Vectors Class 12 Maths 1 Exercise 5.1 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Ex 5.1 Questions and Answers.

12th Maths Part 1 Vectors Exercise 5.1 Questions And Answers Maharashtra Board

Question 1.
The vector $\bar{a}$ is directed due north and $|\bar{a}|$ = 24. The vector $\bar{b}$ is directed due west and $|\bar{b}|$ = 7. Find $|\bar{a}+\bar{b}|$.
Solution:

Let $\overline{\mathrm{AB}}$ = $\bar{a}$, $\overline{\mathrm{BC}}$ = $\bar{b}$
Then $\overline{\mathrm{AC}}$ = $\overline{\mathrm{AB}}$ + $\overline{\mathrm{BC}}$ = a + b
Given : $|\bar{a}|$ = $|\overline{\mathrm{AB}}|$ = l(AB) = 24 and
$|\bar{b}|$ = $|\overline{\mathrm{BC}}|$ = l(BC) = 7
∴ ∠ABC = 90°
∴ [l(AC)]² = [l(AB)]² + [l(BC)]²
= (24)² + (7)² = 625
∴ l(AC) = 25 ∴ $|\overline{\mathrm{AC}}|$ = 25
∴ $|\bar{a}+\bar{b}|$ = $|\overline{\mathrm{AC}}|$ = 25.

Question 2.
In the triangle PQR, $\overline{\mathrm{PQ}}$ = 2$\bar{a}$ and $\overline{\mathrm{QR}}$ = 2$\bar{b}$. The mid-point of PR is M. Find following vectors in terms of $\bar{a}$ and $\bar{b}$.
(i) $\overline{\mathrm{PR}}$
Solution:

Given : $\overline{\mathrm{PQ}}$ = 2$\bar{a}$, $\overline{\mathrm{QR}}$ = 2$\bar{b}$
(i) $\overline{\mathrm{PR}}$ = $\overline{\mathrm{PQ}}$ + $\overline{\mathrm{QR}}$
= 2$\bar{a}$ + 2$\bar{a}$.

(ii) $\overline{\mathrm{PM}}$
Solution:
∵ M is the midpoint of PR
∴ $\overline{\mathrm{PM}}$ = $\frac{1}{2} \overline{\mathrm{PR}}$ = $\frac{1}{2}$[2$\bar{a}$ + 2$\bar{b}$]
= $\bar{a}$ + $\bar{b}$.

(iii) $\overline{\mathrm{QM}}$
Solution:
$\overline{\mathrm{RM}}$ = $\frac{1}{2}(\overline{\mathrm{RP}})$ = $-\frac{1}{2} \overline{\mathrm{PR}}$ = $-\frac{1}{2}$(2$\bar{a}$ + 2$\bar{b}$)
= –$\bar{a}$ – $\bar{b}$
∴ $\overline{\mathrm{QM}}$ = $\overline{\mathrm{QR}}$ + $\overline{\mathrm{RM}}$
= 2$\bar{b}$ – $\bar{a}$ – $\bar{b}$
= $\bar{b}$ – $\bar{a}$.

Question 3.
OABCDE is a regular hexagon. The points A and B have position vectors $\bar{a}$ and $\bar{b}$ respectively, referred to the origin O. Find, in terms of $\bar{a}$ and $\bar{b}$ the position vectors of C, D and E.
Solution:

Given : $\overline{\mathrm{OA}}$ = $\bar{a}$, $\overline{\mathrm{OB}}$ = $\bar{a}$ Let AD, BE, OC meet at M.
Then M bisects AD, BE, OC.
$\overline{\mathrm{AB}}$ = $\overline{\mathrm{AO}}$ + $\overline{\mathrm{OB}}$ = –$\overline{\mathrm{OA}}$ + $\overline{\mathrm{OB}}$ = –$\bar{a}$ + $\bar{b}$ = $\bar{b}$ – $\bar{a}$
∵ OABM is a parallelogram

Hence, the position vectors of C, D and E are 2$\bar{b}$ – 2$\bar{a}$, 2$\bar{b}$ – 3$\bar{a}$ and $\bar{b}$ – 2$\bar{a}$ respectively.

Question 4.
If ABCDEF is a regular hexagon, show that $\overline{\mathrm{AB}}$ + $\overline{\mathrm{AC}}$ + $\overline{\mathrm{AD}}$ + $\overline{\mathrm{AE}}$ + $\overline{\mathrm{AF}}$ = 6$\overline{\mathrm{AO}}$, where O is the center of the hexagon.
Solution:

ABCDEF is a regular hexagon.
∴ $\overline{\mathrm{AB}}$ = $\overline{\mathrm{ED}}$ and $\overline{\mathrm{AF}}$ = $\overline{\mathrm{CD}}$
∴ by the triangle law of addition of vectors,

Question 5.
Check whether the vectors $2 \hat{i}+2 \hat{j}+3 \hat{k}$, + $-3 \hat{i}+3 \hat{j}+2 \hat{k}$, + $3 \hat{i}+4 \hat{k}$ form a triangle or not.
Solution:
Let, if possible, the three vectors form a triangle ABC
with $\overline{A B}$ = $2 \hat{i}+2 \hat{j}+3 \hat{k}$, $\overline{B C}$ = $3 \hat{i}+3 \hat{j}+2 \hat{k}$, $\overline{A C}$ = $3 \hat{i}+4 \hat{k}$
Now, $\overline{A B}$ + $\overline{B C}$
= $(2 \hat{i}+2 \hat{j}+3 \hat{k})$ + $(-3 \hat{i}+3 \hat{j}+2 \hat{k})$
= $-\hat{i}+5 \hat{j}+5 \hat{k} \neq 3 \hat{i}+4 \hat{k}$ = $\overline{\mathrm{AC}}$
Hence, the three vectors do not form a triangle.

Question 6.
In the figure 5.34 express $\bar{c}$ and $\bar{d}$ in terms of $\bar{a}$ and $\bar{b}$. Find a vector in the direction of $\bar{a}$ = $\hat{i}-2 \hat{j}$ that has magnitude 7 units.

Solution:
$\overline{\mathrm{PQ}}$ = $\overline{\mathrm{PS}}$ + $\overline{\mathrm{SQ}}$
∴ $\bar{a}$ = $\bar{c}$ – $\bar{d}$ … (1)
$\overline{\mathrm{PR}}$ = $\overline{\mathrm{PS}}$ + $\overline{\mathrm{SR}}$
∴ $\bar{b}$ = $\bar{c}$ + $\bar{d}$ … (2)
Adding equations (1) and (2), we get
$\bar{a}$ + $\bar{b}$ = ($\bar{c}$ – $\bar{d}$) + ($\bar{c}$ + $\bar{d}$) = 2$\bar{c}$

Question 7.
Find the distance from (4, -2, 6) to each of the following :
(a) The XY-plane
Solution:
Let the point A be (4, -2, 6).
Then,
The distance of A from XY-plane = |z| = 6

(b) The YZ-plane
Solution:
The distance of A from YZ-plane = |x| = 4

(c) The XZ-plane
Solution:
The distance of A from ZX-plane = |y| = 2

(d) The X-axis
Solution:
The distance of A from X-axis
= $\sqrt{y^{2}+z^{2}}$ = $\sqrt{(-2)^{2}+6^{2}}$ = $\sqrt{40}$ = $2 \sqrt{10}$

(e) The Y-axis
Solution:
The distance of A from Y-axis
= $\sqrt{z^{2}+x^{2}}$ = $\sqrt{6^{2}+4^{2}}$ = $\sqrt{52}$ = $2 \sqrt{13}$

(f) The Z-axis
Solution:
The distance of A from Z-axis
= $\sqrt{x^{2}+y^{2}}$ = $\sqrt{4^{2}+(-2)^{2}}$ = $\sqrt{20}$ = $2 \sqrt{5}$

Question 8.
Find the coordinates of the point which is located :
(a) Three units behind the YZ-plane, four units to the right of the XZ-plane and five units above the XY-plane.
Solution:
Let the coordinates of the point be (x, y, z).
Since the point is located 3 units behind the YZ- j plane, 4 units to the right of XZ-plane and 5 units , above the XY-plane,
x = -3, y = 4 and z = 5
Hence, coordinates of the required point are (-3, 4, 5)

(b) In the YZ-plane, one unit to the right of the XZ-plane and six units above the XY-plane.
Solution:
Let the coordinates of the point be (x, y, z).
Since the point is located in the YZ plane, x = 0. Also, the point is one unit to the right of XZ-plane and six units above the XY-plane.
∴ y = 1, z = 6.
Hence, coordinates of the required point are (0, 1, 6).

Question 9.
Find the area of the triangle with vertices (1, 1, 0), (1, 0, 1) and (0, 1, 1).
Solution:
Let A = (1, 1, 0), B = (1, 0, 1), C = (0, 1, 1)

Question 10.
If $\overline{\mathrm{AB}}$ = $2 \hat{i}-4 \hat{j}+7 \hat{k}$ and initial point A ≡ (1, 5, ,0). Find the terminal point B.
Solution:
Let $\bar{a}$ and $\bar{b}$ be the position vectors of A and B.
Given : A = (1, 5, 0) .’. $\bar{a}$ = $\hat{i}+5 \hat{j}$
Now, $\overline{\mathrm{AB}}$ = $2 \hat{i}-4 \hat{j}+7 \hat{k}$
∴ $\bar{b}$ – $\bar{a}$ = $2 \hat{i}-4 \hat{j}+7 \hat{k}$
∴ $\bar{b}$ = $(2 \hat{i}-4 \hat{j}+7 \hat{k})$ + $\bar{a}$
= $(2 \hat{i}-4 \hat{j}+7 \hat{k})$ + $(\hat{i}+5 \hat{j})$
= $3 \hat{i}+\hat{j}+7 \hat{k}$
Hence, the terminal point B = (3, 1, 7).

Question 11.
Show that the following points are collinear :
(i) A (3, 2, -4), B (9, 8, -10), C (-2, -3, 1).
Solution:
Let $\bar{a}$, $\bar{b}$, $\bar{c}$ be the position vectors of the points.
A = (3, 2, -4), B = (9, 8, -10) and C = (-2, -3, 1) respectively.

∴ $\overline{\mathrm{BC}}$ is a non-zero scalar multiple of $\overline{\mathrm{AB}}$
∴ they are parallel to each other.
But they have the point B in common.
∴ $\overline{\mathrm{BC}}$ and $\overline{\mathrm{AB}}$ are collinear vectors.
Hence, the points A, B and C are collinear.

(ii) P (4, 5, 2), Q (3, 2, 4), R (5, 8, 0).
Solution:
Let $\bar{a}$, $\bar{b}$, $\bar{c}$ be the position vectors of the points.
P = (4, 5, 2), Q = (3, 2, 4), R = (5, 8, 0) respectively.

= 2.$\overline{\mathrm{AB}}$ …[By (1)]
∴ $\overline{\mathrm{BC}}$ is a non-zero scalar multiple of $\overline{\mathrm{AB}}$
∴ they are parallel to each other.
But they have the point B in common.
∴ $\overline{\mathrm{BC}}$ and $\overline{\mathrm{AB}}$ are collinear vectors.
Hence, the points A, B and C are collinear.

Question 12.
If the vectors $2 \hat{i}-q \hat{j}+3 \hat{k}$ and $4 \hat{i}-5 \hat{j}+6 \hat{k}$ are collinear, then find the value of q.
Solution:
The vectors $2 \hat{i}-q \hat{j}+3 \hat{k}$ and $4 \hat{i}-5 \hat{j}+6 \hat{k}$ are collinear
∴ the coefficients of $\hat{i}, \hat{j}, \hat{k}$ are proportional

Question 13.
Are the four points A(1, -1, 1), B(-1, 1, 1), C(1, 1, 1) and D(2, -3, 4) coplanar? Justify your answer.
Solution:
The position vectors $\bar{a}$, $\bar{b}$, $\bar{c}$, $\bar{d}$ of the points A, B, C, D are

By equality of vectors,
y = -2 ….(1)
2x – 2y = 2 … (2)
3y = 0 … (3)
From (1), y = -2
From (3), y = 0 This is not possible.
Hence, the points A, B, C, D are not coplanar.

Question 14.
Express $-\hat{i}-3 \hat{j}+4 \hat{k}$ as linear combination of the vectors $2 \hat{i}+\hat{j}-4 \hat{k}$, $2 \hat{i}-\hat{j}+3 \hat{k}$ and $3 \hat{i}+\hat{j}-2 \hat{k}$.
Solution:

By equality of vectors,
2x + 2y + 3 = -1
x – y + z = -3
-4x + 3y – 2z = 4
We have to solve these equations by using Cramer’s Rule
D = $\left|\begin{array}{rrr} 2 & 2 & 3 \\ 1 & -1 & 1 \\ -4 & 3 & -2 \end{array}\right|$
= 2(2 – 3) – 2(-2 + 4) + 3(3 – 4)
= -2 – 4 – 3 = -9 ≠ 0

= 2(-4 + 9) – 2(4 – 12) – 1(3 – 4)
= 10 + 16 + 1 = 27

Class 12 Maharashtra State Board Maths Solution

• Exercise 5.1 Class 12 Maths 1
• Exercise 5.2 Class 12 Maths 1
• Exercise 5.3 Class 12 Maths 1
• Exercise 5.4 Class 12 Maths 1
• Exercise 5.5 Class 12 Maths 1
• Miscellaneous Exercise 5 Class 12 Maths 1