdot \bar{c}\) and \(\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\overline{\mathrm{a}} \times \overline{\mathrm{c}}\) then is \(\bar{b}=\bar{c}\)?
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-70

Question 37.
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:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-71
∴ opposite sides AB and DC of ABCD are parallel and equal.
∴ ABCD is a parallelogram.

Maharashtra-Board-Solutions

(ii) find its area.
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-72
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-73

Question 38.
Let A, B, C, D be any four points in space. Prove that \(|\overline{A B} \times \overline{C D}+\overline{B C} \times \overline{A D}+\overline{C A} \times \overline{B D}|\) = 4 (area of ∆ABC)
Solution:
Let A, B, C, D have position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{d}\) respectively.
Consider \(|\overline{A B} \times \overline{C D}+\overline{B C} \times \overline{A D}+\overline{C A} \times \overline{B D}|\)
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-74

Question 39.
Let \(\hat{a}, \hat{b}, \hat{c}\) be unit vectors such that \(\hat{a} \cdot \hat{b}=\hat{a} \cdot \hat{c}=0\) and the angle between \(\hat{b}\) and \(\hat{c}\) be\(\frac{\pi}{6}\).
Prove that \(\hat{a}=\pm 2(\hat{b} \times \hat{c})\)
Solution:
\(\hat{a} \cdot \hat{b}=\hat{a} \cdot \hat{c}=0\)
∴ \(\hat{a}\) is perpendicular to \(\hat{b}\) and \(\hat{c}\) both
∴ \(\hat{a}\) is parallel to \(\hat{b}\) × \(\hat{c}\)
∴ \(\hat{a}\) = m(\(\hat{b}\) × \(\hat{c}\)), m is a scalar.
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-75

Question 40.
Find the value of ‘a’ so that the volume of parallelopiped a formed by \(\hat{i}+\hat{j}+\hat{k}+a \hat{k}\) aand \(a j+\hat{k}\) becomes minimum.
Question is modified.
Find the value of ‘a’ so that the volume of parallelopiped formed by \(\hat{i}+a \hat{j}+\hat{k}, \hat{j}+a \hat{k}\) and \(a \hat{i}+\hat{k}\) becomes minimum.
Solution:
Let \(\bar{p}\) = \(\hat{i}+a \hat{j}+\hat{k}\), \(\bar{q}\) = \(\hat{j}+a \hat{k}\), \(\bar{r}\) = \(a \hat{i}+\hat{k}\)
Let V be the volume of the parallelopiped formed by \(\bar{p}, \bar{q}, \bar{r}\).
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-76
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-77

Maharashtra-Board-Solutions

Question 41.
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:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-78
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
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-79

Question 42.
If \(\bar{a}, \bar{b}, \bar{c}\) are three non-coplanar vectors, then show that \(\frac{\bar{a} \cdot(\bar{b} \times \bar{c})}{(\bar{c} \times \bar{a}) \cdot \bar{b}}+\frac{\bar{b} \cdot(\bar{a} \times \bar{c})}{(\bar{c} \times \bar{a}) \cdot \bar{b}}\) = 0
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-80

Question 43.
Prove that \((\bar{a} \times \bar{b}) \cdot(\bar{c} \times \bar{d})\left|\begin{array}{ll}
\bar{a} \cdot \bar{c} & \bar{b} \cdot \bar{c} \\
\bar{a} \cdot \bar{d} & \bar{b} \cdot \bar{d}
\end{array}\right|\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-81
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-82

Question 44.
Find the volume of a parallelopiped whose coterminus edges are represented by the vector \(\hat{j}+\hat{k} \cdot \hat{i}+\hat{k}\) and \(\hat{i}+\hat{j}\). Also find volume of tetrahedron having these coterminous edges.
Solution:
Let \(\bar{a}\) = \(\hat{j}+\hat{k}\), \(\bar{b}\) = \(\hat{i}+\hat{k}\) and \(\bar{c}\) = \(\hat{i}+\hat{j}\) be the co-terminus edges of a parallelopiped.
Then volume of the parallelopiped = \([\bar{a} \bar{b} \bar{c}]\)
= \(\left|\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right|\)
= 0(0 – 1) – 1(0 – 1) + 1(1 – 0)
= 0 + 1 + 1 = 2cu units.
Also, volume of tetrahedron = \(\frac{1}{6}[\bar{a} \bar{b} \bar{c}]\)
= \(\frac{1}{6}(2)=\frac{1}{3}\) cubic units.

Maharashtra-Board-Solutions

Question 45.
Using properties of scalar triple product, prove that \(\left[\begin{array}{llll}
\bar{a}+\bar{b} & \bar{b}+\bar{c} & \bar{c}+\bar{a}
\end{array}\right]=2\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\).
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-83

Question 46.
If four points A(\(\bar{a}\)), B(\(\bar{b}\)), C(\(\bar{c}\)) and D(\(\bar{d}\)) are coplanar then show that \(\left[\begin{array}{lll}
\bar{a} \bar{b} \bar{d}]+\left[\begin{array}{lll}
\bar{b} & \bar{c} & \bar{d}
\end{array}\right]+\left[\begin{array}{lll}
\bar{c} & \bar{a} & \bar{d}
\end{array}\right]=[\overline{\bar{a}} \bar{b} & \bar{c}
\end{array}\right]\)
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-84
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-85

Question 47.
If \(\bar{a}\) \(\bar{b}\) and \(\bar{c}\) are three non coplanar vectors, then \((\bar{a}+\bar{b}+\bar{c}) \cdot[(\bar{a}+\bar{b}) \times(\bar{a}+\bar{c})]=-\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\).
Solution:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-86

Maharashtra-Board-Solutions

Question 48.
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:
Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-87
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 \(\bar{a}\) \(\bar{b}\) and Maharashtra-Board-12th-Maths-Solutions-Chapter-5-Vectors-Miscellaneous-Exercise-5-88
∴ the third pair (OC, AB) is perpendicular.