CBSE Class 12

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-9-ex-9-3/

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.3

Question 1.
\(\frac{x}{a}+\frac{y}{b}\) = 1
Solution:
\(\frac{x}{a}+\frac{y}{b}\) = 1
i.e., bx + ay = ab … (1)
Differentiating (1) w.r.t. x, we get
b + ay’ = 0
i.e., ay’ = – b
y = \(\frac { – b }{ a }\)
Differentiating again w.r.t. x, we get y’ = 0
which is the required differential equation.

Question 2.
y² = a(b² – x²)
Solution:
y² = a(b² – x²)
i.e., y² = ab² – x²a
Differentiating w.r.t. x, we get 2yy’ = – 2ax
i.e., yy’ = – ax
\(\frac { yy’ }{ x }\) = – a
Differentiating again w.r.t. x, we get
\(\frac{x\left[y y^{\prime \prime}+\left(y^{\prime}\right)^{2}\right]-y y^{\prime}}{x^{2}}\) = 0
i.e., x yy” + x(y’)² – yy’ = 0
which is the required differential equation.

Question 3.
y = ae^3x + be^-2x
Solution:
Given that
y = ae^3x + be^-2x … (i)
Differentiating w.r.t. x two times, we get
y’ = 3ae^3x – 2be^-2x
y” = 9ae^3x + 4be^-2x
i.e., y” = 6ae^3x + 3ae^3x + 6be^-2x – 2be^-2x
= 3ae^3x – 2be^-2x + 6(ae^3x + be^-2x)
= y’ + 6y
i.e., y” – y’ – 6y = 0 is the required differential equation.

Question 4.
y = e^2x (a + bx)
Solution:
y = e^2x (a + bx)
y’ = be^2x + (a + bx)e^2x x 2
y’ = be^2x + 2y
Differentiating again w.r.t. x, we get
y” = 2be^2x + 2y’
i.e., y” = 2(y’ – 2y) + 2y’ [∵ y’ – 2y = be^2x]
y” = 4y’ – 4y
y” – 4y’ + 4y = 0 is the required differential equation.

Question 5.
y = e^x(a cosx + b sinx)
Solution:
The curve y = e^x(a cosx+b sinx) …(i)
Differentiating (1) w.r.t. x, we get
y’ = e^x(- a sinx + b cosx) + e^x(a cosx + b sinx)
y’ = e^x(b cos x – a sin y) + y
i.e., y’ – y = e^x(b cosx – a sinx) … (2)
Differentiating (2) w.r.t. x, we get
y”- y’ = – e^x(b sinx + a cosx) + e^x(b cosx – a sinx)
y” – y’ = – y + y’- y [from (1) and (2)]
y” – 2y’ + 2y = 0 is the required differential equation.

Question 6.
Form the differential equation of the family of circles touching the y axis at origin
Solution:

Circles touching y axis at the orgin will have its centre on x-axis and will pass through the origin.
So the centre of circle will be (a, 0) and radius a.
∴ Equation of the circle is (x – a)² + (y – 0)² = a²
i.e., x² + y² – 2ax = 0
i.e., \(\frac{x^{2}+y^{2}}{x}\) = 2a
\(\frac{x\left(2 x+2 y y_{1}\right)-\left(x^{2}+y^{2}\right)}{x^{2}}\) = 0
Differentiating w.r.t. x, we get
2x² + 2xyy₁ – x² – y² = 0
x² – y² + 2xyy₁ = 0 or y² – x² – 2xyy₁ = 0

Question 7.
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Solution:

Consider the family of parabolas having focus (0, a) at the positive y-axis, where a is an arbitrary constant.
∴ The equation of family of parabolas is x² = 4ay … (1)
Differentiating both sides w.r.t. x, we get
2x = 4 ay’
i.e., 4a = \(\frac { 2x }{ y’ }\) … (2)
Substituting (2) in (1) we get x² = (\(\frac { 2x }{ y’ }\))y
x²y’ – 2xy = 0
i.e., xy’ – 2y = 0 is the required differential equation.

Question 8.
Form the differential equation of family of ellipses having foci on y-axis and centre at origin.
Solution:
The equation of family ellipses having foci at y- axis is

Question 9.
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
Solution:
The equation of family of hyperbolas having foci on x axis is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1, a and b are the parameters.

i.e., xyy” + x(y’)² – yy’ = 0 is the required differential equation.

Question 10.
Form the differential equation of the family of circles having centre on y-axis and radius 3 units
Solution:
Consider a circle of radius 3 unit and centre on (0, a). The equation of the circle is
(x – 0)² + (y – a)² = 3²
x² + (y – a)² = 9 … (1)
Differentiating both sides w.r.t. x, we get
2x + 2(y – a)y’ = 0
x + yy’ – ay’ = 0
ay’ = x + yy’

⇒ (x² – 9)(y’)² + x² = 0 is the required differential equation.

Question 11.
Which of the following differential equation has y = \({ c }_{ 1 }{ e }^{ x }+{ c }_{ 2 }{ e }^{ -x }\) as the general solution?
(a) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0\)
(b) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -y=0\)
(c) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +1=0\)
(d) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -1=0\)
Solution:
(b) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -y=0\)

Question 12.
Which of the following differential equations has y = x as one of its particular solution ?
(a) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -{ x }^{ 2 }\frac { dy }{ dx } +xy=x\)
(b) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ x }\frac { dy }{ dx } +xy=x\)
(c) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -{ x }^{ 2 }\frac { dy }{ dx } +xy=0\)
(d) \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ x }\frac { dy }{ dx } +xy=0\)
Solution:
(c) y = x

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-9-miscellaneous-exercise/

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Miscellaneous Exercise

Question 1.
For each of the differential equations given in Exercises 1 to 12, find the general solution.
i. \(\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x\)
ii. \(\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}+7 y=\sin x\)
iii. \(\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0\)
Solution:
i. \(\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x\)
∴ Order 2 and degree = 1

ii. \(\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}+7 y=\sin x\)
∴ Order 1 and degree = 3

iii. \(\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0\)
∴ Order 4 and degree not defined.

Question 2.
For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
i. y = a e^x + be^-x + x²:
\(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}\) – xy + x² – 2 = 0

ii. y = e^x(a cos x + b sin x) :
\(\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}\) + 2y = 0

iii. y = x sin 3x :
\(\frac{d^{2} y}{d x^{2}}\) + 9y – 6 cos 3x = 0

iv. x² = 2y² log y :
(x² + y²) \(\frac { dy }{ dx }\) – xy = 0
Solution:
i. y = a e^x + be^-x + x²:
Differentiating w.r.t. x, we get

ii. y = e^x(a cos x + b sin x) :
Differentiating w.r.t. x, we get

iii. y = x sin 3x :
Differentiating w.r.t. x, we get
\(\frac { dy }{ dx }\) = 3x cos3x + sin 3x
\(\frac{d^{2} y}{d x^{2}}\) = 3x(- 3sin 3x) + 3cos 3x + 3cos 3x
= – 9x sin 3x + 6cos 3x
\(\frac{d^{2} y}{d x^{2}}\) + 9y – 6cos 3x
= – 9x sin 3x + 6cos 3x + 9x sin 3x – 6cos 3x = 0
∴ y = xsin 3x is the solution of the differential equation \(\frac{d^{2} y}{d x^{2}}\) + 9y – 6cos 3x = 0

iv. x² = 2y² log y :
Differentiating w.r.t. x, we get

Question 3.
Form the differential equation representing the family of curves given by (x – a)² + 2y² = a², where a is an arbitrary constant.
Solution:
(x – a)² + 2y² = a² … (1)
Differentiating w.r.t. x, we get
2(x – a) + 4yy’ = 0
x – a + 2yy’ = 0
∴ a = x + 2yy’
(1) → (- 2yy’)² + 2y² =[x + 2yy’]²
4(yy’)² + 2y² = x² + 4 xyy’ + 4 (yy’)²
2y² = x² + 4xyy’
∴ y’ = \(\frac{2 y^{2}-x^{2}}{4 x y}\) is the required differential equation.

Question 4.
Prove that x² – y² = c (x² + y²)² is the general solution of differential equation (x³ – 3xy²) dx = (y³ – 3x²y) dy, where c is a parameter.
Solution:

x² – y² = C(x² + y²)² where C = C²₁ is the general solution.

Question 5.
Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.
Solution:

The equation of the required circles is
(x – r)² + (y – r)² = r² … (1)
x² + y² – 2rx – 2ry + r² = 0
Differentiating (1) w.r.t. x, we get
2x + 2yy₁ – 2r – 2ry₁ = 0
r = \(\frac{x+y y_{1}}{1+y_{1}}\)
Substituting r in (1), we get

is the required differential equation.

Question 6.
Find the general solution of the differential equation \(\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0\).
Solution:

∴ sin^-1y + sin^-1x = C, is the general solution.

Question 7.
Show that the general solution of the differential equation \(\frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}\) = 0 is given by
(x + y + 1) = A(1 – x – y – 2xy), where A is the parameter.
Solution:

Question 8.
Find the equation of the curve passing through the point (0, \(\frac { π }{ 4 }\)) whose differential equation is sin x cos y dx + cos x siny dy = 0.
Solution:
sin x cos y dx + cos x sin y dy = 0.
Divide by cos x cos y, we get
\(\frac{\sin x}{\cos x} d x+\frac{\sin y}{\cos y} d y=0\)
tan x dx + tan y dy = 0
Integrating both sides, we get
∫tan x dx + ∫tan y dy = ∫o dx
log|sec x| + log|sec y| = log|C|
log|sec x secy| = log|C|
sec x sec y = C … (1)
(1) Passes through the point
we get C = \(\sqrt{2}\)
∴ (1) → sec x secy = \(\sqrt{2}\)
⇒ cos y = \(\frac{\sec x}{\sqrt{2}}\), is the equation of the curve.

Question 9.
Find the particular solution of the differential equation
(1 + e^2x) dy + (1 + y²)e^x dx = 0. Given that y = 1 when x = 0.
Solution:
(1 + e^2x) dy + (1 + y²)e^x dx = 0
Divide by (1 + e^2x)(1 + y²), we get

Question 10.
Solve the differential equation
\(y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y\) (y ≠ 0).
Solution:

i.e., \(e^{\frac{x}{y}}\) = y + C is the general solution.

Question 11.
Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy. Given that y = – 1, when x = 0.
Solution:
(x – y)(dx + dy) = dx – dy
xdx – ydx + xdy – ydy = dx – dy
(x – y – 1)dx = (y – x – 1)dy

t + log|t| = 2x + C
x – y + log|x – y|= 2x + C … (2)
When x = 0, y = – 1
∴ (2) → 1 + log 1 = 0 + C
C = 1
∴ (2) → x – y + log |x – y| = 2x + 1
Hence x + y + 1 = log|x – y| is the particular solution

Question 12.
Solve the differential equation \(\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}=1(x \neq 0)\).
Solution:

i.e., \(y e^{2 \sqrt{x}}=2 \sqrt{x}+C\) is the general solution.

Question 13.
Find a particular solution of the differential equation \(\frac { dy }{ dx }\) + y cot x = 4x cosec x (x ≠ 0).
Given that y = 0 when x = \(\frac { π }{ 2 }\)
Solution:
\(\frac { dy }{ dx }\) + y cot x = 4x cosec x is a linear differential equation.
∴ P = cot x and Q = 4x cosec x
∫Pdx = ∫cot x dx = log sinx
∴ I.F. = e^{∫p dx} = e^{log sinx} = sin x
The solution is y(I.F) = ∫Q(I.F)dx + C
y sinx = ∫4x cosec x sin xdx + C
= ∫4x dx + C
y sin x = 2x + C … (1)
When x = \(\frac { π }{ 2 }\), y = 0
(1) → o = \(\frac { π² }{ 2 }\) + C
∴ C = \(\frac { π² }{ 2 }\)
Hence the particular solution is
ysin x = 2x² – \(\frac { π² }{ 2 }\)

Question 14.
Find a particular solution of the differential equation (x + 1) \(\frac { dy }{ dx }\) = 2e^-y – 1, given that y = 0 when x = 0.
Solution:

Hence y = log\(\left|\frac{2 x+1}{x+1}\right|\), is the particular solution.

Question 15.
The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20,000 in 1999 and 25,000 in the year 2004, what will be the population of the village in 2009?
Solution:
Let P be the population at time t
Given \(\frac { dP }{ dt }\) ∝ P
i.e, \(\frac { dP }{ dt }\) = kP, k is a constant
i.e., \(\frac { dP }{ dt }\) = kdt
Integrating both sides, we get
∫\(\frac { dP }{ dt }\) = k ∫dt
log|P| = kt + C … (1)
The initial population in 1999 is 20,000
i.e., when t = 0, P = 20,000
Hence (1) → log 20,000 = k(0) + C
∴ C = log 20,000
After 5 years, i.e., in 2004, the population is 25,000
i.e., when t = 5, P = 25000
Hence (1) → log 25000 = 5k + log 20000
log 25000 – log 20000 = 5k

Question 16.
The general solution of the differential equation \(\frac{y d x-x d y}{y}\) = 0 is
a. xy = C
b. x = Cy²
c. y = Cx
d. y = Cx²
Solution:
c. y = Cx
\(\frac{y d x-x d y}{y}\) = 0
∴ ydx – xdy = 0
ydx = xdy
\(\frac { dy }{ y }\) = \(\frac { dx }{ x }\)
Integrating both sides, we get
∫\(\frac { dy }{ y }\) = ∫\(\frac { dx }{ x }\)
log|y| = log|x| + log|C|
log|y| = log|Cx|
y = Cx

Question 17.
The general solution of a differential equation of the type \(\frac { dx }{ dy }\) + P₁, x = Q₁ is

Solution:
\(e^{\int {P}_{1} d y}\) is the integrating factor
∴ The general solution is
\(x \cdot e^{\int {P}_{1} d y}=\int\left({Q}_{1} e^{\int {P}_{1} d y}\right) d y+ {C}\)

Question 18.
The general solution of the differential equation e^xdy + (ye^x + 2x) dx = 0 is
a. x e^y + x² = C
b. x e^y + y² =C
c. y e^x + x² = C
d. y e^y + x² = C
Solution:

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-9-ex-9-2/

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.2

Ex 9.2 Class 12 NCERT Solutions Question 1.
y = e^x + 1 : y”- y’ = 0
Solution:
y = e^x + 1
Differentiating w.r.t. x, we get
y’ = e^x
i.e., y” = y’
⇒ y” – y’ = 0
Hence y = e^x is a solution of y” – y’ = 0.

Ex 9.2 Class 12 Maths Ncert Solutions Question 2.
y = x² + 2x + C : y’ – 2x – 2 = 0
Solution:
y = x² + 2x + C
Differentiating w.r.t. x, we get y’ = 2x + 2
i.e., y’ – 2x – 2 = 0
Hence y = x² + 2x + C is a solution of y’ – 2x – 2 = 0

Question 3.
y = cos x + C : y’ + sin x = 0
Solution:
y = cos x + C
Differentiating w.r.t., x, we get y’ = – sin x
i.e., y’ + sin x = 0
Hence y = cos x + C is a solution of y’ + sin x = 0

Question 4.
y = \(\sqrt{1+x^{2}} \quad: \quad y^{\prime}=\frac{x y}{1+x^{2}}\)
Solution:
\(\sqrt{1+x^{2}}\)
Differentiating w.r.t. x, we get

Question 5.
y – Ax : xy’ = y (x ≠ 0)
Solution:
y = Ax … (1)
Differentiating w.r.t., x, we get y’ = A
i.e., y’ = \(\frac { y }{ x }\) [From(1), A = \(\frac { y }{ x }\)]
i.e., xy’ = y (x ≠ 0)
∴ y = Ax is a solution of xy’ = y.

Question 6.
y = x sin x : xy’= y + x \(\sqrt{x^{2}-y^{2}}\)
(x ≠ 0 and x > y or x < – y)
Solution:
y = x sinx … (1)
Differentiating (1) w.r.t. x, we get
y’ = x cos x + sin x … (2)
From (1) sin x = \(\frac { y }{ x }\)

∴ y = x sinx is a solution of xy’ = x\(\sqrt{x^{2}-y^{2}}\) + y

Question 7.
Ay = logy + C : y’ = \(\frac{y^{2}}{1-x y}(x y \neq 1)\)
Solution:
xy = logy + C
Differentiating w.r.t. x, we get
xy’ + y = \(\frac { 1 }{ y }\) y’
i.e., xyy’ + y² = y’
y’ – xyy = y²
y'(1 – xy) = y²
i.e., y’ = \(\frac{y^{2}}{1-x y}\)
Hence xy = logy + C is a solution of y’ = \(\frac{y^{2}}{1-x y}\)

Question 8.
y – cos y = x : (ysiny + cosy + x)y’ = y.
Solution:
y – cosy = x … (1)
Differentiating (1) w.r.t. x, we get
y’ + sin y . y’ = 1
i. e., yy’ + yy’sin y = y
(x + cos y)y’ + yy’sin y = y
[since from (1) y = x + cosy]
xy’ + y’cosy+ yy’siny = y
(y sin y + cos y+ x)y’ = y
Hence y – cosy = x is a solution of
(y sin y + cos y + x)y’ = y

Question 9.
x + y = tan^-1 y : y²y’ + y² + 1 = 0
Solution:
x + y = tan^-1
Differentiating w.r.t. x, we get,
1 + y’ = \(\frac{1}{1+y^{2}}\)
i.e., (1 + y²)(1 + y’) = y ‘
1 + y’ + y² + y²y’ = y’
i.e., y²y’ + y² + 1 = 0
Hence x + y = tan^-1 is a solution of y²y’ + y² + 1 = 0

Question 10.
\(y=\sqrt { { a }^{ 2 }-{ x }^{ 2 } } x\in (-a,a);x+y\frac { dy }{ dx } =0,(y\neq 0)\)
Solution:
y = \(\sqrt{a^{2}-x^{2}}\) , x ∈ (- a, a)
Differentiating w.r.t. x, we get

Hence y = \(\sqrt{a^{2}-x^{2}}\) is a solution of x + y\(\frac { dy }{ dx }\) = 0

Question 11.
The number of arbitrary constants in the general solution of a differential equation of fourth order are
a. 0
b. 2
c. 3
d. 4
Solution:
d. 4
The general solution of a differential equation contains as many arbitrary constants as the order of the differential equation. Since the differential equation is of order 4, its solution contains 4 arbitrary constants.

Question 12.
The number of arbitrary constants in the particular solution of a differential equation of third order are
a. 3
b. 2
c. 1
d. 0
Solution:
d. 0
The particular solution is free from the arbitrary constants.

These NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.3 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-10-ex-10-3/

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.3

Class 12 Maths Ncert Solutions Chapter 10.3 Question 1.
Find the angle between two vectors \(\vec{a}\) and \(\vec{b}\) with magnitudes \(\sqrt{3}\) and 2 respectively, and such that \(\vec{a}\).\(\vec{b}\) = \(\sqrt{6}\)
Solution:
|\(\vec{a}\)|= \(\sqrt{3}\)
and \(\vec{b}\) = 2, \(\vec{a}\).\(\vec{b}\) = \(\sqrt{6}\)
Let θ be the angle between \(\vec{a}\) and \(\vec{b}\). Then
\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{\sqrt{6}}{(\sqrt{3})(2)}=\frac{1}{\sqrt{2}}\)
∴ θ = \(\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}\)

Question 2.
Find the angle between the vectors \(\hat{i}-2 \hat{j}+3 \hat{k} \text { and } 3 \hat{i}-2 \hat{j}+\hat{k}\)
Solution:
Let \(\vec{a}\) = \(\hat{i}-2 \hat{j}+3 \hat{k}\), \(\vec{b}\) = \(\hat{3i}-2 \hat{j}+ \hat{k}\)
\(\vec{a}\).\(\vec{b}\) = \((\hat{i}-2 \hat{j}+3 \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+\hat{k})\)
= 1(3) + (- 2)(- 2) + 3(1) = 3 + 4 + 3 = 10

Question 3.
Find the projection of the vector \(\overrightarrow { i } -\overrightarrow { j }\), on the line represented by the vector \(\overrightarrow { i } +\overrightarrow { j }\).
Solution:

Question 4.
Find the projection of the vector \(\hat{i}+3 \hat{j}+7 \hat{k}\) on the vector \(7 \hat{i}-\hat{j}+8 \hat{k}\)
Solution:

Question 5.
Show that each of the given three vectors is a unit vector \(\frac { 1 }{ 7 } \left( 2\hat { i } +3\hat { j } +6\hat { k } \right) ,\frac { 1 }{ 7 } \left( 3\hat { i } -6\hat { j } +2\hat { k } \right) ,\frac { 1 }{ 7 } \left( 6\hat { i } +2\hat { j } -3\hat { k } \right)\)
Also show that they are mutually perpendicular to each other.
Solution:
Let \(\vec{a}\) = \(\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k})\), \(\vec{b}\) = \(\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})\) and \(\vec{c}\) = \(\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})\)

Here \(\vec{a}\).\(\vec{b}\) = 0, \(\vec{b}\).\(\vec{c}\) = 0 and \(\vec{a}\).\(\vec{c}\) = 0
∴ The vectors \(\vec{a}\).\(\vec{b}\) and \(\vec{c}\) are mutually per-pendicular vectors.

Question 6.
\(Find\left| \overrightarrow { a } \right| and\left| \overrightarrow { b } \right| if\left( \overrightarrow { a } +\overrightarrow { b } \right) \cdot \left( \overrightarrow { a } -\overrightarrow { b } \right) =8\quad and\left| \overrightarrow { a } \right| =8\left| \overrightarrow { b } \right| \)
Solution:

Question 7.
Evaluate the product :
\((3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})\)
Solution:
\(\left( 3\overrightarrow { a } -5\overrightarrow { b } \right) \cdot \left( 2\overrightarrow { a } +7\overrightarrow { b } \right) \)
\(=6\overrightarrow { a } .\overrightarrow { a } -10\overrightarrow { b } \overrightarrow { a } +21\overrightarrow { a } .\overrightarrow { b } -35\overrightarrow { b } .\overrightarrow { b } \)
\(=6{ \left| \overrightarrow { a } \right| }^{ 2 }-11\overrightarrow { a } \overrightarrow { b } -35{ \left| \overrightarrow { b } \right| }^{ 2 }\)

Question 8.
Find the magnitude of two vectors \(\vec{a}\) and \(\vec{b}\) having the same magnitude and such that the angle between them is 60° and their scalar product is \(\frac { 1 }{ 2 }\)
Solution:

Question 9.
Find |\(\vec{a}\)| , if for a unit vector \(\vec{a}\), \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})\) = 12
Solution:

Question 10.
If \(\overrightarrow { a } =2\hat { i } +2\hat { j } +3\hat { k } ,\overrightarrow { b } =-\hat { i } +2\hat { j } +\hat { k } and \overrightarrow { c } =3\hat { i } +\hat { j } \) such that \(\overrightarrow { a } +\lambda \overrightarrow { b } \bot \overrightarrow { c } \) , then find the value of λ.
Solution:

Question 11.
Show that \(|\vec{a}| \vec{b}+|\vec{b}| \vec{a} \), is per-pendicular to \(|\vec{a}| \vec{b}-|\vec{b}| \vec{a} \) for any two non-zero vectors \(\vec{a} \text { and } \vec{b}\)
Solution:

Question 12.
If \(\overrightarrow { a } \cdot \overrightarrow { a } =0\quad and\quad \overrightarrow { a } \cdot \overrightarrow { b } =0\), then what can be concluded about the vector \(\overrightarrow { b } \) ?
Solution:
\(\vec{a}\).\(\vec{a}\) = 0 ⇒ \(\vec{a}\) is a zero vector.
since \(\vec{a}\) = \(\vec{0}\), \(\vec{a}\).\(\vec{b}\) = 0 for any vector \(\vec{b}\)
Vector \(\vec{b}\) be any vector.

Question 13.
If \(\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } \) are the unit vector such that \(\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =0\) , then find the value of \(\overrightarrow { a } .\overrightarrow { b } +\overrightarrow { b } .\overrightarrow { c } +\overrightarrow { c } .\overrightarrow { a } \)
Solution:

Question 14.
If either vector \(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0}\), then \(\vec{a} \cdot \vec{b}\). But the converse need not be true. Justify your answer with an example.
Solution:

Thus two non-zero vectors \(\vec{a}\) and \(\vec{b}\) may have \(\vec{a}\).\(\vec{a}\) zero.

Question 15.
If the vertices A,B,C of a triangle ABC are (1, 2, 3) (-1, 0, 0), (0, 1, 2) respectively, then find ∠ABC.
Solution:

Question 16.
Show that the points A (1, 2, 7), B (2, 6, 3) and C (3, 10, -1) are collinear.
Solution:
A (1, 2, 7), B (2, 6, 3) and C (3, 10, -1) are the points.

∴ \(\overrightarrow{AB}\)\(\overrightarrow{AC}\) are parallel and A is a common point. Therefore A, B, C are collinear.

Question 17.
Show that the vectors \(2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k} \text { and } \quad 3 \hat{i}-4 \hat{j}-4 \hat{k}\) from the vertices of a right angled triangle.
Solution:
Let A, B and C be the vertices of the triangle with the vectors \(2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k} \text { and } \quad 3 \hat{i}-4 \hat{j}-4 \hat{k}\)
∴ \(\overrightarrow{AB}\) = p.v. of B – p.v. of A

∴ A, B, C are the vertices of ∆ ABC
\(\overrightarrow{BC}\).\(\overrightarrow{CA}\) = (2)(-1) + (-1)(3) + (1)(5)
= – 2 – 3 + 5 = 0
∴ \(\overrightarrow{BC}\)⊥\(\overrightarrow{CA}\)
Hence triangle ABC is a right triangle

Question 18.
If \(\overrightarrow { a } \) is a non-zero vector of magnitude ‘a’ and λ is a non- zero scalar, then λ \(\overrightarrow { a } \) is unit vector if
(a) λ = 1
(b) λ = – 1
(c) a = |λ|
(d) a = \(\frac { 1 }{ \left| \lambda \right| } \)
Solution:
\(\left| \overrightarrow { a } \right| =a\)
Given : \(\lambda \overrightarrow { a } \) is a unit vectors.
\(|\lambda \vec{a}|=1 \Rightarrow|\lambda||\vec{a}|=1 \Rightarrow|\lambda| a=1 \Rightarrow a=\frac{1}{|\lambda|}\)

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-9-ex-9-1/

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.1

Question 1.
\(\frac { { d }^{ 4 }y }{ { dx }^{ 4 } } +({ sin }y^{ “‘ })\) = 0
Solution:
The highest order derivatives is \(\frac { { d }^{ 4 }y }{ { dx }^{ 4 } }\)
∴ Order of the equation is 4
The differential equation is not a polynomial equation in its derivatives.
Hence its degree is not defined.

Question 2.
y’ + 5y = 0
Solution:
The highest order derivative is y’
∴ Order is 1
This is a polynomial equation in y’ and the
Hence its degree is 1.

Question 3.
\({ \left( \frac { ds }{ dt } \right) }^{ 4 }+3s{ \left( \frac { { d }^{ 2 }s }{ { dt }^{ 2 } } \right) }\) = 0
Solution:
The highest order derivative is \(\frac{d^{2} s}{d t^{2}}\)
∴ Order is 2
This is a polynomial equation in \(\frac{d^{2} s}{d t^{2}}\) and \(\frac{ds}{dt}\) and the power of \(\frac{d^{2} s}{d t^{2}}\) is one
Hence its degree is one.

Question 4.
\({ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 2 }+cos\left( \frac { dy }{ dx } \right) =0\)
Solution:
The highest order derivatives is \(\frac{d^{2} y}{d x^{2}}\)
∴ Order of the equation is 2
The differential equation is not a polynomial equation in its derivatives.
Hence its degree is not defined.

Question 5.
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } }\) = cos3x + sin3x
Solution:
The highest order derivative is \(\frac{d^{2} y}{d x^{2}}\)
∴ Order is 2
This is a polynomial equation in \(\frac{d^{2} y}{d x^{2}}\) and the power of \(\frac{d^{2} y}{d x^{2}}\) is one
Hence its degree is one.

Question 6.
\({ { (y }^{ ”’ }) }^{ 2 }+{ { ( }y^{ ” }) }^{ 3 }+{ { (y }^{ ‘ }) }^{ 4 }+{ y }^{ 5 }\) = 0
Solution:
The highest order derivative is y”‘
∴ Order is 3
This is a polynomial equation in y”‘, y” and y’ the power of y” is two.
Hence its degree is 2.

Question 7.
y”‘ + 2y” + y’ = 0
Solution:
The highest order derivative is y”‘
∴ Order is 3
This is a polynomial equation in y”‘, y” and y’ and the power of y”‘ is one.
Hence its degree is 1.

Question 8.
y’ + y = e^x
Solution:
The highest order derivative is y’.
∴ Order is 1
This is a polynomial equation in y’ and the power of y’ is one.
Hence its degree is 1.

Question 9.
y” + (y’) +2y = 0
Solution:
The highest order derivative is y”
∴ Order is 2
This is a polynomial equation in y” and
y’ and the degree of y” is one.
Hence its degree is 1.

Question 10.
y” + 2y’ + sin y = 0
Solution:
The highest order derivative is y”.
∴ Order is 2
This is a polynomial equation in y”, y’ and the power of y” is one.
Hence its degree is 1.

Question 11.
The degree of the differential equation
\({ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 3 }{ +\left( \frac { dy }{ dx } \right) }^{ 2 }+sin{ \left( \frac { dy }{ dx } \right) }+1=0\)
(a) 3
(b) 2
(c) 1
(d) not defined
Solution:
The differential equation is not a polynomial equation in derivatives. Hence the degree is not defined.

Question 12.
The order of the differential equation
\({ 2x }^{ 2 }\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -3\frac { dy }{ dx } +y=0\)
(a) 2
(b) 1
(c) 0
(d) not defined
Solution:
The highest order derivative is \(\frac{d^{2} y}{d x^{2}}\)
∴ Order is 2.

These NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-10-ex-10-2/

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.2

10.2 Class 12 Question 1.
Compute the magnitude of the following vectors:
\(\overrightarrow { a } =\hat { i } +\hat { j } +\hat { k } ,\overrightarrow { b } =\hat { 2i } -\hat { 7j } -\hat { 3k } \)
\(\overrightarrow { c } =\frac { 1 }{ \sqrt { 3 } } \hat { i } +\frac { 1 }{ \sqrt { 3 } } \hat { j } -\frac { 1 }{ \sqrt { 3 } } \hat { k } \)
Solution:

Question 2.
Write two different vectors having same magnitude.
Solution:
Let \(\overrightarrow { a } =\hat { i } +\hat { 2j } +\hat { 3k } ,\overrightarrow { b } =\hat { 3i } +\hat { 2j } +\hat { k } \)
\(|\vec{a}|=\sqrt{1+1+9}=\sqrt{11}\)
\(|\vec{b}|=\sqrt{9+1+1}=\sqrt{11}\)
∴ \(\overline{a}\) and \(\overline{b}\) are examples for two vectors having the same magnitude.
There are infinitely many Vectors having same magnitude.

Question 3.
Write two different vectors having same direction.
Solution:
\(\hat{i}+\hat{j}+\hat{k}\) and \(\hat{3i}+\hat{3j}+\hat{3k}\) are examples for two vectors having the same direction. Generally, if \(\vec{a}\) is a nonzero vector, then \(\vec{a}\) and λ\(\vec{a}\) have the same direction whenever λ is positive.

Question 4.
Find the values of x and y so that the vectors \(2 \hat{i}+3 \hat{j} \text { and } x \hat{i}+y \hat{j}\) are equal.
Solution:
We are given \(2 \hat{i}+3 \hat{j}=x \hat{i}+y \hat{j}\)
If vectors are equal, then their respective components are equal. Hence x = 2, y = 3.

Question 5.
Find the scalar and vector components of the vector with initial point (2,1) and terminal point (-5,7).
Solution:
Let A(2, 1) be the initial point and B(-5,7) be the terminal point \(\overrightarrow { AB } =\left( { x }_{ 2 }-{ x }_{ 1 } \right) \hat { i } +\left( { y }_{ 2 }-{ y }_{ 1 } \right) \hat { j } =-\hat { 7i } +\hat { 6j } \)
∴ The vector components are \(\vec{-7i}\), \(\vec{6j}\) and scalar components are – 7 and 6.

Question 6.
Find the sum of three vectors:
\(\overrightarrow { a } =\hat { i } -\hat { 2j } +\hat { k } ,\overrightarrow { b } =-2\hat { i } +\hat { 4j } +5\hat { k } \quad and\quad \overrightarrow { c } =\hat { i } -\hat { 6j } -\hat { 7k } ,\)
Solution:
\(\overrightarrow { a } =\hat { i } -\hat { 2j } +\hat { k } ,\overrightarrow { b } =-2\hat { i } +\hat { 4j } +5\hat { k } \quad and\quad \overrightarrow { c } =\hat { i } -\hat { 6j } -\hat { 7k }\)

Question 7.
Find the unit vector in the direction of the vector \(\overrightarrow { a } =\hat { i } +\hat { j } +\hat { 2k } \).
Solution:

Question 8.
Find the unit vector in the direction of vector \(\overrightarrow { PQ }\), where P and Q are the points (1, 2, 3) and (4, 5, 6) respectively.
Solution:

Question 9.
For given vectors \(\overrightarrow { a } =2\hat { i } -\hat { j } +2\hat { k } \quad and\quad \overrightarrow { b } =-\hat { i } +\hat { j } -\hat { k }\) find the unit vector in the direction of the vector \(\overrightarrow { a } +\overrightarrow { b }\)
Solution:

Question 10.
Find a vector in the direction of \(5\hat { i } -\hat { j } +2\hat { k }\) which has magnitude 8 units.
Solution:

Question 11.
Show that the vector \(2 \hat{i}-3 \hat{j}+4 \hat{k}\) and \(-4 \hat{i}+6 \hat{j}-8 \hat{k}\) are collinear.
Solution:
\(\overrightarrow { a } =2\hat { i } -3\hat { j } +4\hat { k } \quad and\quad \overrightarrow { b } =-4\hat { i } +6\hat { j } -8\hat { k } \)
\(=-2(2\hat { i } -3\hat { j } +4\hat { k } ) \)
vector \(\vec{a}\) and \(\vec{b}\) have the same direction they are collinear.

Question 12.
Find the direction cosines of the vector \(\hat { i } +2\hat { j } +3\hat { k }\)
Solution:

Question 13.
Find the direction cosines of the vector joining the points A (1, 2, – 3) and B(- 1, – 2, 1), directed from A to B.
Solution:

∴ Direction cosines of \(\overline{AB}\) = Scalar components of \(\overline{AB}\)
= \(\frac{-1}{3}, \frac{-2}{3}, \frac{2}{3}\)

Question 14.
Show that the vector \(\hat { i } +\hat { j } +\hat { k }\) are equally inclined to the axes OX, OY, OZ.
Solution:
Let \(\vec{r}\) = \(\hat { i } +\hat { j } +\hat { k }\)
The direction ratios of \(\vec{r}\) are 1, 1, 1.
The direction cosines of \(\vec{r}\) are
\(\frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}, \frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}, \frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}\)
\(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)
∴ \(\vec{r}\) is equally inclined to the axes.

Question 15.
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are \(\hat{i}+2 \hat{j}-\hat{k}\) and \(-\hat{i}+\hat{j}+\hat{k}\) respectively, in the ratio 2:1
i. internally
ii. externally
Solution:
i. Internal division
Let \(\vec{a}\) = position vectorof P = \(\hat{i}+2 \hat{j}-\hat{k}\)
Let \(\vec{b}\) = position vectorof Q = \(-\hat{i}+\hat{j}+\hat{k}\)
Let R divides PQ internally in the ratio 2 : 1

ii. External division
Let S divides PQ externally in the ratio 2 : 1

Question 16.
Find position vector of the mid point of the vector joining the points P (2, 3, 4) and Q (4, 1, – 2).
Solution:
Let R be the midpoint of PQ

Question 17.
Show that the points A, B and C with position vector \(\overrightarrow { a } =3\hat { i } -4\hat { j } -4\hat { k } ,\overrightarrow { b } =2\hat { i } -\hat { j } +\hat { k } and\quad \overrightarrow { c } =\hat { i } -3\hat { j } -5\hat { k }\) respectively form the vertices of a right angled triangle.
Solution:

Question 18.
In triangle ABC (fig.), which of the following is not

(a) \(\overrightarrow { AB } +\overrightarrow { BC } +\overrightarrow { CA } =\overrightarrow { 0 } \)
(b) \(\overrightarrow { AB } +\overrightarrow { BC } -\overrightarrow { AC } =\overrightarrow { 0 } \)
(c) \(\overrightarrow { AB } +\overrightarrow { BC } -\overrightarrow { CA } =\overrightarrow { 0 } \)
(d) \(\overrightarrow { AB } -\overrightarrow { CB } +\overrightarrow { CA } =\overrightarrow { 0 } \)
Solution:
By the triangle law of vector addition,
\(\overrightarrow { AB } +\overrightarrow { BC } +\overrightarrow { CA } =\overrightarrow { 0 } \)
\(\overrightarrow { AB } +\overrightarrow { BC } -\overrightarrow { AC } =\overrightarrow { 0 } \), is not true.

Question 19.
If \(\overrightarrow { a } ,\overrightarrow { b } \) are two collinear vectors then which of the following are incorrect:
(a) \(\overrightarrow { b } =\lambda \overrightarrow { a } \), for some scalar λ.
(b) \(\overrightarrow { a } =\pm \overrightarrow { b } \)
(c) the respective components of \(\overrightarrow { a } ,\overrightarrow { b } \) are proportional.
(d) both the vectors \(\overrightarrow { a } ,\overrightarrow { b } \) have same direction, but different magnitudes.
Solution:
If \(\vec{a}\) and \(\vec{b}\) are collinear, then they need not be in the same direction, d and b may have opposite directions.